# How to find current of resistor in complex circuit?

## Homework Statement

http://dl.getdropbox.com/u/119186/Picture%201.png [Broken]

I = I2+I3
I= e/R

## The Attempt at a Solution

http://dl.getdropbox.com/u/119186/snapshot-1253898419.920882.jpg [Broken]
As you can see I got all of it, but part C. I can't figure out what the process is. I tried making equivalent resistors, but there must be some rule I don't understand about resistors and being equivalent or not when there's a junction?

If someone could point me in the right direction I'd appreciate it.

## The Attempt at a Solution

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## Answers and Replies

Delphi51
Homework Helper
Oh, you've done all the hard work!
Just to get you started, work out the voltage drop across the 1 ohm resistor due to the 5.8 A current. That leaves about 24.2 V across the rest of the circuit. You got 3.25 ohms for the parallel combo of the 4 and 17.5; add the 3.3 and you've got 24.2 V across the 6.55 ohm leg so you can find that current easily. Use that to find the drop in the 3.3 ohm resister and you'll know the potential remaining across the 4 and 17.5 combo. Knowing the V across the 17.5 ohm resistor, calc the current.

I'm still stumped. Voltage drop is given as V=-IR yes?

Voltage drop from 1ohm resistor is 5.8 so I have 24.2V left to go into the lower branch. I take:

24.2/3.3 = 7.33A current through the 3.3ohm resistor, right?

Now I do 7.33A*3.25ohm = 23.8225, which I subtract from 24.2, giving me 0.3775A left for R2?

Doesn't work...

Nevermind, got it! Didn't follow your instructions properly :P

Why or what rule let you subtract 5.8Amps from 30 Volts? I don't get that...

Nevermind again. It's Voltage Drop = (5.8Amps)(1ohm) = 5.8volts drop so 24.2 volts left for the rest.

Divide that by 6.55 and you have the total lower leg current of 3.694.

Now find the voltage drop due to the 3.3ohm resistor:

Vdrop = (3.694)(3.3) = 12.19.

24.2-12.19 = 12.0098V left for the 3.25combo resistors.

12.0098/17.5 = 0.68627Amps.

Thanks!