# Baffling Circut problem

1. Mar 14, 2005

### sys.fail

In the figure provided, all the resistors have a resistance of 4 ohms and all the ideal batteries have an emf of 4 V. What is the current through resistor R?

This question had me baffled for a while, I can't quite trace through the circut. Even my physics teacher got lost with the unbelievable loops in the circut. Any help would be greatly appreciated!

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2. Mar 16, 2005

### sys.fail

ANYONE? Come on, this can't be that hard.

3. Mar 16, 2005

### robphy

I think it's 2A.
Use Kirchoff's Voltage Law.
Start at the junction below the lower-left battery.
Travel up to the upper left corner, passing along three batteries along the way.
Then travel right to the battery near the right-side, passing it "backwards".
Then travel to your resistor, then down to the bottom wire, which is

(+4V) + (+4V) + (+4V) + (-4V) + (-IR) = 0
or
(+8V)=IR... so I=(8 V)/R = (8 V)/(4 ohms)= 2A.

4. Mar 17, 2005

### Chi Meson

Aha! I knew there had to be a trick! Here I was starting to set up a 19 x 19 matrix. After bogging down in all the junctions I thought I'd just wait for the answer to show up. So, to summarize: find a loop that only passes through the batteries and the resistor that you want, eh? Brilliant! I love it!