# Applying Kirchoff's rule

## Homework Statement

In the circuit shown in fig. 27-29, ε1 = 35.0, R1 = 6.00 , and R2 = 1.00 . a. Find the current in resistor R.
b. Find the resistance R
c. Find the unknown emf. ε.
d. If the circuit is broken at point x, what is the magnitude of the current in the 35.0 V battery?

sigmaI=0
sigmaV=0

## The Attempt at a Solution

I got that the current is 2 A, based on the sum of currents at a junction. But I can't solve for the resistance of R. I tried V=IR, with I=2, but I guess I was using the wrong V, since 6 is not the answer to b. Then I figured that I needed b to solve for c, so I came here looking for some help. Thanks in advance.

## Answers and Replies

You can (and pretty much have to) use V = IR. As you already figured out, you need to find the correct value of V. Think how you might use Kirchoff's voltage law (that the sum of voltage drops around a loop in a circuit is zero) to do this. Which loop would you pick?

I'd do it by saying that one point in the circuit is at 'zero' voltage (it doesn't matter where you pick) and work out what the voltage of all the other points relative to this (you can use V = IR across resistors - remembering that the V in this equation is the _difference_ in voltage across the resistor) and of course the voltage difference across each battery is the voltage of the battery.

You should be able to work out a voltage for either side of that resistor like that, _then_ you can use V=IR.

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