Help working out resistance and supply current please.

[Jacko]

hi,

I’m doing some revision and I can’t quite understand working out the resistance of a circuit or the supply current. Could some one please help and explain how it would be done please?

I have attached a photo of the network and the question.

Thanks

 

[Spoon]

Hi mate.

1. Work out the resistance. Look at the resistors in series and parallel.
2. What is the voltage?

 

[Lucien Nunes]

We must assume, although it does not explicitly state, that they want the resistance 'seen by' the battery, i.e. across its terminals with the battery disconnected. This particular network is easy to simplify step by step, using only the formulae for resistors in series and in parallel.

There are some groups of resistors in series - calculate these first, i.e. replace each series group with a single resistor of equal resistance. Then note how those are connected together, and calculate a suitable resistor to replace them. Then note how that is connected to the last remaining resistor, and calculate.

Show your working and we'll correct you if you go wrong...

 

[Spoon]

I will try and clarify my post above... I come across as a muppet..

1. Total resistance? Look at the resistors. E.g.
1a. R1 & R2 are in series.
1b. R3, R4 & R5 are in series.
The answer for 1a is in parallel with the answer for 1b.
That just leaves R6 that needs adding to the equation.

2. What is the voltage of the circuit. Hard to work out the supply current when you only have the resistance..

 

[telectrix]

first, add the 20 + 30 + 50 (series). call that R3

then the 40 + 60 (series). call that R2.

then you have R2 in parallel with R3. work that out.

then you have that result in series with the 10.

once you have the network resistance, you need the voltage to determine the current. I = V/R.

with those resistance values, you can do it in your head. i did.

 

[telectrix]

and the answer is NOT 42.

 

[Wilko]

and the answer is NOT 42.

But it has the right to be 42 if it wants (and is a paid up member of the Judaean People's Front).

 

[telectrix]

XXXXII. what have the romans done for us.

 

[DPG]

If the diagram is drawn correctly then the supply voltage is 3VDC, ie. 2 single cells in series. Although I suspect the actual figure has been missed out when it was copied down.

 

[Jacko]

Sorry..... it is a 12v battery

Thanks

 

[DPG]

Sorry..... it is a 12v battery

Thanks

So now you can give us your calculations...

 

[Jacko]

 

[Jacko]

Is that correct? Thanks in advance

 

[DPG]

Yes it is. You have 2 lots of 100R in parallel with each other, and then 10R in series with that. This gives the 60R total
Very neat writing as well. You need to sort that before filling any certs in!

 

[Spoon]

The answers are correct. You might want to mark up the diagram with your R1,R2 & R3.

 

[The Ghost]

@Jacko are you sure in that product over sum calculation for resistors in series you have the right values???

 

[The Ghost]

i.e. 40 x 60 /40 + 60 product over sum for two resistors in parallel

 

[Spoon]

Are you saying the answer isn't 60 ohms @Vortigern ?
That's what I've worked it out as.

 

[David Prosser]

I use the one over method for parallel resistors and make it 60 ohm.
My question is make sure you check the voltage as it shows two batteries in that diagram, is it 2x6v for your 12v or should it be 2x12v ?

 

[The Ghost]

I see R1 and R2 as resistors in parallel so product over sum would be my method for working those out first then add the the series. But I was interested to see if the sum used by the example worked is as it should be with regard to the OP.