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Davey1988

Hello Guys

This is a right noob question but its really been bugging me.

ok so i have a pc which is 500watt (labelled) but runs on 110v and 230v.

V/ I X R :
230/38 = 6.05A
110/38 = 3.10A

But using the formula P VI i get the following

i wanna work our i so p/v

500/230 = 2.17
500/110 = 4.54

im changing the voltage on the pc to 110volts and want to know is it the 3.10a reading i use or the 4.54 reading.

i always thought lowering voltage lowers amps as less amps (v/i) but the power traingle says different.

thank you
 
no.lowering the voltage i.e.halving it, means doubling the current to get the same wattage.
 
Ok but why does the power triangle overrule the ohms law triangle.

Isit because a power generation unit that generates wattage is inside the circuit. Or because of resistance of the pc circuit.

I want to know when to use the power triangle and when to use ohms law
 
You would select the appropriate equation based on the information you have available and what you are trying to find, and that may mean using both.

So if you're trying to size cables and you know the power of the equipment it's P=IV (or more precisely I=P/V) time.

If you're trying to find out whether an immersion heater element is still good by measuring it's resistance, and you know it's power, it's I=P/V and then R=V/I. Or you can combine the two, so using this example R=(VxV)/P.

But in essence, you need to know what you want to find and pick the equation that uses what you know.
 
What i was thinking i always thought if you increase the voltage the current increases.

But not in pvi table which is a curve ball.

I heard tou cannot use ohms law on ressistive loads. Is there no criteria for each triangle to use

Still wierd one goes up and one goes down (amps)
 
What does the PC manufacturer instructions say about running on 110v in respect of the supply specifications? I don't understand what you are trying to calculate. Surely if the PC is manufactured to work on either you just plug in the 110v supply. Am I missing something?
 
Think about it like this....

V = I x R

In that case, for a given voltage, if resistance goes up, the current flowing will go down. It's the same with power.

P = I x V

If voltage goes up, you require less current to achieve the same power. However, if the load is purely resistive, the resistance won't change so actually it will yield less power, so an immersion element with a resistance of around 19 ohms will yield a power of 3kW when connected across 240v and only 2.7kW when connected across 230v.

Would write more, but just been called out. Laters.
 
Design spec. I am installing 2 pcs running on a 110 transformer along with alot of control gear.

On the back of the pc it says 230v 3amp. 110v 6 amp.

I was confused as vir the load (amps) would decrease in vir. So surely pvi is what was needed. But my question is why. Why not vir

Reason im wanting to know as vir decreases amps if lower volt and pvi increases
 
Could it be because a power generation source requires 500watt no matter what. If so is this only in power supplys to use pvi. (Creates power&/kinetic energy)
 
a pc is not a resistive load. as the supply usually goes directly to a transformer, the load is mainly inductive. if you get your meter set on ohms and measure the resistance of the transformer primary winding, you'll get a low reading.maybe as low as 1 ohm. using V=IR, that would indicate I=V/R = 230/1 =230A. now that can't be right, can it. this is because as soon as you apply A.C to the winding, it's reactive impedance comes into play.
 
my workings

19ohms 3KW ELEMENT
V / I * R
SO = V / R;
240v = 12.63
230 = 12.10
110 = 5.78a

P / V * I
SO
v * I = WATTS
240*12.63 = 3031.2WATTS
230*12.10 = 2783WATTS
110*5.78a = 635.8WATTS

But if stick to 3kw as element is rated:
watt / voltage = current
3000*240=12.5a
3000*230=13.044a
3000*110=27.273a


So my new question is, is the wattage of say power supplys / heating elements always 3k wattage rated or is that just a power consumption assumed on the voltage.
But if so the element will not work if under voltage.
 
a pc is not a resistive load. as the supply usually goes directly to a transformer, the load is mainly inductive. if you get your meter set on ohms and measure the resistance of the transformer primary winding, you'll get a low reading.maybe as low as 1 ohm. using V=IR, that would indicate I=V/R = 230/1 =230A. now that can't be right, can it. this is because as soon as you apply A.C to the winding, it's reactive impedance comes into play.

Ok, but there is a label on the back of the PC 100v - 240v, 6amp - 3amp.

100v = 6 amp
240v = 3amp

which is reduced amp by increasing voltage. just like PVI table shown if fixed wattage.
 
19ohms 3KW ELEMENT
V / I * RSO = V / R;

240v = 12.63
230 = 12.10
110 = 5.78a

CORRECT As you reduce the voltage, the power reduces

P / V * I
SO
v * I = WATTS
240*12.63 = 3031.2WATTS
230*12.10 = 2783WATTS
110*5.78a = 635.8WATTS

CORRECT

But if stick to 3kw as element is rated:
watt / voltage = current
3000*240=12.5a
3000*230=13.044a
3000*110=27.273a

WRONG the wattage is not fixed. it is a variable dependent on the voltage. the higher the volts, the more current it can puysh through the load, thus increasing the watts.
 
You have to separate power supplies and heating elements. They are very different beasts.

A heating element is typically a purely resistive load. Purely resistive loads have a fixed resistance and consequently the current flowing through them decreases as the voltage decreases, there is a direct linear correlation between the three elements (V=IR is a linear equation, i.e. it's a straight line). Consequently, the power dissipated will go down. So yes, a heating element of 19 ohms connected across 110v will only consume 5.78A meaning it will be using an instantaneous power of 645W. It will still be working, but it will only be dissipating 645W.

PC Power supplies on the other hand are complex beasties and will be a combination of capacitance, inductance and resistance, meaning there is no direct linear relationship between voltage and current (which is where the power factor comes in).

Typically a power supply will be rated based on the power it can provide to the system it is connected to. So in the case of a PC power supply, a 500W unit will be able to deliver 500W in total across it's various output voltages to the PC it is powering. Looking at the figures above you will see that the power consumed at 240v is 720W (assuming a full load of 500W), so you have a loss of 220W due to the conversion. At 100v supply, you're consuming 6A which is an instantaneous power of 600W, so the loss is only 100W. These losses are largely a result of resistance in the switching element junctions (and other inefficiencies) and correlate directly to the input voltage because for a given junction, the higher the voltage, the higher the current flowing through it and consequently the power lost = current flowing through junction x voltage drop across it... P = I x V (or at least that's my understanding of them).

In short, for a non-linear power supply (i.e. one which does not rely purely on a transformer for the conversion) the instantaneous input power will vary based on input voltage and load because the output voltages will be fixed as will the maximum available output current (i.e. the total maximum output power).

Power is an instantaneous value based on other design factors, so for your heating elements, the power they consume will be varying almost constantly based on the slight variations in supply voltage (because their resistance is fixed - as voltage goes up, so too will current and as P=IxV, so too will the power they consume), and for your PC power supply, the power they consume will be based on input voltage and the load placed on it by the PC (because they are designed to provide their output voltages at a specified maximum current - i.e. their total rated output power - as load on the output goes up, power consumed will go up, as input voltage goes down, power consumed [ignoring losses] will stay the same but the current drawn will go up because it's a constant load so I=P/V).

Hope that helps and doesn't add confusion :)
 
it's confused me. more than 2 lines on a post is just too much for the alcohol to dissemminate through my system. need another pint to re=read.
 

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