***Useful Information for Apprentices***

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AC motor theory
Question 1:
Electromechanical watt-hour meters use an aluminium disk that is spun by an electric motor. To generate a constant "drag" on the disk necessary to limit its rotational speed, a strong magnet is placed in such a way that its lines of magnetic flux pass perpendicularly through the disk's thickness:
the disk itself need not be made of a ferromagnetic material in order for the magnet to create a "drag" force. It simply needs to be a good conductor of electricity.
Explain the phenomenon accounting for the drag effect, and also explain what would happen if the roles of magnet and disk were reversed: if the magnet were moved in a circle around the periphery of a stationary disk.
This is an example of Lenz' Law. A rotating magnet would cause a torque to be generated in the disk.
Notes:
Mechanical speedometer assemblies used on many automobiles use this very principle: a magnet assembly is rotated by a cable connected to the vehicle's driveshaft. This magnet rotates in close proximity to a metal disk, which gets "dragged" in the same direction that the magnet spins. The disk's torque acts against the resistance of a spring, deflecting a pointer along a scale, indicating the speed of the vehicle. The faster the magnet spins, the more torque is felt by the disk.

Question 3:
Explain what slip speed is for an AC induction motor, and why there must be such as thing as slip" in order for an induction motor to generate torque.
The difference between the speed of the rotating magnetic field (fixed by line power frequency) and the speed of the rotor is called slip speed". Some amount of slip is necessary to generate torque because without it there would be no change in magnetic flux , seen by the rotor, and thus no induced currents in the rotor.
concept, because it is at the heart of induction motor operation.

Question 5:
What would we have to do in order to reverse the rotation of this three-phase induction motor?
Explain your answer. Describe how the (simple) solution to this problem works. Reverse any two lines. This will reverse the phase sequence (from ABC to CBA).
Notes:
One of the reasons three-phase motors are preferred in industry is the simplicity of rotation reversal. However, this is also a problem because when you connect a three-phase motor to its power source during maintenance or installation procedures, you often do not know which way it will rotate until you turn the power on!

Question 6:
If a copper ring is brought closer to the end of a permanent magnet, a repulsive force will develop between the magnet and the ring. This force will cease, however, when the ring stops moving. What is this effect called?
N .. Magnet .. S ( reaction → force ← Motion )
Also, describe what will happen if the copper ring is moved away from the end of the permanent magnet.
The phenomenon is known as Lenz' Law. If the copper ring is moved away from the end of the permanent magnet, the direction of force will reverse and become attractive rather than repulsive.
Follow-up question: trace the direction of rotation for the induced electric current in the ring necessary to produce both the repulsive and the attractive force.
Challenge question: what would happen if the magnet's orientation were reversed (south pole on left and north pole on right)?

Notes:
This phenomenon is difficult to demonstrate without a very powerful magnet. However, if you have such apparatus available in your lab area, it would make a great piece for demonstration!
One practical way I've demonstrated Lenz's Law is to obtain a rare-earth magnet (very powerful!), set it pole-up on a table, then drop an aluminium coin (such as a Japanese Yen) so it lands on top of the magnet. If the magnet is strong enough and the coin is light enough, the coin will gently come to rest on the magnet rather than hit hard and bounce off.
A more dramatic illustration of Lenz's Law is to take the same coin and spin it (on edge) on a table surface. Then, bring the magnet close to the edge of the spinning coin, and watch the coin promptly come to a halt, without contact between the coin and magnet.
Another illustration is to set the aluminium coin on a smooth table surface, then quickly move the magnet over the coin, parallel to the table surface. If the magnet is close enough, the coin will be "dragged" a short distance as the magnet passes over.
In all these demonstrations, it is significant to show to your students that the coin itself is not magnetic. It will not stick to the magnet as an iron or steel coin would, thus any force generated between the coin and magnet is strictly due to induced currents and not ferromagnetism.

Question 13:
Synchronous AC motors operate with zero slip, which is what primarily distinguishes them from induction motors. Explain what slip" means for an induction motor, and why synchronous motors do not have it.

Synchronous motors do not slip because their rotors are magnetized so as to always follow the rotating magnetic field precisely. Induction motor rotors are become magnetized by induction, necessitating a difference in speed (slip") between the rotating magnetic field and the rotor.

Step-up, step-down, and isolation transformers
Question 1:

Calculate the voltage output by the secondary winding of a transformer if the primary voltage is 35 volts, the secondary winding has 4500 turns, and the primary winding has 355 turns. V/secondary = ( V/ secondary = 443.7 volts )
Notes:
Transformer winding calculations are simply an exercise in mathematical ratios.

Question 3:

Calculate the number of turns needed in the secondary winding of a transformer to transform a primary voltage of 300 volts down to a secondary voltage of 180 volts, if the primary winding has 1150 turns of wire.
N/secondary = ( N/secondary = 690 turns )

Question 5:
Suppose 1200 turns of copper wire are wrapped around one portion of an iron hoop, and 3000 turns of wire are wrapped around another portion of that same hoop. If the 1200-turn coil is energized with 15 volts AC (RMS), how much voltage will appear between the ends of the 3000-turn coil ? ( 37.5 volts AC, RMS. )

Question 17:

In a typical step-up or step-down transformer, the higher-voltage winding usually uses finer gauge wire than the lower-voltage winding. Explain why this is.
The higher-voltage winding handles less current than the lower-voltage winding.

Question 24:

Explain how the construction of a step-down transformer differs from that of a step-up transformer.
Step-down transformers have fewer secondary turns than primary turns, while step-up transformers have more secondary turns than primary turns.

Question 26:
When calculating power in transformer circuits, how do the primary and secondary circuit powers (Pprimary = VprimaryIprimary and Psecondary = VsecondaryIsecondary) compare with each other? Is one greater than the other? If so, which one, and why?

Ideally, P/secondary = P/primary, although this equivalence is never quite exact. In practice, P/secondary will always be a little bit less than P/primary.

Question 27:

Explain why transformers are used extensively in long-distance power distribution systems. What advantage do they lend to a power system?

Transformers are used to step voltage up for efficient transportation over long distances, and used to step the high voltage down again for point-of-use circuits.

Question 31:

Suppose a step-down transformer fails due to an accidental short-circuit on the secondary (load) side of the circuit:
That the transformer actually failed as a result of the short is without any doubt: smoke was seen coming from it, shortly before current in the circuit stopped. A technician removes the burned-up transformer and does a quick continuity check of both windings to verify that it has failed open. What she finds is that the primary winding is open but that the secondary winding is still continuous. Puzzled at this finding, she asks you to explain how the primary winding could have failed open while the secondary winding is still intact, if indeed the short occurred on the secondary side of the circuit. What would you say? How is it possible that a fault on one side of the transformer caused the other side to be damaged?

A short-circuit would cause current in both windings of the transformer to increase.
Notes:
It is important for students to realize that a transformer "reflects" load conditions on the secondary side to the primary side, so that the source "feels" the load in all respects. What happens on the secondary (load) side will indeed be reflected on the primary (source) side.

 

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Electric circuits

You might have been wondering how electrons can continuously flow in a uniform direction through wires without the benefit of these hypothetical electron Sources and Destinations. In order for the Source-and-Destination scheme to work, both would have to have an infinite capacity for electrons in order to sustain a continuous flow! Using the marble-and-tube analogy, the marble source and marble destination buckets would have to be infinitely large to contain enough marble capacity for a "flow" of marbles to be sustained.

The answer to this paradox is found in the concept of a circuit: a never-ending looped pathway for electrons. If we take a wire, or many wires joined end-to-end, and loop it around so that it forms a continuous pathway, we have the means to support a uniform flow of electrons without having to resort to infinite Sources and Destinations:

electrons can flow in a path without beginning or end , continuing forever ( hula - hoop - circuit )

Each electron advancing clockwise in this circuit pushes on the one in front of it, which pushes on the one in front of it, and so on, and so on, just like a hula-hoop filled with marbles. Now, we have the capability of supporting a continuous flow of electrons indefinitely without the need for infinite electron supplies and dumps. All we need to maintain this flow is a continuous means of motivation for those electrons,

It must be realized that continuity is just as important in a circuit as it is in a straight piece of wire. Just as in the example with the straight piece of wire between the electron Source and Destination, any break in this circuit will prevent electrons from flowing through it:

( break ) no flow , continuous electron flow cannot occur anywhere in a " broken circuit " no flow

An important principle to realize here is that it doesn't matter where the break occurs. Any discontinuity in the circuit will prevent electron flow throughout the entire circuit. Unless there is a continuous, unbroken loop of conductive material for electrons to flow through, a sustained flow simply cannot be maintained.

no flow ! continuous electron flow cannot occur anywhere in a " ( break )

REVIEW:
A circuit is an unbroken loop of conductive material that allows electrons to flow through continuously without beginning or end.
If a circuit is "broken," that means its conductive elements no longer form a complete path, and continuous electron flow cannot occur in it.
The location of a break in a circuit is irrelevant to its inability to sustain continuous electron flow. Any break, anywhere in a circuit prevents electron flow throughout the circuit.

DC circuit equations and laws

Ohm's and Joule's Laws
E = IR : I = E/ R : R = E/ I

Joules law
P = IE : P = E2/R : P = I2/R

WHERE ,
E = Voltage in volts .
I = Current in amperes ( amps )
R = Resistance in ohms
P = Power in watts

NOTE: the symbol "V" ("U" in Europe) is sometimes used to represent voltage instead of "E". In some cases, an author or circuit designer may choose to exclusively use "V" for voltage, never using the symbol "E." Other times the two symbols are used interchangeably, or "E" is used to represent voltage from a power source while "V" is used to represent voltage across a load (voltage "drop").

Kirchhoff's Laws
"The algebraic sum of all voltages in a loop must equal zero."

Kirchhoff's Voltage Law (KVL)

"The algebraic sum of all currents entering and exiting a node must equal zero."

Kirchhoff's Current Law (KCL)

Series circuit rules
Uninterruptible Power , UPS ( European )

Components in a series circuit share the same current. I,total = I1 = I2 = . . . In
Total resistance in a series circuit is equal to the sum of the individual resistances, making it greater than any of the individual resistances. R.total = R1 + R2 + . . . Rn
Total voltage in a series circuit is equal to the sum of the individual voltage drops. Etotal = E1 + E2 + . . . En

Ohmeter

Resistance is the measure of electrical "friction" as electrons move through a conductor. It is measured in the unit of the "Ohm," that unit symbolized by the capital Greek letter omega (Ω).

Set your multimeter to the highest resistance range available. The resistance function is usually denoted by the unit symbol for resistance: the Greek letter omega (Ω), or sometimes by the word "ohms." Touch the two test probes of your meter together. When you do, the meter should register 0 ohms of resistance. If you are using an analog meter, you will notice the needle deflect full-scale when the probes are touched together, and return to its resting position when the probes are pulled apart. The resistance scale on an analog multimeter is reverse-printed from the other scales: zero resistance in indicated at the far right-hand side of the scale, and infinite resistance is indicated at the far left-hand side. There should also be a small adjustment knob or "wheel" on the analog multimeter to calibrate it for "zero" ohms of resistance. Touch the test probes together and move this adjustment until the needle exactly points to zero at the right-hand end of the scale.

Although your multimeter is capable of providing quantitative values of measured resistance, it is also useful for qualitative tests of continuity: whether or not there is a continuous electrical connection from one point to another. You can, for instance, test the continuity of a piece of wire by connecting the meter probes to opposite ends of the wire and checking to see the the needle moves full-scale. What would we say about a piece of wire if the ohmmeter needle didn't move at all when the probes were connected to opposite ends?

Digital multimeters set to the "resistance" mode indicate non-continuity by displaying some non-numerical indication on the display. Some models say "OL" (Open-Loop), while others display dashed lines.

What is risk assessment ?
A risk assessment is simply a careful examination of what, in your work, could cause harm to people, so that you can weigh up whether you have taken enough precautions or should do more to prevent harm. Workers and others have a right to be protected from harm caused by a failure to take reasonable control measures.

How to assess the risks in your workplace
Step 1 : Identify the hazards
Step 2 : Decide who might be harmed and how
Step 3 : Evaluate the risks and decide on precautions Step 4 : Record your findings and implement them Step 5 : Review your assessment and update if necessary

When thinking about your risk assessment, remember:


a hazard is anything that may cause harm, such as chemicals, electricity,working from ladders, an open drawer etc; the risk is the chance, high or low, that somebody could be harmed by these and other hazards, together with an indication of how serious the harm could be.




Basic Electronic Components
All components used in electronic circuits have three basic properties, known as resistance, capacitance, and inductance. In most cases, however, one of these properties will be far more prevalent than the other two. Therefore we can treat components as having only one of these three properties and exhibiting the appropriate behavior according to the following definitions:

Resistance
The property of a component to oppose the flow of electrical current through itself.

Capacitance
The property of a component to oppose any change in voltage across its terminals, by storing and releasing energy in an internal electric field.

Inductance
The property of a component to oppose any change in current through itself, by storing and releasing energy in a magnetic field surrounding itself.

As you might expect, components whose main property is resistance are called resistors; those that exhibit capacitance are called capacitors, and the ones that primarily have inductance are called inductors.

 

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Transformer ratios :
Up/ Us = Np/ Ns = Is / Ip
Up is the primary voltage , Ip is the primary current , Np is the number of turns on the primary windings , Us is the secondary voltage , Is is the secondary current , and Ns is the number of turns on the secondary winding ,
a step-down transformer is one in which the secondary voltage is less than the primary voltage , a transformer has a step-down ratio of 20/1 ( i.e. Up / Us = 20/1 ) primary winding consists of 2000 turns

Calculate : a) the number of secondary turns , b) the secondary voltage when the primary is supplied at 230v , a Up / Us = Np / Ns ( 20 / 1 = 2000 / Ns ) therefore … 1 / 20 = Ns / 2000 , therefore … Ns = 2000 / 20 = 100 turns ,
b) since this is a step-down transformer , the voltage is reduced in the ratio 20/1 , therefore … secondary voltage = 1/20 x 230 = 11.5v ( 1 ÷ 20 x 230 = 11.5 )
a single-phase transformer , 230 / 50v , supplies 150A from its secondary side ,
Calculate : the primary current , Up / Us = Is / Ip ( 230 / 50 = 150 / Ip ) therefore … 50 / 230 = Ip / 150 therefore … Ip = 50 x 150 / 230 = 32.6A ( 50 x 150 ÷ 230 = 32.6A )

The 100A H.B.C service fuse BS-1361 revision

At the origin of an installation has a fusing factor of 1.4 ,
The voltage at the supply transformer ( Uo ) is 230V , and the tested value of ( Ze ) At the origin of an installation is 0.36Ω ,
a) how much current will be required to blow the fuse ?
b) what earth leakage current will flow if the Line conductor comes into direct contact with the protective conductor ( the service-cable sheath )
At the origin of an installation ? c ) will the fuse operate under the conditions of b)
a) Minimum fusing current ( I2 ) = 100 x 1.4 = 140A
b) Earth leakage current ( If ) = Uo / Ze = 230 ÷ 0.36 =638.9A
c) the fuse will therefore operate . ( the speed of operation may be estimated by reference to the characteristic graphs contained in the Regs ,
3.8 sec

Unit 201 Working effectively and safely in the electrotechnical environment : Revision ,

Q1 Which of the following documents is non-statutory
a Health and Safety at Work Act
b Electricity at Work Regulations
c COSHH
d BS 7671 Requirements for Electrical Installations.
Q2 Prior to using an electric saw on a construction site, a user check finds that the insulation on the supply flex is damaged. The correct procedure would be to
a replace the cord with a new one
b report the damage to a supervisor after use
c repair the cord with insulation tape
d report the damage to a supervisor before use.
Q3 When carrying out repairs to the base of a street lighting column it is essential to wear
a a safety harness
b high visibility clothes
c gauntlets
d high voltage clothing.
Q4 First aid points are indicated using signs bearing a white cross on a
a yellow background
b blue background
c red background
d green background.
Q5 The type of fire extinguisher which would not be suitable for flammable liquids is
a dry powder
b water
c carbon dioxide
d foam.
2330 Level 2 Electrotechnical Technology 5
Q6 CO2 fire extinguishers are indicated by the colour code
a black
b red
c beige
d blue.
Q7 An independent regulatory body responsible for monitoring standards of electrical installation contractors is the
a Electrical Institute Council
b Institute of Electrical Engineers
c National Electrical Contractors Institute Inspection Council
d National Inspection Council for Electrical Installation Contractors.
Q8 To ensure that a particular item of electrotechnical equipment meets a particular British Standard or BSEN Harmonised Standard, the best source of information would be the
a manufacturer of the equipment
b British Standards Institute
c Institute of Electrical Engineers
d supplier of the equipment.
Q9 Using a scale of 1:50, a 10 mm measurement taken from a plan would be equal to an actual measurement of
a 5 mm
b 5 cm
c 0.5 m
d 5 m.

Unit 201
Q1 – D ,Q2 – D ,Q3 – B ,Q4 – D ,Q5 – B ,Q6 – A , Q7 – D , Q8 – A , Q9 – C

Unit 202 Principles of electrotechnology

Q1 The Tesla is the unit of
a magnetic flux
b molecular flux
c magnetic flux density
d molecular flux density.
Q2 A single rotation of an alternator, intended to provide a 50 Hz supply frequency, will take
a 2 ms
b 20 ms
c 50 ms
d 5000 ms.
Q3 An increase in current through a conductor will lead to
a a decrease in conductor temperature
b a decrease in conductor resistance
c an increase in insulation resistance
d an increase in conductor temperature.
Q4 Four resistors having values of 2 Ω, 2 Ω, 5 Ω and 20 Ω are connected in a parallel circuit arrangement. The total resistance of this circuit is
a 0.8 Ω
b 1.25 Ω
c 29 Ω
d 400 Ω.
Q5 Where P = V I. The value V can be determined using
a V = I / P
b V = P I
c V = P - I
d V = P / I
Q6 A mass of 20 kg is to be raised by a hoist 2 m in 30 seconds. Assuming no losses, the power required to raise this load is
a 13.08 Watts
b 196.2 Watts
c 392.4 Watts
d 1200 Watts.
Q7 The white or grey pvc outer layer of a twin and cpc flat thermoplastic (pvc) cable is the
a conductor
b insulation
c conductor
d sheath.
Q8 The purpose of a bonding conductor is to provide
a an earth fault path
b an equal potential zone
c short circuit protection
d overload protection.
Q9 A 110 V, centre tapped earth, reduced low voltage supply for power tools provides a voltage of
a 25 V between live conductors
b 55 V to earth
c 110 V to earth
d 12 V SELV.
8 2330
Q10 A particular extension lead used on a construction site is coloured yellow to
a indicate its mechanical stress properties
b enable it to be seen in the dark
c indicate the supply voltage to it d enable it to be to be identified as suitable for site use.

Unit 202
Q1 – C ,Q2 – B ,Q3 – D ,Q4 – A ,Q5 – D ,Q6 – A , Q7 – D ,Q8 – B ,Q9 – B ,Q10 – C

c indicate the supply voltage to it : Unit 202 d enable it to be to be identified as suitable for site use. :

 

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Unit 203 Application of health and safety and electrical principles :

Q1 The five main stages of the risk assessment procedure are:
1 identify
2 evaluate
3 record
4 implement
5 review
The order in which they should be carried out is
a 1, 3, 2, 4, 5
b 1, 2, 3, 4, 5
c 2, 1, 4, 3, 5
d 2, 3, 1, 4, 5.
Q2 Before any work is done within an electrical installation, the first procedure would be to
a carry out a risk assessment
b turn off the main switch
c remove all components
d install temporary supplies.
Q3 On a large construction site, inductions are carried out for new members of staff in order to inform them of the
a location of the canteen
b requirements within BS 7671
c fire safety procedure
d location of the nearest wholesaler.
Q4 A suitable means of recording the number of visitors on a large construction site is by the use of a
a day work sheet
b timesheet
c take off sheet
d visitors book.
Q6 In order to prove safe isolation of an electrical circuit, it is essential to use
a a multi-meter
b an insulation resistance tester
c an approved voltage indicator
d a low reading ohm meter.
Q7 In order to determine the amount of accessories required for a particular contract, the best method would be to use the layout drawings and a
a site diary
b take off sheet
c daywork sheet
d time sheet
Q8 The ratio of the true power to apparent power in an a.c circuit is the
a power consumption
b harmonic current
c power factor
d reactive power
Q9 Which of the following is a transmission voltage?
a 400 kV
b 33 kV
c 400 V
d 230 V.
Q10 If a circuit protective device requires 200 A in order to disconnect in the required time, the overall impedance of the earth fault path for a 230 V circuit protected by the device must not exceed
a 0.86 Ω
b 1.15 Ω , 230 ÷ 200 = 1.15
c 2.15 Ω d 2.30 Ω.

Unit 203 Q1 – B , Q2 – A , Q3 – C , Q4 – D, Q6 – C , Q7 – B , Q8 – C , Q9 – A , Q10 – B ,

Unit 205 Installation (buildings and structures)

Q1 Which of the following is a non-statutory regulation
a Electricity at Work Regulations
b Health and Safety at Work Act
c Electricity Safety, Quality and Continuity Regulations
d BS 7671 Requirements for Electrical Installations.
Q2 For a drawing having a scale of 1:50, a wall 10 m long would be drawn to a length of
a 10 mm
b 20 cm
c 50 cm
d 10050 mm.
Q3 The maximum operating temperature for a thermoplastic (pvc) insulated cable with copper conductors is
a 60 OC
b 70 OC
c 105 OC
d 160 OC
Q4 A circuit wired in 1.5 mm2 thermoplastic (pvc) twin with cpc cable is protected by a 16 A device and is used to supply a 230 V 3 kW water heater. If the cable has a rated voltage drop of 29 mV/A/m and the circuit is 24 m long, the actual voltage drop will be
a 2.08V
b 9.07V
c 11.14V
d 69V.
Q5 A non maintained emergency light is classified as NM3. This means that the luminaire will illuminate during
a normal conditions then automatically switch off after three hours
b any power failure for up to three hours
c both normal and power failure conditions then automatically switch off after three hours
d power failures that last longer than three hours but not shorter than three hours.
Q6 An earthing arrangement that has a PEN conductor is
a TN-C-S
b TN-S
c TT
d IT.
Q7 A residual current device will disconnect under
a short circuit conditions only
b both earth fault and short circuit conditions
c earth fault conditions only
d overload conditions only.
Q8 The most suitable item of equipment to mark a straight line in order to install a horizontal conduit over a distance of 4 m is a
a plumb line
b spirit level
c steel tape
d chalk line
Q9 A 50 mm x 50 mm steel trunking has a tabulated space factor of 1037. If the tabulated factor for PVC cable having a cross sectional area of 6 mm2 is 21.2, the maximum number of these cables that can be installed into the trunking is
a 47
b 48
c 49
d 50
Q10 The four electrical tests that should be carried out on a new ring final circuit before it is energised are
a continuity of protective conductors, continuity of ring final circuits, insulation resistance and polarity
b continuity of ring final circuits, insulation resistance, polarity and earth fault loop impedance
c insulation resistance, polarity, earth fault loop impedance and continuity of protective conductors
d polarity, earth fault loop impedance, continuity of protective conductors and continuity of ring final circuits.

Unit 205 Q1 – D , Q2 – B , Q3 – B , Q4 – B , Q5 – B , Q6 – A , Q7 – C , Q8 – D , Q9 – B , Q10 – A

Unit 207 Electrical maintenance

Q1 The Electricity at Work Regulations state that electrical equipment must be maintained to prevent danger. Electrical equipment is defined as all small items of battery-powered equipment up to and including overhead power lines rated at
a 230V
b 400V
c 33kV
d 400kV.
Q2 In order to prevent danger during maintenance operations, a voltage warning notice is required on all electrical accessories where
a voltage is present
b the voltage exceeds 230V and such voltage would not be expected
c the voltage exceeds 400V
d the voltage is below 230V.
Q3 In order to facilitate maintenance of a complex electrical system, the sequence of control for isolation can be best shown using a
a block diagram
b layout drawing
c circuit diagram
d bar chart.
Q4 Balancing loads over three phases will reduce
a phase currents
b earth faults
c neutral currents
d overloads.
Q5 Once a low pressure mercury vapour lamp has lit, the purpose of the choke/ballast unit is to
a discharge voltage
b correct power factor
c suppress radio interference
d limit lamp current.
Q6 Certain discharge luminaires mounted above rotating machine can give the appearance that the machine is at a standstill. This effect is known as
a stroboscopic effect
b robotic effect
c rotor effect
d telescopic effect.
Q7 Guidance on regular inspection and testing of portable appliances can be found in
a BS 7671 Requirements for Electrical Installation
b IEE Guidance Note 3 Inspection and testing
c IEE Code of Practice for In-Service Inspection and Testing of Electrical Equipment
d IEE On Site Guide.
Q8 In order to understand the operating procedure for a particular item of equipment, the best source of information would be the
a company sales representative
b manufacturers’ manual
c manufacturers’ catalogue
d circuit diagram.
Q9 When carrying out maintenance work on a distribution board in a busy walkway, it is advisable to protect others by the use of a
a warning sign
b safety barrier
c temporary bollard
d audible warning device.
Q10 Low pressure mercury vapour lamps should be disposed of by
a throwing in a skip
b putting in a glass recycle point (clear glass only)
c using a suitable lamp crusher d putting in the refuse bins which are collected weekly.

Unit 207 Q1 – D , Q2 – B, Q3 – A, Q4 – C Q5 – D, Q6 – A , Q7 – C , Q8 – B , Q9 – B , Q10 – C,

Double Pole and Single Pole : Service Connector Blocks , for Tails 25mm2 100 Amp 5 way DP Connector Block , Width: - 82mm Height: - 40mm Depth: - 52mm 100 Amp 5 Way Single Pole Connector Block with 1 x 5 Way Connector Blocks ,Height: -64mm Width: - 47mm Depth: - 41mm

Connection Blocks and SERVICE Connection Blocks, SP & DP!
Can anyone help in explaining the differences to me so I can make sure I am choosing the right thing ?
SP and DP means single pole and double pole. Single pole can connect one pole only (say just live to live or just neutral to neutral) whereas double pole can connect two poles (i.e. they can connect live to live and neutral to neutral in one box) They have a separator in the box to keep the live and neutrals well apart!

5 way means up to 5 cables can be attached together. So if you just cut a cable and re-join it through the block, you have used 2 of the 5 ways (cut cable in and cut cable out). This leaves 3 further 'ways' available for 3 further cables.

Therefore you will need one of the two way blocks (or else you will need two of the one way blocks)

 

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Unit 209 Highway electrical systems

Q1 Which of the following normally form part of a highway electrical system?
a Current transformers and photo electrical control units
b Voltage transformers and RCDs
c Ignitors and transformers
d Power factor meter and isolators
Q2 Which of the following is a non-statutory document?
a The New Roads and Street Works Act
b The Electricity Safety, Quality and Continuity Regulations 2002
c BS 7671
d The Electrical Equipment (Safety) Regulations
Q3 Which of the following is a statutory document applicable to operatives carrying out installation and maintenance of highway electrical equipment?
a The Laying of Cables in Public Highways
b H.S.E guidance H.S.G 47 (Avoiding danger from underground and overhead services)
c The Personal Protective Equipment at Work Regulations
d Electricity Association G39/1
Q4 It is necessary to avoid skin contact with quartz lamps, as it
a reduces the life of the lamp
b could cause burns
c generates excessive heat
d lengthens lamp working life.
Q 5 Voltage indication devices need to be proved to ensure
a that the lamp has not gone
b that the highway furniture has been installed correctly
c safe working conditions
d the correct operation of street furniture.
Q6 Relevant information for re commissioning street furniture can be found in
a IEE Guidance Notes 7
b Works instructions
c The Electricity Supply Regulations
d The Petroleum (Consolidation) Act 1928.
Q7 Instruments should be regularly calibrated to
a support local industry
b ensure accuracy when testing
c ensure operative is up to date
d provide a training exercise for operatives.
Q8 To facilitate safe isolation, voltage indicating devices
a must comply with BS 7671
b are used to prove the circuit was dead only
c are used to prove the circuit was alive only
d are “proved” prior to and “re-proved” after isolation.
Q9 Which legislation states the need to avoid live working unless unreasonable in all circumstances?
a BS 7671
b Guidance Notes 3
c On-Site Guide
d The Electricity at Work Regulations
Q10 Which of the following lamps would not normally form part of a highway electrical system?
a Low pressure mercury (T8-T12)
b Low pressure mercury (MBF)
c Metal halide (MBI)
d Incandescent Lamp (100W)
Unit 209 Q1 – C , Q2 – C , Q3 – C , Q4 – A , Q5 – C , Q6 – B ,Q7 – B , Q8 – D , Q9 – D , Q9 – D , Q10 – D

What is a Method Statement ?

A method statement is a sequence of steps taken to complete a work task in a safe manner. The method statement should be written by a person that is competent in the task.

Under no circumstances should a purely generic "ready to print" method statement be used because if there are any accidents or near misses, a non-specific document will be quickly identified and prosecutions may follow.

When a method statement is prepared, the risks are identified during the work sequence. Steps taken to reduce the risk are then determined. Next a series of steps are written down that are to be followed by the person or persons carrying out the works. This sequence of steps should include all health and safety aspects, such as personal protective equipment requirements, tools and equipment, and importantly, safety related equipment such as scaffolding.

All control measures that have been determined whilst preparing a method statement and/or risk assessment should be used as a "tool box" talk prior to the works being carried out. By performing an overview of the method statement and/or risk assessment during the "tool box" talk, everyone involved will have a clear understanding of the work to be carried out, as well as the safe work method sequences and safety equipment required.

What is a Risk Assessment?

In very general terms, a risk assessment is where a task or action is studied to determine what hazards may occur, who could possibly be harmed and how they may be harmed, as well as what current or proposed steps are being and should be taken to reduce any risks. In addition the person or persons responsible for the actions are noted.

The UK Health and Safety Executive require all employers to carry out a risk assessment, and importantly, employers with 5 or more employees are required to record their risk assessments.

Building and Construction Risk Assessment ?

• Work at height
• Safe isolation of live electricity
• Slips and trips
• Manual handling
• Cutting and grinding
• Excavators and excavations
• Mixing cement
• Buried and overhead services
• Confined spaces
• Brick and blockwork

Safety method statements

A method statement is, fundamentally, a written safe system of work, or series of safe systems of work.
Method Statements are agreed between:
* a client and principal contractor; or
* a principal contractor and contractor,
and are produced where work with a foreseeable high hazard content is to be undertaken.* a principal contractor and contractor,

Method Statements

This guidance is for anyone who is asked to provide a method statement as part of their job. It could apply to a range of BBC activities from recording in an airport to the procedure for servicing complex technical equipment. A method statement is a written document which details how you intend to carry out a job or task including all the control measures ( see Risk Assessment Guidelines ) which will be applied. Health and safety method statements are not required by law, but they have proved to be an effective and practical management tool, especially for higher risk work. Assessment Guidelines
Key points
* It's a statement of how you are going to do the job.
* It is closely linked to the risk assessment which says how you are going to do the job safely.
* It should follow a step by step process of logistical arrangements, requirements, monitoring etc, combined with the risk assessment.
* Although not required by law, the person in charge of a work area or work has the right to request one.
* It is not a complicated document, it is based on the who, what, why, when, how principle.
* The complexity of the statement should reflect the complexity of the work and working environment.

Unit 211 Installation, instrumentation and associated equipment

Q1 Which of the following is a statutory piece of legislation?
a BS 7671
b IEE Guidance Notes
c Electricity at Work Regulations
d IEE On-Site Guide
Q2 Which of the following has an impact upon earthing?
a BS EN 60439-01
b BS 4491
c BS 4444
d BS EN 60898
Q3 When an ammeter is connected in a circuit it is essential that it
a has a very low resistance
b has a very high resistance
c is connected across the supply
d is connected across the load.
Q4 When a voltmeter is connected in a circuit it is essential that it
a has a very low resistance
b has a very high resistance
c is connected in series with the supply
d is connected in series with the load.
Q5 To facilitate ease of installation and assembly of equipment reference is made to
a BS 7430
b manufacturer’s catalogues
c GS 38
d data charts.
Q6 Electrostatic sensitive equipment, in transit, need not be protected against damage from
a high temperature
b dust and fibres
c moisture ingress
d day light.
Q7 The purpose of a method statement is to
a ensure compliance with BS 7671
b identify a safe working practice
c provide a training document for staff
d provide instructions to be followed at all times.
Q8 The purpose of a visual inspection is to ensure compliance with BS 7671, Section
a 610 ( chapter 61 )
b 712
c 413
d 314.
Q9 Which of the following has an adverse effect on installed equipment?
a Eddy current damping
b Air damping
c Ambient temperature
d Operating temperature
Q10 Which of the following is not a factor which would affect the type of termination?
a Circuit design current
b Physical space around terminations
c Presence of solid foreign bodies
d Size of conductor

Unit 211 Q1 – C , Q2 – C , Q3 – A , Q4 – B , Q5 – D, Q6 – D , Q7 – B , Q8 – A Q9 – C , Q10 – C,

 

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Unit 213 Panel building

Q1 To ensure safe isolation, voltage indicating devices
a must comply with BS 7671
b are used to prove the circuit dead only
c are used to prove the circuit was alive only
d must be “proved” prior to and “re-proved” after isolation.
Q2 A moving coil meter is not used on an ac circuit because
a it cannot read very small variations
b the direction of the deflection depends upon the direction of the current
c it is non linear
d it does not use the damping effect.
Q3 A common type of cable termination used for ribbon cables in panel building is
a pin
b lug
c insulation displacement
d screw.
Q4 The purpose of a switch is to open or close a circuit
a in the event of a fault current
b under load conditions
c under overload conditions
d automatically after a fault has been repaired.
Q5 Which of the following are appropriate tests for a completed panel?
a Flash testing of components
b Insulation resistance and polarity
c Inspection of conductors for current carrying capacity
d Identification of conductors
Q6 Which of the following is statutory?
a BS 7671
b On-Site Guide
c Electricity at Work Regulations
d GS 38
Q7 What degree of protection is specified for protection against a BS finger?
a IP4XB
b IP5XB
c IP6XB
d IPXXB
Q8 BS 7671 provides appropriate advice on
a heights of panels
b design of panels
c environmental conditions
d instrumentation.
Q9 A device with a BS EN number has been
a agreed for use only within the UK
b agreed for operational use within the EU
c standardised for all operational uses only in the UK
d standardised for use in the EU.
Q10 An isolator built into a panel is used for
a normal load switching
b fault load switching
c short circuit protection
d no load switching.

Unit 213 Q1 – D , Q2 – B ,Q3 – C ,Q4 – B ,Q5 – B ,Q6 – C ,Q7 – D ,Q8 – C ,Q9 – D ,Q10 - D

Unit 215 Electrical machines repair and rewind

Q1 Particular starting arrangements are used when a motor has a rating greater than
a 370 micro watts
b 0.37 watts
c 37 watts
d 370 watts.
Q2 With a star-delta starter the windings are brought out to a terminal box. The voltage applied to the windings at starting is
a VL ÷ √3
b VL x √3
c V Phase ÷ √3
d V Phase x √3 .
Q3 When the field windings of an electrical machine are not connected to its own armature, it is known as
a Self- excited
b Polyphase
c Separately-excited
d Synchronous.
Q4 In a shunt-wound motor the field coil is connected in
a series with the armature
b series with the motor
c parallel with the armature
d parallel with the motor.
Q5 If a motor is required to start against a large starting current it is usual to use
a direct on line starter
b face plate starter
c step down centre tapped starter
d rotor resistance starter.
Q6 The usual method of insulating core laminations is
a low reluctance silicon steel
b surface oxidization
c high frequency air cores
d cellulose paper.
Q7 One method of insulating windings is to use
a shelac
b high reluctance silicon steel
c low frequency dust cores
d pvc.
Q8 A single phase double-wound transformer consists of
a a single core mounted winding
b a winding that carries the difference between Ip and Is
c a single solid core
d two electrically separated coils.
Q9 In a star connected three phase transformer the
a three phases are connected together at the start point
b three phases are separate
c neutral conductor is connected to a single phase
d neutral conductor is electrically separated.

Unit 215 Q1 – D , Q2 – A , Q3 – C , Q4 – C , Q5 – D , Q6 – B , Q7 – A , Q8 – D , Q9 – A ,

Low smoke-emitting cables Normal p.v.c. insulation emits dense smoke and corrosive gases when burning. If cables are to be run in areas of public access, such as schools, supermarkets, hospitals, etc, the designer should consider the use of special cables such as those with thermo-setting or elastomeric insulation which do not cause such problems in the event of fire. This action is most likely to be necessary in areas expected to be crowded, along fire escape routes, and where equipment is likely to suffer damage due to corrosive fumes.

Alternating – current circuit calculations :

In a.c circuit the current is limited by the Impedance ( Z ) Impedance is measured in Ohms , and
Voltage = current ( amperes ) x impedance ( Ohms
U = I x Z
Example 1 : the current through an impedance of 24Ω is 6A
Calculate the voltage drop .
U = I x Z
= 6 x 24 ( = 144V )
Example 2 : the current in a fluorescent – lamp circuit is 1.5A when the voltage is 230V .
Determine the impedance of the circuit ,
U = I x Z
Therefore 230 =1.5 x Z
Therefore Z = 230 ÷ 1.5 = 153.3Ω
Example 3 : An a.c contactor coil has an impedance of 300Ω
Calculate the current it will take from a 400V supply ,
U = I x Z
Therefore 400 = I x 300
Therefore I = 400 ÷ 300 = 1.333
= 1.33A ( correct to three significant figures )

Inductive Reactance :

In alternating – current circuits which set up significant magnetic fields , there is opposition to the current in addition to that caused by the résistance of the wires , this additional opposition is called Inductive Reactance . is so low that may ignore it ; the only opposition to the current is then the Inductive Reactance ,
Inductive Reactance XL = 2πf L ohms
Where L is the Inductance of the coil or circuit in henrys ( H ) and f is the supply frequency in hertz ( Hz ) also ,
UL = I x XL ( I .. L .. UL )
Example 1 : Calculate the inductive reactance of a coil of 0.01H
Inductance when connected to a 50Hz supply .
XL = 2πf L
= 2 x 3.14 x 50 x 0.01
= 314 x 0.01 = 3.14Ω
Example 2 : An inductor is required which will cause a voltage drop of 200V when the current through it is 2A at 50Hz
Calculate the value of the inductor ,
UL = I x XL
Therefore 200 = 2 x XL
Therefore XL = 200 ÷ 2 = 100Ω
XL = 2πf L
Therefore 100 = 314 x L
Therefore L = 100 ÷ 314 = 0.319 ( H )

Capacitive Reactance :

When a capacitor is connected to an a.c. supply , the current is limited by the reactance of the capacitor ( Xc )
Capacitive Reactance Xc = 10-6* ------- 2πf C …. ( small 6* on the 10 )
Where C is the capacitance of the capacitor measured in microfarads (µƒ ) and ƒ is the supply frequency in hertz ( Hz ) also
Example 1 : Calculate the reactance of a 15µF capacitor to a 50Hz supply
Xc = 10/6* - 2πƒC
= 10/6* ----- 2 x 3.14 x 50 x 15
= 10/6* ----- 314 x 15 = 212Ω

Example 2 : A power-factor-improvement capacitor is required to take a current of 10A from a 230V 50Hz supply .
Determine the value of this capacitor . Uc = I x Xc
Therefore 230 = 10 x Xc
Therefore Xc = 230 ÷ 10 = 23Ω
Xc = 10/6* --- 2πƒC

Where 2πƒ = 314 when ƒ = 50 and π = 3.14
Therefore 23 = 10/6* --- 314 x C
1-23 = 314 x C --- 10/6*
C = 10/6* --- 314 x 23 = 138.5 µF

Example 1 : A coil has a résistance of 8Ω and an inductance of 0.08H . Calculate its impedance to a 50Hz supply .
Inductive reactance XL = 2πƒL
= 2πƒ x 0.08
= 314 x 0.08 = 25.12Ω
Z2 = R2 + XL2
=8/2* ( small 2 on the 8 ) + ( 25.12 ) 2* ↔ small 2* on 25.12
= 64 + 630 = 694
Z = √ 694 = 26.34 = 26.3Ω

 

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The kilo-watt-hour ( kWh ) Electricity meters :

The kilowatt-hour ( kWh ) is not a unit of power it is a unit of energy ,
The kW is a unit of power , it is 1000W or 1000 J/s , by definition :
Power = energy / time
Rearranging this we get energy = power x time
If we choose to measure power in kW and time in hours , we have energy ( kWh ) = power ( kW ) x time ( h )
The kilowatt-hour ( kWh ) is a unit of energy , but how much ?
We have seen that , energy ( kWh ) = power ( kW ) x time ( h )
Normally we want energy in Joules ( J ) and time in seconds ( S ) the ( kW ) above is 1000 J/s )
The hour is 3.600 seconds , so
1 kWh = 1000 J/s x 3.600 s
1 kWh = 3.600.000 ( J )
1 kWh = 3.6 MJ ( mega-Joules )
The kWh is a large unit of energy used to measure how much energy is used in homes ,
Electricity meters :
* One place where you can easily see the kWh in use is an electricity meter , an old meter with dials ,
* modern meters are digital , as they are easier to read , knowing how many kWh ( or Units ) of electricity are used allows the calculation
Of its cost ,
Electricity meters readings :
below is a close-up of the dials on the meter , each dial represents a number ,
reading the dials clockwise from the top ,
the Red dial reads the tenths of a unit : 0 in this case
the next dial reads : 2
then : 30 ( three 10s )
and : 200
next : 8.000 ( eight 1.000s )
lastly : 60.000 ,
this gives a total of ( 68.232.0 kWh )
Electricity Bills :
We have worked out that the meter below reads , ( 68.232.0 kWh )
Lets work out the bill , would be if , a year earlier , the meter had read ( 64.822.5 )
Assume each unit of electricity cost ( 0.08 )
Number of units used = 68.232.0 – 64.822.5 = 3.409.5 kWh
Total cost = 0.08 x 3.409.5 = £ 272.76 pounds
Running Cost :
You should now be able to work out how much it costs to run an electrical appliance as long as you know its power rating ,
The time you use it for , and the cost of electricity ,
Energy = power x time , can be used as long as power is in kW and time in hours ,
The formula becomes :
Units used ( kWh ) = power ( kW ) x times ( h )
A , 2.5 kW kettle is used for 2 hours a week , units cost £ 0.08
Units used = each ! 2.5 kW x 2h = 5 kWh
Total cost = units used x cost
= 5 kWh x £ 0.08 = £ 0.40 ,

How does a generator work ?

An electric generator is a device that converts mechanical energy obtained from an external source into electrical energy as the output.

It is important to understand that a generator does not actually ‘create’ electrical energy. Instead, it uses the mechanical energy supplied to it to force the movement of electric charges present in the wire of its windings through an external electric circuit. This flow of electric charges constitutes the output electric current supplied by the generator. This mechanism can be understood by considering the generator to be analogous to a water pump, which causes the flow of water but does not actually ‘create’ the water flowing through it.

The modern-day generator works on the principle of electromagnetic induction discovered by Michael Faraday in 1831-32. Faraday discovered that the above flow of electric charges could be induced by moving an electrical conductor, such as a wire that contains electric charges, in a magnetic field. This movement creates a voltage difference between the two ends of the wire or electrical conductor, which in turn causes the electric charges to flow, thus generating electric current.

Alternator
The alternator, also known as the ‘genhead’, is the part of the generator that produces the electrical output from the mechanical input supplied by the engine. It contains an assembly of stationary and moving parts encased in a housing. The components work together to cause relative movement between the magnetic and electric fields, which in turn generates electricity.

(a) Stator – This is the stationary component. It contains a set of electrical conductors wound in coils over an iron core.

(b) Rotor / Armature – This is the moving component that produces a rotating magnetic field in any one of the following three ways:
(i) By induction – These are known as brushless alternators and are usually used in large generators.
(ii) By permanent magnets – This is common in small alternator units.
(iii) By using an exciter – An exciter is a small source of direct current (DC) that energizes the rotor through an assembly of conducting slip rings and brushes.
The rotor generates a moving magnetic field around the stator, which induces a voltage difference between the windings of the stator. This produces the alternating current (AC) output of the generator.
The following are the factors that you need to keep in mind while assessing the alternator of a generator:
(a) Metal versus Plastic Housing – An all-metal design ensures durability of the alternator. Plastic housings get deformed with time and cause the moving parts of the alternator to be exposed. This increases wear and tear and more importantly, is hazardous to the user.

(b) – Ball bearings are preferred and last longer.

(c) Brushless Design – An alternator that does not use brushes requires less maintenance and also produces cleaner power.

* If a heater takes 12A from a 230V supply , what current will take the voltage falls to 220V ?
The lower voltage will give less current ,
Therefore : current at 220v = 12 x 220 ( smaller ) ------ 230 ( larger ) = 11.4A

* An electric sign takes 25A from a 230V supply . what current will it take if the voltage is raised to 240V ?
The higher voltage will cause more current to flow .
Therefore : current at 240V = 25 x 240 ( larger ) ------- 230 ( smaller ) = 26.1A ( to three significant figures )

* The current through a heating element is 5A when the voltage is 100V , what voltage must be applied to obtain a current of 4.5A ?
The small current will required a lower voltage ,
Therefore : voltage to produce 4.5A = 100 x 4.5 ÷ 5 = 90V

* The current through the field coils of a motor is 2A when the résistance is 250Ω. Due to rise in temperature , the résistance
Increases to 275Ω , if the voltage remains the same , what will be the value of the current ?
Greater résistance gives less current ,
Therefore : new current = 2 x 250 --- 275 = 1.82A

* A heater element having a résistance of 54Ω takes a current of 4.5A , what must its résistance be for it to take 5A at the same voltage ?
A larger current requires less résistance ,
Therefore : new résistance = 54 x 4.5 --- 5 = 48.6Ω

Kilo-Watt example :
Transfer of energy can mean may things-from electrical energy in a circuit To work done repeatedly carrying bricks up a ladder , lets see an example . An electrical heater uses 450.000 ( J ) in 5 minutes , what is its power output ?
Time must be in seconds , so 5 minutes = 300 seconds , ( 5 x 60 = 300 )
Power = 450.000 ----- 300 = 1.500W ( of course , we could write this answer as 1.5 kW .

Kilo-Watt question :
Have a go at this example ,
A man loading of ice cream into a van , he manages to lift 120 each minute , If each box requires 15 ( J ) on average , to move it , calculate his power output , Power = energy transferred ----- time taken = 15 x 120 ÷ 60 = 30W

* Cell :
An energy source providing an electrical potential difference between its two terminals such that a current can flow between them. A cell's energy is stored internally as chemicals that react with each other.
* Circuit :
An electrical circuit consists of an energy source connected by conductors to electrical components.
* Conductor :
A conductor is a medium through which an electric current will flow.
* Deceleration :
Deceleration is acceleration in the opposite direction to the direction of motion - in other words, slowing down, or negative acceleration.
* Force :
Forces are pushes and pulls that make things move or change shape.
* kWh :
The kilowatt-hour is a unit of energy It is equivalent to 3•6 mega joules.
* Magnetism :
Magnetism is a force that can be attractive or repulsive.
* Mass :
A measure of the amount of matter in an object. (Do not confuse mass with weight .)
* motor effect :
The motor effect is the result of two things. Firstly, a current is passed through a conductor: this forms a magnetic field around it. Secondly, if an external magnetic field is present, the fields will repel and a force will be experienced by the conductor. This results in movement.
* Ohm :
The unit of electrical résistance .
* Parallel :
in a parallel electrical circuit, the components share an energy source but do not share the same current.
* Power :
The amount of energy transferred per second. Power is measured in watts and is represented by the symbol, P.
* resistance (R) :
Resistance is the extent to which a conductor , hinders the flow of an electric current. Resistance is measured in Ohms* and is represented by the symbol, R.
* Series :
A series electrical circuit is a circuit where the components share the current flowing around the circuit.
* Speed :
Speed is the rate of change in distance. In other words, how quickly something moves through a given distance. It is normally measured in m/s and its formula is speed = distance ÷ time.
* Volt :
The unit of electromotive force * or potential difference *
* voltage (V) :
The measure of energy available to drive a current *. Voltage is measured in volts * and is represented by the symbol, V.
* Watt :
The unit of power. 1 watt is defined as the consumption of energy at the rate of 1 joule per second.
* Weight :
The force of gravity on a mass. It is given by the equation, weight = mass × gravity where, on Earth, gravity = 9.8 N/kg. (Do not confuse weight with mass * .)

 

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Materials costs , discounts and value-added tax : ( revision )

Example 1 , the list price of a certain item of equipment is £ 17.50 but the supplier allows a trade discount of 35%
Calculate the trade price of the item .
35% of £17.50 = 35 ÷ 100 x 17.50 = £6.125 .
Therefore : trade price = £17.50 - £6.125 = £11.375

Example 2 , if 50mm x 50mm light gauge trunking has a list price of £18.10 per length plus 17.5% Vat ,
With a trade discount of 35% , Calculate the trade price of 14 lengths .
List price of 14 lengths = 14 x £18.10 =£253.40
Trade discount offered is 35%
Therefore : discount = £ 253.40 x 35 ÷ 100 = £88.69
Trade price of 14 lengths = £253.40 - £88.69 = £164.71p
= £164.71 x 17.5 ÷ 100 = £28.82
Cost including Vat = £164.71 + £28.82 = £193.53p

Example 3 , an extension to an existing installation is required ,
A ) conduit with drawn-in cable ,
12.5m of conduit costing £137 per 90m
One through box costing £1.95
Three couplings costing £16.35 per 100
Three locknuts costing £10.23 per 100
Two bushes costing £36.10 per 100
28m of cable costing £16.24 per 100m
24 saddles costing £42.80 per 100
Screws and plugs costing £3.20

( Ignore Vat and trade discount )
A ) conduit £ 137 x 12.5 ÷ 90 = £ 19.03
One through box @ £1.97 = £ 1.97
Couplings £ 16.35 x 3 ÷ 100 = £ 0.49
Locknuts £ 10.23 x 3 ÷ 100 = £ 0.31
Male bushes £ 36.10 x 2 ÷ 100 = £ 0.72
Saddles £ 42.80 x 24 ÷ 100 = £ 10.27
Cable £ 16.24 x 28 ÷ 100 = £ 4.55
Screws & plugs @ £ 3.20 = £ 3.20 + total £ 40.54

One Day you’ll have to sit down and do all this , PS it a start !!

* two alternative tariffs are offered by a supplier for domestic consumers :

either a two-part tariff consisting of a standing charge of 23.5p per day plus a unit charge of 7.3p per kWh , or a block rate comprising
a fixed charge of 29.1p per day plus a unit charge of 5.2p per kWh for daytime use and 3.9p per kWh for night use .
on the assumption that 40% of the units used will be daytime and the remainder night-time use , and assuming 5% VAT is chargeable , find ;
a) the total cost of each tariff for 90 days for 1000 units consumed ;
standing charge = 90 x 23.5p = £21.15
unit charge = 1000 x 7.3p = £ 73.00
total £94.15 x 1.05 : inc . VAT = £98.86
cost of 1000 units on block tariff ;
standing charge = 90 x 29.1p = £26.19
daytime units = 400 x 5.2p = £20.80
night-time units = 600 x 3.9p = £23.40
total £70.39 x 1.05 : inc, VAT = £73.91
b) overall cost per unit :
on two-part tariff 9886 ÷ 1000 = 9.886p
on block tariff 7391 ÷ 1000 = 7.391p

The electricity provided by the source has two basic characteristics, called voltage and current. These are defined as follows:
Voltage
The electrical "pressure" that causes free electrons to travel through an electrical circuit. Also known as electromotive force (emf). It is measured in volts.
Current
The amount of electrical charge (the number of free electrons) moving past a given point in an electrical circuit per unit of time. Current is measured in amperes.
The load, in turn, has a characteristic called resistance. By definition:
Resistance
That characteristic of a medium which opposes the flow of electrical current through itself. Resistance is measured in ohms.
The relationship between voltage, current, and resistance in an electrical circuit is fundamental to the operation of any circuit or device. Verbally, the amount of current flowing through a circuit is directly proportional to the applied voltage and inversely proportional to the circuit resistance. By explicit definition, one volt of electrical pressure can push one ampere of current through one ohm of resistance. Two volts can either push one ampere through a resistance of two ohms, or can push two amperes through one ohm. Mathematically, ( E = I × R,

where
E = The applied voltage, or EMF
I = The circuit current
R = The resistance in the circuit

Allowance for diversity : o.s.g. 1B , old notes

A shop has the following single-phase loads, which are balanced as evenly as possible across the 400V three-phase supply.

2 x 6 kW and 7 x 3kw thermostatically controlled water heaters ( 2 x 6 = 12000 / 7 x 3 = 21000 )
2 x 3 kW instantaneous water heaters
2 x 6 kW and 1 x 4 kW cookers
12 kW of discharge lighting (Sum of tube ratings)
8 x 30 A ring circuits feeding 13 A sockets. ( old fuses 3036 )

Calculate the total demand of the system, assuming that diversity can be applied.

Lighting : 66% total demand : 90% total demand : 75% total demand :
Cookers : 10 A + 30% balance + 5 A for socket : 100% + 80% + 60% : 100% + 80% + 60% :
Motors ( but not lifts ) Not applicable : 100% + 80% + 60% : 100% + 50% :
Instantaneous water heaters : 100% + 100% + 25% : 100% + 100% + 25% : 100%X + 50%
Floor warming installations : 100% No Diversity allowable
Standard arrangements of final circuits : 100%X + 40% : 100% + 50% :

Water heaters (thermostatic)
No diversity is allowable, so the total load will be:
( 2 x 6 ) + ( 7 x 3 ) kW = 12 + 21kw = 33kw
This gives a total single-phase current of I = 33 x 4.17 = 137.6 A
Water heaters (instantaneous) 100% of largest plus 100% of next means that in effect there is no allowable diversity. Single-phase current thus = 2 x 3 x 4.17 = 25.0 A
Cookers 100% of largest = 6 x 4.17A = 25.0 A : 80% of second = 80 ÷ 100 x 6 x 4.17A = 20.0 A : 60% of remainder = 60 ÷ 100 x 4 x 4.17 A = 10.0 A : Total for cookers = 55.O A
Discharge lighting 90% of total which must be increased to allow for power factor and control . ( c/f . 1.8 ) Lighting current = 12kW x 4.17 x 1.8 x 90% ÷ 100 = 81.1 A
Ring circuits First circuit 100%, 50 current is 30 A - 75% of remainder : = 7 x 30 x 75 ÷ 100 = 157.5 A Total current demand for ring circuits = 187.5 A Total single phase current demand = 486.2A balance is assumed, three phase line current = 486.2A ÷ 3 = 162.A
The single-phase voltage for a 400V three-phase system is 400 ÷ √ 3 = 230 V. All loads with the exception of the discharge lighting can be assumed to be at unity power factor, so current may be calculated from I = P/ U ( Thus the current per kilowatt will be ( I = 1000 ÷ 240 = 4.17A ) or ↔ I = 1000 ÷ 230 = 4.37A

* 1kw light bulb how many amps do you need if the voltage is 240v ? Watts = Amps x Volts. 1kw = 1000 watts. amps = watts/volts. 1000 ÷ 240 = 4.17 amps.
( 1000 ÷ 230 = 4.37 )

 

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All large plant ( motors , transformers , e.t.c ) are rated in kVA , unlike most domestic appliances which are rated in kW , the reason for this is best explained by example , if a heating appliance has a power rating of 1 kW at 230V it will take a current of : I = P ---- V = 1000 ÷ 230 = 4.37A but if a motor has a power rating of 1kW at 230V and the motor windings cause of PF of 0.6 , then as ↔ PF = kW ---- kVA ( 1 ÷ 0.6 = 1.667 kVA = 1.667 VA and since : current = volt amperes ---- volts ( I = 1667 ÷ 230 = 7.2A ) had the cable supplying the motor been rated on the kW value it would clearly have been undersized ,

* A single-phase step- down transformer has 796 turns on the primary and 365 turns on the secondary windings . if the primary voltage is 240V Calculate the secondary voltage , also Calculate the secondary current if the primary current is 10A , Vp – Vs = Np -- Ns ( therefore : Vs = Vp x Ns – Np = 240 x 365 ÷ 796 = 110V ) also Vp – Vs = Is – Ip ( therefore : Is = Ip x Vp – Vs ( 10 x 240 ÷ 110 = 21.82A ) note : the larger secondary current , the secondary windings would need to have a larger conductor size than the primary windings to carry this current . if the transformer were of the step-up type the secondary current would be smaller ,

* ( 8.74 kWA : 15.83 kW : kVA total . resistance , inductance & capacitance in installation work : therefore : total reactive component = 0 + 3.3 + 5.44 = 8.74kVA , tan 0 = 8.74 ÷ 15.83 = 0.55 ( therefore : 0 = 28.9° therefore : PF = cos Ø = 0.875 also , cos Ø = 15.83 – kVA ( therefore : kVA = 15.83 - cos Ø = 15.83 ÷ 0.875 = 18kVA at 0.875 PF lagging

* Trigonometry : active component of 8kW = 8kW active component of 5kVA x cos 41.4° = 5 x 0.75 =3.75kW , active component of 6.8kVA = 6.8 x cos 53.1° = 6.8 x 0.6 = 4.08kW , total of active component = 8 + 3.75 + 4.08 = 15.83kW , reactive component of 8kW = 0 reactive component of 5kVA = 5 sin 41.4° = 5 x 0.66 = 3.3kVA reactive component of 6.8kVA = 6.8 sin 53.1° = 6.8 x 0.8 = 5.44kVA as both 5kVA and 6.8kVA have been lagging power factors , their reactive components are added ,

* Cos Ø = W – VA = PF : power may be added by phasor diagram or calculated by Trigonometry ; example , the following loads are connected to a factory supply , 5kVA at 0.77PF lagging , 8kW at a PF of unity , 6.8kVA at 0.6 PF lagging . determine the total load taken from the supply and the overall PF ← method 1 . by phasor diagram , Cos Ø = 0.75 ( therefore a, = 41.4° : Cos B, = 0.6 : ( therefore B, = 53.1°

( total kVA = 18.1kVA : 0 = 29° therefore cos 0 = PF = 0.874 )

Examples:
• An appliance drawing 1000 Watts (1 kW) for 1 hour is said to have consumed 1 kWh of electricity.
• An appliances drawing 2 kW for a period of 2 hours has consumed 4 kWh of electricity.
• A clock radio rated at 2 W will draw 48 Wh or 0.048 kWh over a 24 hour period.
• A 1000 Watt toaster takes 3 minutes to cook toast. The electricity consumed will be 1000 Watts * (3 minutes / 60 minutes) = 50 Wh or 0.05 kWh.

candela (cd)
The SI base unit for measuring the luminous intensity of light. Candela is the Latin word for "candle." The candela is defined as the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 x 1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.

faraday (Fd)
A unit of electric charge. The British electrochemist and physicist Michael Faraday (1791-1867) determined that the same amount of charge is needed to deposit one mole of any element. This amount of charge, equal to about 96 485 coulombs, became known as Faraday's constant. Later, it was adopted as a convenient unit for measuring the charges used in electrolysis. One faraday is equal to the product of Avogadro's number (see mole) and the charge (1 e) on a single electron.

gray (Gy)
The SI derived unit of absorbed dose. Radiation carries energy, and when it is absorbed by matter the matter receives this energy. The absorbed dose is the amount of energy deposited per unit of mass. One gray is defined to be the dose of one joule of energy absorbed per kilogram of matter, or 100 rad. The unit was named after the British physician L. Harold Gray (1905-1965).

horsepower (hp) convert
An old unit of power originating from power exerted by a horse. The horsepower was defined by James Watt (1736-1819) who determined that a horse is typically capable of a power rate of 550 foot-pounds per second. Today the SI unit of power is named for Watt, and one horsepower is equal to approximately 746 watts. (Slightly different values have been used in certain industries.)

joule (J)
The SI derived unit of energy. Energy is said to exist in a variety of forms, each of which corresponds to a separate energy equation, but all resulting with the same energy unit of the joule. Some of the more common forms of energy are: kinetic energy, heat, potential energy, chemical energy, electrical energy, electromagnetic radiation, matter and antimatter. One joule is defined as the amount of work or energy exerted when a force of one newton is applied over a displacement of one metre. One joule is the equivalent energy of one watt of power radiated or dissipated for one second. The joule was named after the British physicist James Prescott Joule (1818-1889).

 

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kelvin (K)
The SI base unit of temperature is 1/273.16 of the thermodynamic temperature of the triple point of water (temperature where water exists as gas, liquid and solid simultaneously). The unit kelvin and its symbol K should be used to express both thermodynamic temperature and an interval or a difference of temperature. In addition to the thermodynamic temperature (symbol T) there is also the Celsius (symbol t) defined by the equation t=T-T0 where T0=273.15 K. Celsius temperature is expressed in degree Celsius (symbol C). The unit 'degree Celsius' is equal to the unit 'kelvin', and a temperature interval or a difference of temperature may also be expressed in degrees Celsius. (The word degree and the sign o must not be used with kelvin or K). Since this temperature is also equal to 0.01 C, the temperature in kelvin is always equal to 273.15 plus the temperature in degrees Celsius. The unit was named after the British mathematician and physicist William Thomson (1824-1907), later known as Lord Kelvin.

kilogram meter (kgfm) convert
A metric unit of torque equal to 9.806 65 newton meters (Nm).

kilogram of force (kgf) convert
A unit of force equal to the gravitational force on a mass of one kilogram. One kilogram of force equals 9.806 65 newton. This unit is also called the kilopond.

lumen (lm)
The SI derived unit for measuring the luminous flux of light being produced by a light source or received by a surface. The luminous intensity of a light source is measured in candela. One lumen represents the total flux of light emitted, equal to the intensity in candela multiplied by the solid angle in steradians (1/(4pi) of a sphere) into which the light is emitted. Thus the total flux of a one-candela light, if the light is emitted uniformly in all directions, is 4pi lumen.

lux (lx)
The SI derived unit for measuring the illumination of a surface. One lux is defined as an illumination of one lumen per square metre or 0.0001 phot.

ohm (Ω )
The SI unit of electric resistance. If a conductor connects two locations having different electric potentials, then a current flows through the conductor. The amount of the current depends on the potential difference and the resistance to the flow of current. This property of a circuit, the electric resistance, is measured in ohms. One ohm is the resistance that requires a potential difference of one volt per ampere of current. The unit was name after the German physicist Georg Simon Ohm (1787-1854).

watt hour (Wh) convert
A unit of work or energy, representing the energy delivered at a rate of one watt for a period of one hour. This is equivalent to exactly 3.6 kilojoule (kJ) of energy.

weber (Wb)
The SI derived unit of magnetic flux. The magnetic flux in webers is equal to the potential difference, in volts, that would be created by collapsing the field uniformly to zero in one second (V s).

* a.c. motor takes 7.5A from a 230V supply and a wattmeter connected in the circuit shows 1380 watts
Calculate the power factor , P = U x I x p.f
Therefore ; 1380 = 230 x 7.5 x p.f . ( Therefore ; p.f = 1380 ---- 230 x 7.5 = 0.8
( there are no units , power factor is a number which is never greater than 1 , )

* The current supplied to a fluorescent circuit is 0.68A at a voltage of 230V and power factor 0.77 ,
Calculate the power supplied , P = U x I x p.f ( = 230 x 0.68 x 0.77 = 120.4W )

* Ammeter – 8A
Voltmeter – 230V
Wattmeter – 1.152kW
Calculate the kVA and the power factor of the load .
kVA = VA ---- 1000 ( = 8 x 230 ÷ 1000 = 1.84kVA )
PF = kW ---- kVA ( 1.152 ÷ 1.84 = 0.6 )

* Power = voltage 2 --- résistance
Calculate the power absorbed by a 40Ω resistor when connected to a 240V d.c. supply .
Power absorbed P = U2* --- R ( = 240 x 240 ÷ 40 = 1440W

* Determine the résistance of a heater which absorbs 3kW from a 240V d.c. supply
P = U2 --- R ( 3000 = 240 :2*: --- R ( therefore : 1 --- 3000 = R --- 240 :2*: )
therefore : R = 240 x 240 ÷ 3000 = 19.2Ω

* Determine the voltage which must be applied to a 9.8Ω resistor to produce 500W of power ,
P = U2 --- R ( 500 = U2* --- 9.8 ) therefore : U2* = 9.8 x 500 = 4900
therefore : U = √ 4900 = 70V

* power = current :2*: x resistance , P = I2* R
Calculate the power absorbed in a resistor of 8Ω when a current of 6A flows ,
P = I2* R ( = 62* X 8 = 36 X 8 = 288w ) 2* small ↔ 2 on 6 ←←←←←

* a current of 12A passes through a resistor of such value that the power absorbed is 50W , what is the value of this resistor ?
P = I 2* R ( 50 = I2:2* x R ) therefore : R = 50 --- 12 x 12 = 0.347

* determine the vale of current which when flowing in a resistor of 400Ω causes a power loss of 1600W ,
P = I 2* R ( therefore : 1600 = I2* x 400 ( therefore : I2* = 1600 ÷ 400 = 4 ) ( therefore : I = √ 4 = 2A )

* A Circuit supplying a shower with a loading of 6kW would have a design current ( Ib ) : Ib = 6 x 1000W ÷ 230 = 26A
The nominal current rating in amps , ( In ) of the protective device ( fuse or circuit-breaker ) for a circuit is selected so that ( In )
is greater than or equal to the design current , ( Ib ) of the circuit ( In ≥ Ib )
so , in the example of 6kW shower ( In ) must be ≥ 26 , select say a 32A circuit-breaker that is ( In ) = 32A
a cable must now be selected so that its rating ( Iz ) in the particular installation conditions exceeds the design current of the load ( Ib )
( Iz ≥ Ib ) 6kW shower load ( Ib = 26 , so ( Iz ≥ 26 )
Where overload protection of the cable is to be provided , as is usual the cable is also selected so that its rating in its installed conditions ( Iz ) exceeds the current rating of the circuit protective device , ( Iz ≥ In )
The 6kW shower circuit with overload protection ( Iz ≥ 32 )
Therefore ( Iz ≥ In ≥ Ib ) 6kW shower with overload protection , this relationship will be satisfied if ( Ib = 26A , In = 32A ) and
The circuit conductors are sized such that ( Iz ) ≥ 32A ,

* 10 down lights , assume 100W demand per lighting point : ( Therefore : I = 10 x 100 ÷ 230 x 1 = 4.3A )
* 3kW immersion heater : ( circuit demand , ( I = 3000 ÷ 230 = 13A )
* 3kW single-phase motor with a power factor of 0.8 , ( current demand of motor circuit ( I = 1000 x 3 = 3000 : ---- ( 230 x 0.8 = 184 )
( 3000 ÷ 184 = 16.3A )

* power : a 10kW machine and a 2kW ( at 230V ) machine ,
Power load = ( 100% – 100 x 10 x 1000 --- 230 ) + ( 75% --- 100 x 2 x 1000 --- 230 ) = 43.4 + 6.52 = 49.92A
( 100 ÷ 100 x 10 x 1000 ÷ 230 = 43.4 ) ( 75 ÷ 100 x 2 x 1000 ÷230 = 6.52 ) …. ( 43.4 + 6.52 = 49.92A )
Diversity of 100% for largest 75% for second

 

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Requirements for Electrical Installations : BS 7671:2008 , 17th Edition An overview of

* Construction and Demolition Site Installations
Reduced low voltage systems are strongly preferred for the supply to portable hand lamps for general use and portable hand tools and local lighting up to 2 kW

For circuits supplying one or more socket-outlets with a rated current exceeding 32 A, an RCD having a rated residual operating current not exceeding 500mA shall be provided


* Earth Loop Fault Impedance
Appendix 14 provides information relating to the interpretation of earth loop impedance test results. It states that compliance is met when
Zs ( m ) ≤ 0.8 x Uo / Ia ( 230 ÷ 24A = 9.58 ( Ze 0.8 x Zs 9.58 =7.66 )

A more precise assessment may be necessary and guidance on how this may be achieved is also given.

* Safety Circuits : 560.7.9
In addition to a general schematic diagram, full details of all electrical safety sources shall be given. The information shall be maintained and displayed adjacent to the relevant distribution board. A single-line diagram is sufficient.

Part 5 Selection and Erection of Wiring Systems
521.8.2 The line and neutral conductors of each final circuit shall be electrically separate from those of every other final circuit.

Disconnection Times
* In a TN system the maximum disconnection time may be extended to 5 s for distribution circuits and other circuits having a rating exceeding 32 A. >
* In a TT system the maximum disconnection time may be extended to 1 s for distribution circuits and other circuits having a rating exceeding 32 A. >

* Disconnection Times
If an earth fault occurs then all final circuits rated at 230 V to earth and having a current rating of 32 A or less must disconnect in the time shown below.
System Disconnection time (s)
TN : 0.4 <
TT : 0.2 <

* Protection Against Electric Shock
The following protective measures are generally permitted:
• Automatic disconnection of supply (ADS) (most commonly used)
• Double or reinforced insulation
• Electrical separation for the supply to one item of current-using equipment
• Extra-low voltage (SELV or PELV)

311.1 : For economic and reliable design, the maximum demand of an installation shall be assessed.
314.1 : Every installation shall be divided into circuits, as necessary to: ……
(iii) Take account of the hazard that may arise from the failure of a single circuit such as a lighting circuit ….

* Part 2 - Definitions
* Line conductor replaces Phase conductor

* The term “Live conductor” still includes the neutral conductor

* Fundamental Principles
134.1.1 Good workmanship by competent persons or persons under their supervision and proper materials …

* Fundamental Principles
131.7 Where danger or damage is expected to arise due to an interruption of supply, suitable provisions shall be made in the installation or installed equipment
132.13 Every electrical installation shall be provided with appropriate documentation

* Object and Effect
Regulation 120.3 : Any intended departure …. requires special consideration by the designer …
The resulting degree of safety of the installation shall be not less than that obtained by compliance with the regulations

* General requirements
*753.411.3.2 : 30 mA RCDs shall be used for automatic disconnection of supply. Heating units manufactured without exposed conductive-parts shall be provided on site with a grid with spacing of not more than 30 mm, or other suitable conductive covering above the floor heating or below the ceiling heating and connected to the protective conductor of the installation.

Part 7 – Special Installations or Locations 753 Floor and ceiling heating systems : General requirements 753.1 Scope: This section applies to the installation of electric floor and ceiling heating systems, either thermal storage or direct heating systems It does not apply to wall heating or outdoor heating systems The risk is one of penetration of the element

* Chapter 41 - Protection against Electric Shock Section 411 - Protective Measure: Automatic Disconnection of Supply 411.3 Requirements for fault protection 411.3.1 Protective earthing and protective equipotential bonding 411.3.1.1 Protective earthing: Exposed-conductive-parts shall be connected by a protective conductor

* Object and Effects 120.3 This Standard sets out technical requirements… Any intended departure…requires special consideration by the designer…The resulting degree of safety of the installation shall be not less than that obtained by compliance with the Regulations.

* Fundamental Principles
131.8 No addition or alteration, temporary or permanent, shall be made to an existing installation, unless…the rating and the condition of any existing equipment, including that of the distributor, will be adequate…. the earthing and bonding arrangements, if necessary for the protective measure applied for the safety of the addition or alteration, shall be adequate.
132.3 The number and type of circuits …knowledge of: (vi) anticipated future demand if specified.
132.13 Every electrical installation shall be provided with appropriate documentation…
134.1.1 Good workmanship by competent persons or persons under their supervision and proper materials …

* Chapter 35, Safety Services, requires that an assessment is made of any need for safety services, such as: - Emergency escape lighting - Fire alarm systems - Fire rescue lifts - Smoke extractors

What is the formula for calculating the full load current of 11 kV generator ? Basically the formula is I = P / V where I = amps, P = power (kV) and V = volts
for a single phase 11 kV genset the formula is 11000 ÷ 400 = 27.5 amps max load. For a 3 - PHASE GEN then : ( 11000 x 0.8 ÷ 400 ÷ 3 = 7.3 amps per phase. )

 

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Three Phase Power and Current : The power taken by a circuit (single or three phase) is measured in watts W (or kW). The product of the voltage and current is the apparent power and measured in VA (or kVA) . The relationship between kVA and kW is the power factor (pf): ( kW = kVA x pf )

Single phase system - this is the easiest to deal with. Given the kW and power factor the kVA can be easily worked out. The current is simply the kVA divided by the voltage. As an example, consider a load consuming 23 kW of power at 230 V and a power factor of 0.86:
* kVA = kW / power factor = 23 ÷ 0.86 = 26.7 kW ( 26700 W ) * Current = W / voltage = 26700 ÷ 230 = 116 A

Three phase system - The main difference between a three phase system and a single phase system is the voltage. In a three phase system we have the line to line voltage ( VLL ) and the phase voltage ( V.LN ), related by: ( V.LL = √3 x V.LN )

I find that the easiest way to solve three phase problems is to convert them to a single phase problem. Take a three phase motor (with three windings, each identical) consuming a given kW. The kW per winding (single phase) has to be the total divided by 3. Similarly a transformer (with three windings, each identical) supplying a given kVA will have each winding supplying a third of the total power. To convert a three phase problem to a single phase problem take the total kW (or kVA) and divide by three.

As an example, consider a balanced three phase load consuming 36 kW at a power factor of 0.86 and voltage of ( 400 V ( V.LL ) and 230 V ( V.LN ):
* three phase power is 36 kW, single phase power = 36 ÷ 3 = 12 kW now follow the above single phase method kVA = kW / power factor = 12 ÷ 0.86 = 13.9 kW ( 13900 W ) Current = W / voltage = 13900 ÷ 230 = 60 A
Easy enough. To find the power given current, multiply by the voltage and then the power factor to convert to W. For a three phase system multiply by three to get the total power.

Using Formulas The above method relies on remembering a few simple principals and manipulating the problem to give the answer. More traditionally formulas may be used to give the same result. These can be easily derived from the above, giving for example:

( I = kW / ( √3 x pf x VLL ), in kA )

Summary By remembering that a three phase power (kW or kVA) is simply three times the single phase power, any three phase problem can be simplified. Divide kW by the power factor to get the kVA. VA is simply the current times the voltage, so knowing this and the voltage can give the current. When calculating the current use the phase voltage which is related to the line voltage by the square root of three. Using these rules it is possible to work out any three phase problem without the need to remember and/or resort to formulas.

Remember When calculating : • x = multiply • / = divide .

Inductive loads , Current Lags the Voltage : Capacitive loads, Current Leads the Voltage.

Billing : it is important to understand why electric utilities bill customers for demand. Let us begin by utilizing a simple formula to compare the electrical energy use of two customers.

The following formula may be used to calculate kilowatt-hours to be billed: kW x Hours of Use = kWH to be billed .

Example 1: Let us compare two customers during a 24-hour period. Customer number one has a steady 25kW load during the entire 24-hour period. Customer number one: 25kW x 24 Hours = 600 kWH accumulated on the kWH register Customer number two has a 600 kW load but only utilizes his equipment for one hour during the entire 24-hour period. Customer number two: 600kW x 1 Hour = 600 kWH accumulated on the kWH register.

If the electric utility billed these two customers for energy alone, they would each receive a bill for the same amount. However, customer number one could be served with a single phase service with self-contained metering. Customer number two would require a large three phase service with instrument rated metering. In addition, customer number two would require larger conductors all the way back to the power plant providing the electrical energy. Demand charges from the Generation and Transmission company may also contribute to higher costs associated with serving customer number two. Therefore, to fairly bill larger customers, electrical meters are often required to record electrical quantities for both energy (kWH) and demand (kW).

Another useful formula that helps explain the functionality of mechanical demand is: kWH ---- H (hours) = kW ( demand )

Example: Assume a customer has a steady load and the utility’s meter records 500 kWH during a 10 hour period. Therefore: 500 kWH ÷ 10 hours = 50 kW demand .

This formula has determined that the customer’s electrical equipment exposed the utility equipment to an average demand of 50 kW during the 10 hour period.

KVA is just kilovolt-amps, or volts times amps divided by 1000:

Kilowatt ( kW ) = Volts x Amperes x Power Factor x 1.732 / 1000

* ( 208V x 22A x 1.732 = 7925.632 ) ↔ ( 7925.632 x pf 0.85% = 6736.7872 ) ↔ ( 6736.7872 ÷ 1000 = 6.736kW ) USA
* ( 230V x 22A x 1.732 = 8763.92 ) ↔ ( 8763.92 x 0.85% = 7449.332 ) ↔ ( 7449.332 ÷ 1000 = 7.449kW ) UK

* Given . we have a medium-sized Sever that draws ( 6.0Amp )

Kilowatt ( kW ) = Volts x Amperes x Power Factor / 1000
( 120V x 6.0A = 720 VA ) ÷ ( 720 x 0.85 = 612 ) ↔ ( 612 ÷ 1000 = .612kW )

( 120V x 6.0A = 720 VA ) ↔ ( 720 x 0.85 = 612 ) ↔ ( 612 ÷ 1000 = .612kW ) this one : ****

Converting VA to Amps How to convert ( VA ) to amps ? : Use the following formula ( A = VA * Pf -- V ) The conversion of VA to Amps is governed by the equation Amps = VA•PF / Volts) For Example 12 VA x 0.6% ÷ (12 volts) = 0.6 amp ( 28VA ÷ 12V = 2.3A ) VA / V = 2.3A

 

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One day you will have to design your on Installation , keep this in mind !!!!!
Whenever amps go running down wires they give off surface heat. The amount of heat given off depends on the wire resistance. Bigger wires have less resistance, and hence give of less heat.

The amount of heat given off leads to a temperature increase, which in turn heats the wire insulation. If the insulation gets too hot, it melts, leading to cable failure. ( If the installation has been designed properly, this will blow a fuse. If the design was wrong, it can start a fire. ) The insulation melt-point depends on the material selected. This is why different cable types can take different amounts of current for the same conductor size.

Fuses are sized so they let enough amps flow down the circuit to service the load, but not enough amps to overheat the cable insulation... which would cause it to fail as mentioned above.

So. First you need to know how many amps the load needs. Next, make sure the fuse size will allow the load to be met. Next, make sure the cable can carry more amps than the fuse.

That’s what they call it → ( * YY ) ↔ PVC / PVC - Number Coded Control Cable : Used as connecting cable, as measuring, checking and control cable in machine tool manufacturing, plant engineering and on assembly lines and production lines to meet stringent safety requirements. Suitable for fixed installation or flexible applications with unrestricted mobility without forced movement control and without exposure to tensile load. installation in dry and moist rooms; outdoor installation not permitted.

That’s what they call it → ( * SY ) ↔ Steel Wire Braid Cable : Power control cable with galvanised steel wire braiding. Suitable for fixed installation or flexible applications. Used as measuring, checking and control cable in machine tool manufacturing, plant engineering and on assembly lines and product lines. For unrestricted mobility without forced movement control and without exposure to tensile load. Used as energy or connecting cable in dry and moist rooms to meet safety requirements. Due to the galvanised steel wire braiding, these cables can even be used under adverse operating conditions or when exposed to high mechanical strain.

SWA Cable : Steel Wired Armoured cable is the accepted standard for underground installations. This enables you to run cables underground or through water filled ducts with confidence. SWA Armoured Cables * Insulation: PVC or XLPE (Cross Linked Polyethylene) * Bedding: Extruded PVC * Armour: Galvanised Steel Wire Armour, Sheath: PVC Black. * Conductor: Stranded or Solid Plain Annealed Copper. * Application: Designed for use in mains supply electricity. These cables are provided with mechanical protection are therefore suitable for external use and direct burial. * Technical Data: Voltage: 600/1000V, Temperature range: 0oC to +90oC . Core Identification: 2 Cores: Blue, Brown
3 Cores: Brown, Black, Grey
4 Cores: Blue, Brown, Black, Grey 5 Cores (6mm and below) : White Insulation With Black Numbers
5 Cores (10mm and above) : Brown, Blue, Black, Grey and G/Y
3 Core ( XLPE 11KV (Copper)
: A 3 core ↔ ( XLPE ↔ screened ) ↔ / SWA / PVC 11 KVIEC and BS6622. Plain annealed circular compacted copper conductors, semi conducting layer XLPE (Cross-linked Polyethylene) semi-conducting tape insulated -individually copper tape screened / PVC extruded bedded / steel wire armoured / PVC sheathed.

In-Line Plugs :
16A . 110V . 2P + E . 50 / 60 Hz ↔ ( Yellow ) 16A . 220 / 320 . 2P + E . 50 / 60 Hz ↔ ( Blue ) 32A . 110V . 2P + E . 50 / 60 Hz . ↔ ( Yellow ) 32A . 220 / 230V . 2P + E . 50 / 60 Hz . ↔ ( Blue ) 16A . 380 / 400V . 3P + N + E . 50 / 60 Hz .↔ ( Red ) 32A . 380 / 400V . 3P + N + E . 50 / 60 Hz .↔ ( Red ) 63A . 220 / 230V . 2P + E . 50 / 60 Hz .↔ ( Blue ) 63A . 380 / 400V . 3P + N + E . 50 / 60 Hz .↔ ( Red ) 125A . 220 / 230V . 2P + E . 50 / 60 Hz .↔ ( Blue ) 125A . 380 / 400V . 3P + N + E . 50 / 60 Hz .↔ ( Red )
In-Line Connectors : are the same , ( the terminology , Male / Female connectors )

 

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The term "RMS Voltage" is often thrown around when the subject of inverters is discussed but how many people actually understand what it is, what it means and what relevance it is to inverters?
In our usual style we will try to answer these questions in simple terms that someone not overly familiar with electronics or electricity can understand. The following explanation is far removed from the usual technical explanations seen in electronic or electrical text books but it is perfectly valid and was chosen so as to be simple for the non technically minded to understand. Some very basic electrical knowledge and some very basic maths is assumed.
Firstly "RMS" stands for "Root Mean Square". The relevance of this term will become apparent later.
Two very simple formulae that you should understand first are:
V=IR = volts = I (amps) * resistance
and
P=VI = Power = Volts * amps
(I is the standard letter used to denote current)
So let's start with a very simple DC circuit using some very simple numbers......
A DC power source of 10 volts feeding into a load resistor of 1 Ohm
The first formula V=IR can be rearranged to I=V/R so we can put these numbers in and we get I=10/1=10 therefore a current of 10 Amps will exist in the load.
Dead simple.
The second formula (P=VI) will show the total power consumed by the load. In this case we will get P=10*10 = 100 Watts. So the load resistor will dissipate 100 watts as heat. That is where all the power goes, into heat.
Now then, if we leave this switched on for one hour, then the load will have produced 100 Watts of heat continuously for 1 hour. That is, obviously, an average power output of 100 Watts for the period of 1 hour.
If we leave it switched on for 30 minutes, then switched off for 30 minutes then it will have produced a total of 100 Watts for the first 30 minutes and 0 Watts for the second 30 minutes. Quite clearly this is an average of 50 Watts over the one hour period.
The power supply of 10 volts was switched on for 30 minutes, then switched off for 30 minutes. So clearly over the 1 hour perdiod, the average voltage was half this ie 5 volts (think about it).
So we know that 10 Volts (average) produced 100 watts (average) and 5 volts (average) produced 50 watts (average) into the same load.
But if we try to calculate this using our two formulae of I=V/R and P=VI we find that the answer is different!
Let's try it. We have an average of 5 volts so the current I=5V/1 Ohm=5 Amps. And using P=VI we therefore get P=5Volts*5Amps=25 Watts. Yet we know from what we just did above that the average power was 50 watts not 25 Watts.
Clearly something is very wrong!
The problem is this. Had we reduced the DC voltage from 10 Volts to 5 Volts then this calculation would work correctly. However we didn't reduce the voltage, we switched it on for half the time, and off for half the time. This apparently gives us an average voltage of 5 Volts.
Look at the formulae again.......

V=IR and P=VI. Rearrange V=IR to I=V/R and it becomes clear that P=V2/R. Therefore power is proportional to the square of the voltage. So if we double the voltage (ie multiply by 2), the power will be increased by 22 ie by 4 times.
Conversely, if we half the voltage (ie divide by 2) the power will be divided by 22 ie 4.
And this is what we found above when we reduced the voltage from 10 volts to 5 volts. The power reduced from 100 watts to 25 watts.
So why didn't this work when we switched the 10 Volt supply on for 30 minutes, then off for 30 minutes?
It's because the power is proportional to the square of the voltage as opposed to being proportional the voltage directly.
So the power is clearly (somehow) related to the average of the voltage (as switching it on and off obviously affects the average voltage and also affects the average power output) and is also related to the square of the voltage as doubling the voltage produces 4 times the power.
Well, let's try this. First square the voltage, so in our first example we get 102 Volts for the first 30 minutes ie 100 Volts, then 02 Volts ie 0 Volts for the next 30 minutes. The average (strictly speaking the arithmetic mean) of this is 50 Volts. But we squared the voltages to start with so let's now take the square root of this = 7.071 Volts.
Firstly take a note of what we did to the voltage above. First we squared it, then we took the arithmetic mean, then we took the square root of the result. So we end up with the square Root, of the Mean of the Square. Hence RMS. Root Mean Square.
And ended up with the very odd looking figure of 7.071 Volts.
However back to our formula:
(remember that this is the RMS voltage of the 10 volts switched on for half of the time and off for half of the time)
V=IR and P=VI
Rearrange V=IR to I=V/R we get I=7.071 Volts/1 Ohm = 7.071 Amps.
And from P=VI we get P=7.071 Volts*7.071 Amps= 50 Watts. So it worked fine.
This is what RMS Voltage is and means. It is the true voltage of a waveform that is not steady. You have seen above how it is distinctly different from the average voltage. The RMS voltage of a changing waveform is that voltage which would produce the same heating effect in a purely resistive load as would a pure DC voltage of the same level.
Now what possible relevance is this to us?
Well, as it happens, most voltmeters and multimeters measure and display the average voltage not the RMS voltage. This can cause serious problems when measuring waveforms that are not steady DC.
In our example of the 10 Volt supply switched on for half the time and switched off for half the time, then clearly the meter would show 10 Volts for 30 minutes, then 0 Volts for the next 30 minutes. However with a quickly changing waveform where the voltage switches on and off at say 50Hz, the meter would show 5 Volts, which is the average voltage and will give erroneous results when trying to compute power.
The shape of the waveform greatly affects the difference between the average voltage and the RMS voltage so there is no simple conversion between the two. The only exceptions to this are with a pure sine wave where the RMS voltage is 1.11 times the average voltage and a perfect squarewave where the RMS and average voltage are the same. With any other waveshape the difference could be literally any number you care to think of.
Special meters are available that measure RMS voltage and they usually state on them "True RMS reading" or similar words. If the meter does not show this (or something similar including the acronym RMS) then it is an average reading meter which will not give the true voltage reading on a changing waveform. They only work on pure sinewaves and steady DC. And in fact when

measuring an AC signal they actually measure the average voltage then the scale is set up to display this multiplied by 1.11 so it displays the RMS voltage but only of a pure sinewave.
One way to calculate the RMS voltage of a waveform is to draw a graph of the waveform on graph paper, divide the waveform up into tiny segments, square each segment, then take an average of these squared segments over one complete cycle, then square root the results. This is, in fact, how modern, microprocessor based, measuring instruments do it.
And to calculate the true power in watts (as opposed to VA) a similar procedure is followed by dividing the voltage waveform up into segments, calculating the current in the load for each segment then averaging the results over one complete cycle. And again this is how modern, microprocessor based, instruments perform power calculations. This also has the advantage of automatically taking into account the power factor and thereby showing the true power as opposed to apparent power.
The main relevance in our field is that of modified sinewave inverters where a normal meter will give grossly inaccurate readings which will always be much lower than the true RMS voltage.
A typical 230 Volt modified sine wave inverter could show anything between 150 and 230 Volts on a normal meter (depending upon the exact waveform).
So next time someone says "What is this RMS thingy all about" you can tell them.

Calculate Amperage :
When calculating the amperage on a branch circuit, you must know if it is a single or a three phase circuit. In case of a 3 phase circuit, you will have a constant multiplier that you'll need to use in the formula. Look at the examples below, and how the formula is used.
Calculating Amperage - 1 Phase : I (Amperage - also known as Current) VA (Volt Amp - also known as Watt) V (Volt) : Formula to Use: I = VA / V Example 1: Find the Amperage of an 2400 VA load on a 120 Volt, 1 phase branch circuit. Use the formula above, and substitute the given values. I = 2400 / 120 = 20 Amps : Example 2: Find the Amperage of an 5600 VA load on a 230 Volt, 1 phase branch circuit. Use the formula above, and substitute the given values. I = 5600 / 230 = 24.34 Amps

 

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Calculating Amperage - 3 Phase
I (Amperage - also known as Current)
VA (Volt Amp - also known as Watt)
V (Volt)
1.732 (Multiplying Factor to use for 3 Phase circuits; Sometimes this is represented as the square root of 3)
Formula to Use: I = VA / (V * 1.732) Example 1:
Find the Amperage of an 2400 VA load on a 240 Volt, 3 phase branch circuit.
Use the formula above, and substitute the given values.
I = 2400 / (240 * 1.732) = 5.78 Amps
Example 2:
Find the Amperage of an 7600 VA load on a 480 Volt, 3 phase branch circuit.
Use the formula above, and substitute the given values.
I = 7600 / (480 * 1.732) = 9.15 Amps

How do photovoltaic (PV) cells work ?
PV cells are panels you can attach to your roof or walls. Each cell is made from one or two layers of semiconducting material, usually silicon. When light shines on the cell it creates an electric field across the layers. The stronger the sunshine, the more electricity is produced.
PV cells come in a variety of shapes and colours, from grey "solar tiles" that look like roof tiles to panels and transparent cells that you can use on conservatories and glass.
The strength of a PV cell is measured in kilowatt peak (kWp) - that's the amount of energy the cell generates in full sunlight.

Solar electricity
• Do you need planning permission? In England and Scotland, you don't need planning permission for most home solar electricity systems, as long as they're below a certain size - but you should check with your local planning officer if your home is a listed building, or is in a conservation area or World Heritage Site.
• Do you have a sunny place to put it? You'll need a roof or wall that faces within 90 degrees of south, and isn't overshadowed by trees or buildings. If the surface is in shadow for parts of the day, your system will generate less energy.

Solar Power Q/A
* Solar cells use semiconductors to turn light into energy.
When light hits a solar cell, semiconductors such as silicon absorb the light and eventually transform it into an electrical current.
* Silicon works as a semiconductor because: Added impurities allow electrons to move freely.
In its ordinary, crystalline form, silicon is not a good conductor because none of its electrons can move freely. But if you add tiny amounts of impurities, electrons can move, creating a tiny current .
* The phosphorous in a solar cell's semiconductor: provides free electrons
Phosphorous has five electrons in its outer shell. Silicon has four. An atom of phosphorous will bond with silicon, leaving one unpaired electron. Phosphorous-laced silicon is called N-type silicon because it carries a negative charge.
* The boron in a solar cell's semiconductors: attracts electrons
Boron has three electrons in its outer shell to silicon's four. So when boron bonds with silicon, there's room for those unpaired electrons coming from N-type silicon's phosphorous. Boron-laced silicon is P-type, or positive, silicon.
* When you combine P-type and N-type silicon in a solar cell: Electrons at the N-P junction eventually form a barrier.
It may seem like all the free electrons from N-type silicon would make a mad dash for the available empty spaces in P-type silicon. In reality, the electrons near the borders of each type mix, gradually creating an effective barrier -- an electrical field.
* Once equilibrium is reached between the N and P sides of a cell, the cell acts as: a diode
Once the N and P sides of a cell reach equilibrium, the system acts like a diode. It allows electrons to move from the P side to the N side, but not the other way around.
* When light hits a solar cell, electrical current comes from: electrons traveling from the N side to the P side along a circuit
When sunlight hits a solar cell, it loosen electrons, creating both free electrons and spaces for the electrons to go. Add a path from N to P, and the electrons will follow it, creating an electrical current along the path.
* A solar cell can only absorb 15 to 25 percent of the sun's energy because: both A and B
When a photon of light hits a solar cell it can free an electron, simultaneously making a place for another electron to go. But if the photon doesn't have enough energy, nothing happens. On top of that, silicon isn't as efficient at moving electrons as a conductor.
* A typical photovoltaic array has 36 cells. If one is shaded, power production drops by: 50 percent
If you put solar cells on your roof, it's important to make sure that the whole array gets full sun. If one cell is shaded, power output is cut in half.
* To use solar energy in your home, you need: an inverter
An inverter converts direct current (DC), which comes from your solar cells , into alternating current (AC), which the appliances in your house can use. Without an inverter, there's not much you can do with your solar panels.

Electrical installation work :
By measurement :
I = 0.77 .
Ø = 30°
Therefore PF = cos Ø ( = 0.866 )
Two 240V fluorescent lamps are arranged to overcome the stroboscopic effect . one unit takes 0.8A at 0.45 PF , leading , the other takes 0.7A at 0.5 PF lagging , determine the total current drawn and the overall PF .
Ia → Unit (A) 0.8A ; 045 PF ,
Ia → Unit (B) 0.7A ; 05 PF ,
I . 240V

Method 1 , by phasors ,
Unit A : cos Ø = 0.45
Therefore Ø = 63.25°
Unit B : cos a = 0.5
Therefore a = 60°

Power in resistive and reactive AC circuits :
Consider a circuit for a single-phase AC power system, where a 120 volt, 60 Hz AC voltage source is delivering power to a resistive load:
( Ac source drives a purely resistive load. 120V – 60Hz . R 60Ω )
ZR = 60 + J0Ω / OR 60Ω < 0°
I = E/Z , I = 120V ÷ 60Ω = I = 2A

In this example, the current to the load would be 2 amps, RMS. The power dissipated at the load would be 240 watts. Because this load is purely resistive (no reactance), the current is in phase with the voltage, and calculations look similar to that in an equivalent DC circuit. If we were to plot the voltage, current, and power waveforms for this circuit,
AC circuit with a purely reactive (inductive) load.
XL = 60.319Ω . ZL = 0 + j60.319Ω or 60.319Ω < 90. I= E/Z ( I = 120V ÷ 60.319Ω . I = 1.989A )
Power is not dissipated in a purely reactive load. Though it is alternately absorbed from and returned to the source.
Note that the power alternates equally between cycles of positive and negative. ( Figure above ) This means that power is being alternately absorbed from and returned to the source. If the source were a mechanical generator, it would take (practically) no net mechanical energy to turn the shaft, because no power would be used by the load. The generator shaft would be easy to spin, and the inductor would not become warm as a resistor would.
Now, let's consider an AC circuit with a load consisting of both inductance and resistance
Load : 120V / 60Hz . L .load 160mH . R.load 60Ω . AC circuit with both reactance and resistance.
XL = 60.319Ω . ZL = 0 + j60.319Ω or 60.319Ω < 90° . ZR . total 60+ j0Ω or 60Ω < 0° . Z total = 60 + j60.319Ω or 85.078Ω < 45.152°
I = E / Z . I = 120V ÷ 85.078 = I = 1.410A
At a frequency of 60 Hz, the 160 mill henrys of inductance gives us 60.319 Ω of inductive reactance. This reactance combines with the 60 Ω of resistance to form a total load impedance of 60 + j60.319 Ω, or 85.078 Ω ∠ 45.152o. If we're not concerned with phase angles (which we're not at this point), we may calculate current in the circuit by taking the polar magnitude of the voltage source (120 volts) and dividing it by the polar magnitude of the impedance (85.078 Ω). With a power supply voltage of 120 volts RMS, our load current is 1.410 amps. This is the figure an RMS ammeter would indicate if connected in series with the resistor and inductor.
We already know that reactive components dissipate zero power, as they equally absorb power from, and return power to, the rest of the circuit. Therefore, any inductive reactance in this load will likewise dissipate zero power. The only thing left to dissipate power here is the resistive portion of the load impedance. If we look at the waveform plot of voltage, current, and total power for this circuit, we see how this combination works

 

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A combined resistive/reactive circuit dissipates more power than it returns to the source. The reactance dissipates no power; though, the resistor does.
As with any reactive circuit, the power alternates between positive and negative instantaneous values over time. In a purely reactive circuit that alternation between positive and negative power is equally divided, resulting in a net power dissipation of zero. However, in circuits with mixed resistance and reactance like this one, the power waveform will still alternate between positive and negative, but the amount of positive power will exceed the amount of negative power. In other words, the combined inductive/resistive load will consume more power than it returns back to the source.
Looking at the waveform plot for power, it should be evident that the wave spends more time on the positive side of the center line than on the negative, indicating that there is more power absorbed by the load than there is returned to the circuit. What little returning of power that occurs is due to the reactance; the imbalance of positive versus negative power is due to the resistance as it dissipates energy outside of the circuit (usually in the form of heat). If the source were a mechanical generator, the amount of mechanical energy needed to turn the shaft would be the amount of power averaged between the positive and negative power cycles.
Mathematically representing power in an AC circuit is a challenge, because the power wave isn't at the same frequency as voltage or current. Furthermore, the phase angle for power means something quite different from the phase angle for either voltage or current. Whereas the angle for voltage or current represents a relative shift in timing between two waves, the phase angle for power represents a ratio between power dissipated and power returned. Because of this way in which AC power differs from AC voltage or current, it is actually easier to arrive at figures for power by calculating with scalar quantities of voltage, current, resistance, and reactance than it is to try to derive it from vector, or complex quantities of voltage, current, and impedance .

• REVIEW:
• In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other.
• In a purely reactive circuit, no circuit power is dissipated by the load(s). Rather, power is alternately absorbed from and returned to the AC source. Voltage and current are 90o out of phase with each other.
• In a circuit consisting of resistance and reactance mixed, there will be more power dissipated by the load(s) than returned, but some power will definitely be dissipated and some will merely be absorbed and returned. Voltage and current in such a circuit will be out of phase by a value somewhere between 0o and 90o.

Circuits :
A circuit is a loop of wire with its ends connected to an energy source such as a cell* or a battery *
One end of the wire is connected to the positive terminal , the other end of the wire is connected to the negative terminal , the wire is connected in this way so a current * can flow through it ,
Advantages of Parallel circuits :

Parallel circuits have two advantages when compared with series circuits .
The first advantage of a parallel circuit is that a failure of one component does not lead to the failure of the other components . this is because a parallel circuit consists of more than one loop and has to fail in more than one place before the other components fail .
The second advantage of parallel circuits is that more components may be added in parallel without the need for more voltage *

It is essential that delegates are familiar with the different types of earthing system likely to be
encountered (TN-C-S, TT and TN-S ).

Electricity System Earthing Arrangements : revision ,
First letter:
T The live parts in the system have one or more direct connections to earth.
I The live parts in the system have no connection to earth, or are connected only through a high impedance.
Second letter:
T All exposed conductive parts are connected via your earth conductors to a local ground connection.
N All exposed conductive parts are connected via your earth conductors to the earth provided by the supplier.

Remaining letter(s):
C Combined neutral and protective earth functions (same conductor).
S Separate neutral and protective earth functions (separate conductors).
TN-C No separate earth conductors anywhere - neutral used as earth throughout supply and installation
TN-S Probably most common, with supplier providing a separate earth conductor back to the substation.
TN-C-S [Protective Multiple Earthing] Supply combines neutral and earth, but they are separated out in the installation.

TT No earth provided by supplier; installation requires own earth rod (common with overhead supply lines).
IT Supply is e.g. portable generator with no earth connection, installation supplies own earth rod.
TN-S The earthing conductor is connected to separate earth provided by the electricity supplier. This is most commonly done by having an earthing clamp connected to the sheath of the supply cable.
TN-C-S The earthing conductor is connected to the supplier's neutral. This shows up as the earthing conductor going onto the connection block with the neutral conductor of the supplier's meter tails. Often you will see a label warning about "Protective Multiple Earthing Installation - Do Not Interfere with Earth Connections" but this is not always present.

TT The earthing conductor goes to (one or more) earth rods, one of them possibly via an old Voltage Operated ELCB (which are no longer used on new supplies). 17th edition
There are probably other arrangements for these systems too. Also, a system may have been converted, e.g. an old TT system might have been converted to TN-S or TN-C-S but the old earth rod was not disconnected.

-&- like to get some off Q/A in : LOOK in the regs , p32/34 , p/187 ( * 32 ) read a must !!!!! earthing

Cables in contact with polystyrene : ←←←←←←
Do not let electrical cables come into contact with polystyrene. It slowly leaches the plasticiser out of the PVC, so that it becomes stiff and brittle. Sometimes it looks like the PVC has melted and run a little.

Question :

Something has failed in this circuit, because the light bulb does not light up when the switch is closed:
Transformer 120V / P : 15V / S .
What type(s) of transformer fault(s) would cause a problem like this, and how might you verify using a multimeter?
A : The most common type of transformer fault causing a problem like this is an open winding. This is very easy to check using a multimeter

Question
Calculate all listed values for this transformer circuit:
48VAC . 13000 turns . ││ 4000 turns . R load . 150Ω .
* V. Primary = / V. Secondary = ,
* I. Primary = / I. Secondary = ,
Explain whether this is a step-up, step-down, or isolation transformer, and also explain what distinguishes the "primary" winding from the secondary" winding in any transformer.
→ A , V. primary = 48Volts ,
V. secondary = 14.77Volts ,
I. primary = 30.3mA ,
I. secondary = 98.5mA ,
This is a step-down transformer.

* the first step in preparing an installation design is to identify the electrical loads, the physical position of the load is required as well as the kVA demand , power factor, voltage , frequency etc ,

Festoon Cable – Internal : Minimum Bending Radius: 10 x cable diameter
SWA Cable - BS-5467 Steel Wire Armoured PVC : Minimum Bending Radius: 1.5mm² - 16mm²: 6 x overall diameter
SWA Cable - BS-6724 Steel Wire Armoured : Minimum Bending Radius: 1.5mm² - 16mm²: 6 x overall diameter
Alarm Cable: Minimum Bending Radius: 12 x overall diameter
Telephone Cable: Minimum Bending Radius 10 x overall diameter
Internal Telecom Cable : Minimum Bending Radius: 8 x overall diameter
( you have to keep in mind the Bending Radius of cables )

 

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Transformer Fault Current Calculation : ( revision )
Transformer Impedance
Transformer Impedance is measured in percent Impedance , this is the percentage of rated Primary Voltage applied to the Transformers’
Primary windings in order for the rated Secondary full load current to flow in the Secondary windings → ( this test is preformed with the Primary winding connected to a variac or variable supply and the Secondary windings shorted-out ) ←
( Transformer Impedance values may vary but typically electrics Transformers will be between 4 to 5% of Primary voltage

Impedance Voltage Vz = Primary Volts (V) x Percent Impedance Z% ----- 100 :
Primary Volts (V) 230 ( Vp
Secondary Volts (V) 110 ( Vs
Transformer Rating (VA) 2000 VA
Transformer Configuration ( CTE )
Percent Impedance ( Z% ) 4
Impedance Voltage ( Vz ) 9.20
Transformer Maximum Earth Fault Current
Maximum Fault Current A = 100 --- Impedance Volts Vz% x Secondary Full Load Current I :
Maximum Earth Fault Current (A) 455.0 * ( Percent Impedance ( Z% ) 4 *
Maximum Earth Fault Current ( kA ) 0.46 * ( Secondary Full Load Current ( I ). 18.18 *
This is the maximum current be achieved at the Transformers Secondary Terminals and doesn’t allow for any Impedance in the
Primary or Secondary circuitry to and from the Transformer .
Transformer Maximum Earth Fault Loop Impedance :
To Calculate the Maximum Earth Fault Current that can be achieved in a circuit fed by a Transformer .
Calculates the Loop Impedance at the end of a circuit fed from the Secondary winding of a Transformer .
When calculating Earth Fault Loop Impedance select the correct Transformer type above
Three-phase Transformers divide the values for ( Vs secondary voltage by √3 and the ( VA by 3 for centre tapped to Earth ( CTE )
Transformers halve the values for ( Vs ) and ( VA ) for RLV / 110V CTE disconnection times & ( Zs figures refer to BS-7671 : 411.8 – table 41.6

Zsec = Zp x }Vs – Vp}2 + {Z% -- 100} x {Vs – VA} + {R1+R2)
0.49Ω ( Zp ) = loop impedance of primary circuit including Source { Ze + 2R1 }
230V ( Vp ) = primary voltage .
55V ( Vs ) = secondary voltage .
1000VA ( VA ) = Transformer rating .
4% ( Z% ) Transformer percentage impedance .
0.2Ω ( R1 + R2 ) = résistance of secondary circuit Phase & Protective conductor .
Primary Circuit Impedance ( Ze + 2R1 ) 0.49 . Zp ( Ze + 2R1 ) Ohms .
Secondary Circuit Impedance ( R1 + R2 ) 0.2 Ω
Total Impedance ( Zsec ) 0.3490 Ωs
Transformer Actual Earth Fault Current :
To calculate actual earth fault current in a transformer we can refer to the below formula taken from BS-7671: 2008 . ( 411.4.5 )
And use the above figure for the transformers total Impedance ( Zsec )

Earth Fault Loop Impedance ( Zs ) Voltage to Earth ( Uo ) ---- Fault Current ( Ia ) :
Fault Current ( Ia ) = Voltage to Earth ( Uo ) ---- Loop Impedance ( Zs ) :
Actual Secondary Fault Current ( Ia ) 157.6 Amps :
Primary Fault Current ( I ) Secondary Fault Current ( Ia ) x Secondary Voltage ( Vs ) ---- Primary Voltage ( Vp ) :
Actual Primary Fault Current ( I ) 37.7 Amps :

Single-Phase Current ( I ) = kVA --- Volts ( V ) 5.0 ÷ 110.0 = ( Current . Amps 45.45 )
Single-Phase ( kVA ) = Volts x Current ( I ) --- 1000 = ( 110.0 x 45.0 ÷ 1000 = ( kVA 4.95 )
Single-Phase Current ( I ) = Power ( kW ) / Power factor ( pf ) --- Volts ( V ) ↔ ( 4.00kW ÷ 0.800pf ÷ 110.0V = 45.45 current (Amps )
Dissipated Power ( W ) = Current ( I2 ) x Résistance ( R ) ↔ 0.010 current (Amps ) 12000.00 Ω = 120* Power ( Watts )
Inductance ( L ) = Volts ( V ) --- ( 2∏ x frequency ( f ) x Current ( I ) ↔ 240V / 50Hz x 18.2A = 0.0420 ( inductance ( Henries ) ←
3-Phase kVA = Volts ( V ) Current ( I ) x √3 ÷ 1000 = 110.0V x 50.0A x √3 ÷ 1000 = 9.53kVA :
3-Phase Current ( I ) kVA / √3 --- Volts ( V ) 10.0 ÷ √3 ÷ 110.0 = 52.49Amps ←
3-Phase Current ( I ) = Power factor ( pf ) --- ( Volts ( V ) x √3 ) 8.00kW ÷ 0.800pf = 10 : 110.0 x √3 = 190.5 : 10 ÷ 190.5 = 52.49 ←

* Frequency : the voltage or current changes from maximum ( plus ) in one direction , through zero to a maximum ( minus ) in the other direction .
This occurs at , f is the frequency in Hertz . ( 1 Hz = 1 cycle / second ,
* Periodic Time :
The time it takes to complete 1 cycle is T seconds ( the periodic time ) it follows that ( T = 1 / f )
* Angular Frequency :
If we think of the voltage and current being generated by a machine that rotates one revolution per cycle , the 1 cycle corresponds to 360° or 2π radian .
( ɷ = 2 π / T rad/s )
* 15 V r.m.s applied across an inductance of 4 µH . calculate the r.m.s . current when the frequency is 200Hz and 200 MHz .
Solution :
20 Hz Xc = 2π fL = 2π x 20 x 4 x 10-6* = 0.5027 mΩ ( Irms = V -- Xc = 15 --- 0.527 x 10-3* = 29.84 kA . )
200 kHz Xc = 2π fL = 2π x 200 x 4 x 10-6* = 5.03Ω ( Irms = V -- Xc = 15 --- 5.03 = 2.984A )
200 MHz Xc = 2π fL = 2π x 2000 x 4 x 10-6* = 5027Ω ( Irms = V -- Xc = 15 --- 5027 = 2.98mA )
* 15 V r.m.s applied across a capacitance of 4.7µf . calculate the r.m.s. current when the frequency is 20Hz & 2000Hz .
Solution :
20Hz Xc = 1 -- 2π fC = 1 -- 2π x 20 x 4.7 x 10-6* = 1693Ω ( Irms = V -- Xc = 15 -- 1693 = 0.00886A )
200Hz Xc = 2π fC = 2π x 200 x 4.7 x 10-6* = 169.3Ω ( Irms = V -- Xc = 15 – 1693 = 0.0886A )
2000Hz Xc = 1 -- 2π fC = 2π x 2000 x 4.7 x 10-6* = 16.93Ω ( Irms = V -- Xc = 15 -- 1693 = 0.886A )
( 6* or 3* small on the number 10 ) ←←←

Phase displacement
Phase displacement\ (Elec.) A charge of phase whereby an alternating current attains its maximum later or earlier. An inductance would cause a lag, a capacity would cause an advance, in phase.

Select Transformer required by multiplying lamp wattage by the number of lamps used i.e. ( 100 x 5w lamps = 500VA transformer )

A string of eighteen identical Christmas tree lights are connected in series to a 120V source, the string dissipates 64.0W.

a) What is the equivalent resistance of the light string ?

P = IV
64W ÷ 120V ( 0.533 )
I = 0.533 A
R = V/I
120V ÷ 0.533A ( 225 )
R = 225 ohms

Equivalent resistance is 225 ohms ?

b) What is the resistance of a single light ?

As it is connected in series 225 ohms ÷ 18 lights = 12.5 ohms

Why is it that when you put two electric lamps into a circuit in parallel with one another, the current through the circuit increases, while when you put those two lamps in series with one another, the current through the circuit decreases ?

When the two lamps are in parallel with one another, they share the current passing through the rest of the circuit. Current arriving at the two lamps can pass through either lamp before continuing its trip around the circuit. The two lamps operate independently and each one draws the current that it normally does when it experiences the voltage drop provided by the rest of the circuit. With both lamps providing a path for current, the current through the rest of the circuit is the sum of the currents through the two lamps.

But when the two lamps are in series with one another, each lamp carries the entire current passing through the circuit. Current arriving at the two lamps must pass first through one lamp and then through the other lamp before continuing its trip around the circuit. There is no need to add the currents passing through the lamps because it is the same current in each lamp. Moreover, the voltage drop provided by the rest of the circuit is being shared by the two lamps so that each lamp experiences roughly half the overall voltage drop. Since lamps draw less current as the voltage drop they experience decreases, these lamps draw less current when they must share the voltage drop. Thus the current passing through the circuit is much less when the two lamps are inserted into the circuit in series than in parallel

Landscape Lighting

Since landscape lights use low voltage, you will need a transformer. How big a transformer you need will be based on how many lights you are planning to use. It's important to know that the farther away lights are from the transformer, the higher the chance of a voltage drop occurring. Voltage drop will cause the lights farther away to be a little less bright than the ones closest to the transformer. How do you prevent the noticeable signs of voltage drop ? By using math, of course! Here’s a formula for figuring out voltage drop:

0.0011 x Total Watts on Cable x Length of Cable = Voltage Drop in %

For example: 8 fixtures x 8 Watts = 64 Watts
0.0011 x 64 watts x 50 feet of cable = 3.52% Voltage Drop

As long as the voltage drop is lower than 8 percent per fixture, you won't notice any dimming. Next, determine what size transformer you will need to power the lights. For instruction purposes, lets say you want to power eight 8-watt path lights, three 50-watt spot lights, and two 12-watt wall lights. First, calculate the total wattage.

8 x 8 = 64 watts
3 x 50 = 150 watts
2 x 12 = 24 watts

64 + 150 + 24 = 238 total watts

You will want to get a 300 watt transformer for this project. Always get a transformer that can handle the addition of any lights down the line. Do not exceed double the load wattage,

Now let’s put it all together. Place the lights where you want them to illuminate. Run the wire from the transformer to the last light in the line

 

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Transformer Calculations
Wattage refers to the amount of power (electricity) consumed by your appliances and equipment. Sounds simple, doesn't it ? Actually it is very simple and it is important to know a little about it because you need to know wattage in order to choose a Converter or a Transformer. The wattage of most appliances and equipment is found on decals or labels of some sort on the appliance or equipment or in the owner's manual. In addition, the decals or labels also usually give the voltage and/or amperage of the appliance.

If the wattage isn't given you can still figure it out if you know the voltage and amperage. If you multiply the voltage times the amperage, the result equals the wattage of the appliance. For example, the decal on the appliance you want to take overseas doesn't list the wattage but gives the voltage as 120 volts and the amperage is 1.1. To find the wattage of that particular appliance, multiply the voltage 120 times the amperage 1.1, ( 120 x 1.1 = 132 watts. ) Convert to UK )

300 watts, times the hours they run per day, times 30 days, divided by 1000, times the cost per kilowatt hour your power company charges. 8 to 18 !!!! depending where you are. ( T wick the pounds around )

300W x 12M x 30D ÷ 1000 = 108 kWH x 10 ? = £ 10.80 per month

First work out the Wattage... that’s easy as a straight V*I on the output side...
You have 12 volts and 1 ampere. Multiplying those together you get 12VA.
First work out the Wattage... that’s easy as a straight V*I on the output side...
12 * 1 = 12 Watts
Now apply the PFC to see what the input side sees by dividing your output wattage by the chosen value of the PFC ( lets use 0.7 )
( 12/0.7 = approximately 17VA ) ↔ 12 ÷ 0.7 = 17.14285714 )

North American 110-120 volt electricity is generated at 60 Hz. (Cycles) ← Alternating Current. Most foreign 220-240 volt electricity is generated at 50 Hz. (Cycles) ← Alternating Current. This difference in cycles may cause the motor in your 60 Hz. North American appliance to operate slightly slower when used on 50 Hz. foreign electricity. This cycle difference will also cause analog clocks and timing circuits that use Alternating Current as a timing base to keep incorrect time. Most modern electronic equipment including battery chargers, computers, printers, stereos, tape and CD players, VCR/DVD players, etc. will not be affected by the difference in cycles.

that being the ~25mA .This figure is the magnetising current, that current being the current required to magnetise the transformer core. The magnetising current depends on the core material, size, shape and makeup.
Then how do you find the total amps out by the other figures ? lets assume we don’t have amps, basically id like to work out the total amount of amps from a 700VA ups ?
V * A = VA
Therefore, A = VA / V

VA = 700 and V = 230, hence A =

700 ÷ 230 = 3A

Your UPS is capable of supplying 3A continually or, in hardware terms, a couple of computers with monitors, your modem and hub/switch without becoming overloaded.

Working on amperes is a bad thing for a UPS as UPSes are rated in VA rather than watts. Working in amps assumes watts and can lead you to overloading your UPS and possibly causing damage.

Solar Panels
Rule of Thumb for sizing Solar Panels: figure out the average current draw of the remote device that you need to power in Milliamps (MA). We're assuming a 12 VDC system. Pick a solar panel with a current rating of at least 10 times this number. Example: your remote Solar Panels device draws 6 MA average. Pick a solar panel of at least 60MA output current, or to calculate power (watts) would be 60MA x 12 VDC = 720 MW which is 7.2 watts so a 10 watt panel will work great for this system.

Rule of Thumb for Batteries: you will want your device to run for 3-5 days without any sun, depending on your weather conditions. (Actually the solar panel will charge a little bit every day, even under sunless conditions.) Example: if your device draws 100MA, figure .1A x 24 hr/day x 5 days = 12 AH (Ampere-Hours). You can only use about 1/2 of the rated battery output so pick a battery rated at least 24 AH.

while marine batteries range from 60 AH to 105 AH. You can connect batteries in parallel ↔ ( + to + and - to - ) to double or triple the available output.

Basic SI units and formulas associated with Electricity :
As mentioned elsewhere, the Watt (W) is the derived SI unit of power, and is defined as one joule per second. When talking about electricity, it is a measure of energy used or generated at a point in time. 1 Kilowatt (kW) = 1000 Watts.
It is often represented by the letter P.
The Ampere (A) is the base SI unit of current. It is normally represented by the letter I.
Voltage (V) is the derived SI unit of electric potential. It is normally represented by the letter V.
Formulas’:
These three units can be tied together in the formula
P = V * I where P is measured in Watts, V is in Volts, & I in Amps. For example. if your current was initially in ( mA ) this must be converted to Amps before using in the formula (e.g. 300mA = 0.3 Amps)
Example 1
A tape player connected to a 12V car battery draws 800mA of current. How much power is it consuming ?
P = V * I = 12 * 0.8 = 9.6 Watts ( 12 x 0.8 = 9.6Watts )
Example 2
A mains operated electric jug draws 9.5 Amps while operating. How much power is it consuming ? Assume the mains is 240V AC.
P = V * I = 240 * 9.5 = 2280 Watts or 2.28 kW ( 240 x 9.5 = 2280. )
Note: The formula for measuring power when dealing with AC is P = V * I * pf. In the above example, ( pf ) would be extremely close to (1) so it can be ignored.
Appliances which plug into standard 240V power outlets are supposed to have a maximum power rating of 2400 Watts. This equates to a maximum current of 10 Amps at 240 Volts.
Measurements of Power :
* The Watt ( symbol W ) is the derived SI unit of power, and is defined as one joule per second. When talking about electricity, it is a measure of energy used or generated at a point in time. 1 Kilowatt (kW) = 1000 Watts.
* The Watt-hour (symbol ↔ Wh ) is a measure of energy used or generated over a period of time. Our electricity providers normally bill us in units of Kilowatt-hours (kWh), one kWh simply being 1000 Wh. Note that the kilowatt-hour is a measure of the instantaneous power multiplied by time and not divided as we are probably more familiar with. e.g. km/h, being kilometres travelled in an hour. The time period is always related to an hour.
- An appliance drawing 1000 Watts (1 kW) for 1 hour is said to have consumed 1 kWh of electricity.
- An appliances drawing 2 kW for a period of 2 hours has consumed 4 kWh of electricity.
- A clock radio rated at 2 W will draw 48 Wh or 0.048 kWh over a 24 hour period.
- A 1000 Watt toaster takes 3 minutes to cook toast. The electricity consumed will be 1000 Watts * (3 minutes / 60 minutes) = 50 Wh or 0.05 kWh. ( 1000 x 3 ÷ 60 = 50 Wh )

 

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Hot Water Power Consumption Formula Derivation : The following two formula are used to derive the resultant third formula: Formula 1 ( Without going into too much detail here ) Q = m x C x T : Where . Q is the amount of heat energy (Joules) . m is the mass of the substance (kg) . C is the specific heat capacity of the substance ( joule / kg / °K ) . T is the change in Temperature ( K or °C ) The mass ( m ) of water is 1 kg per litre, so the number of litres to be heated can be substituted here. The specific heat ( c ) of water is 4200 . Formula 2 ( 1 KWh = 3.6 MJ ) ( By definition, 1 Joule = 1 Watt per second. Knowing that there are 1000 Watts in a kW and 3600 seconds in an hour, the above formula can be derived ). Resultant formula . By combining these two formulas and substituting the specific heat of water value for the constant, the following formula is arrived at: P = ( 4.2 * L * T ) / 3600 . Where . P is the power used in KWh . L is the number of litres of water heated . T is the Temperature difference between the hot water ended up with and the cold water started with in °C. Voltage & Frequency : Abbreviation ( AC ) : Definition ( Alternating Current ) Common Sources ( Household –Mains ) . Abbreviation ( AC ) : Definition ( Direct Current ) Common Sources ( Batteries , Solar Photovoltaic Cells / Panels AC mains in UK is usually specified as 230 or , with a frequency of 50 Hz. ( cycles per second ). This specified voltage is what is called an RMS ( Root Mean Square ) voltage. It's not a constant voltage like DC. Voltage and current are constantly varying around a reference point, following a sinewave pattern. (The waveform on the mains is rarely a pure sinewave. It is often distorted or 'dirty', due in part to some equipment drawing power in spurts rather than at a constant rate). 50 cycles of this sinewave are completed every second. The peak-to-peak voltage of this sinewave is the RMS voltage multiplied by 1.414 ( the square root of 2 ). If we assume the RMS voltage to be 230 Volts, the voltage from one peak of the sine wave to the other is 230 * 1.414 or ??? Volts. This actual RMS voltage from the mains is not usually very steady. It varies with loads on the circuit, location and the time of day. The voltage at my house varies from around 230V to over 248V.

 

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Calculating the amount of electricity consumed to heat water : The following formula can be used to calculate the amount of power required to heat a quantity of water. It assumes 100% efficiency, with no losses. * The start and finish temperatures of the water in °C must be known . If heating water in a jug, adding around 10% to the total should compensate for inefficiencies to see how the formula was derived . Pt = (4.2 * L * T ) / 3600 . where . Pt is the power used in kWh . L is the number of litres of water heated . T is the Temperature difference between the hot water ended up with and the cold water started with in °C . Example 1: Calculate the amount of electricity required to boil 1.5 litres of water in an electric jug if the starting temperature of the water is 20°C. ( P = (4.2 * 1.5 * ( 100-20 )) / 3600 = 140 Wh. ) If we add ~10% to overcome inefficiencies then the figure becomes 154 Wh . Example 2: Calculate the amount of electricity consumed by an electric hot water service to provide 45 litres of water at 50°C for a shower, assuming the water started off at 20°C . P = ( 4.2 * 45 * (50-20)) / 3600 = 1575 Wh or 1.575 kWh. ) This figure isn't allowing for any heat losses in the storage tank which can be quite significant (40% or more) if the water is heated and then stored for later use. - It is also possible to work out how much time it should take to heat water: After coming back from holidays, the 250 Litre hot water system was turned on after being off for a few weeks. How long would you have to wait for the water to heat to say, 50°C so you could enjoy a hot shower ? The starting temperature of the water is 20°C, and the heater element is rated at 3.6 kW (15 Amps). First calculate the kWh required to heat the water from the formula above: Pt = (4.2 * 250 * (50-20)) / 3600 = 8.75 kWh. To work out the time taken, divide Pt (kWh) by the element rating (in kW ) Heating Time = 8.75 kWh / 3.6 kW = 2.43 hours. Again, this figure is assuming 100% efficiency. A more realistic guess might be around 2.8 hours. Deriving instantaneous power (Pi): This method involves timing how long the spinning disc on the meter takes to rotate through a given number of divisions or rotations. The formula given below can be used: Pi = ( 3600 * N) / (T * R ) where: Pi = Real power being used at that point in time in kW . N = The Number of full rotations counted. When measuring smaller loads, it's more appropriate to calculate this as a fraction of a rotation. Normally one rotation consists of 100 divisions . T = Time (in seconds) for the disc to rotate through the N rotations or part of a rotation . R = The number of revolutions per Kilowatt hour (rev/kWh) of the meter being used. This is normally printed on the meter. A few values I've seen are 133.3, 266.6 & 400.

Calculate the power being consumed in the house at present if the meter takes 2 min 15 sec (135 seconds) to rotate through 50 divisions, using a meter with a rev/kWh value of 133.3, and total number of divisions per revolution of 100. P = (3600 * 50/100) / (135 * 133.3) = 0.1 KW or 100 Watts. ( 3600 x 50 ÷ 100 = 1800 ) ↔ ( 135 x 133.3 = 17995.5 ) ↔ ( 1800 ÷ 17995.5 = 0.1 )