# Finding the Potential (Kirchoff's Law)

1. Feb 24, 2012

### Xarvist

1. The problem statement, all variables and given/known data
http://puu.sh/i8Bg

2. Relevant equations
V=IR
ƩV = 0

3. The attempt at a solution
Starting behind the left EMF source I sum the potential changes using Kirchoff's Voltage Law
150V - R2I - 50V - R1I = 0
Solving for I, I get 20Amps

However I know this can't be right because the potential at point Q is 97V
and after traversing the circuit, the potential according ot the calculated I is 110 Volts..

Can someone perhaps show me step-by-step to do problems like these?

2. Feb 24, 2012

### PeterO

Note that there is no Earth connected to this circuit, so no point on the circuit is necessarily at 0 V.

If the current flows anti-clockwise, the potential goes up 150 as you pass through the left battery, drop 40 or 60 through the bottom resistor [having trouble which resistor is which - one you start to reply the picture of the circuit is gone], drop another 50 going "backwards" through the ideal battery on the right, then another 60 or 40 through the top resistor.
So I think P should be at -13 or 7 depending which resistor was the 2Ω and which was the 3Ω

3. Feb 24, 2012

### Xarvist

Okay I understand everything but the very end. What calculations did you do to get those two numbers?

4. Feb 24, 2012

### lazyaditya

Using loop laws you can get the above answer ! join PQ by arbitrary line and can calculate the voltage difference between points P and Q using Loop law.

5. Feb 24, 2012

### PeterO

As I said, as soon as I start answering the post, I no longer have access to the circuit diagram. I remembered that the 150V battery was on the left and the 50V battery was on the right. I didn't remember which resistor was on the top and which was on the bottom. I do know there is a 60v drop across the 3Ω resistor and 40V drop across the 2Ω. Depending which one was top/bottom determines whether Q is at -13 [110 below], or 7 [90 below] point Q.
I figured you would be able to work out which one - as you ahve the original circuit, and a greater need to solve the problem.

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