# Change in Electric Potential

1. Mar 4, 2008

### daimoku

[SOLVED] Change in Electric Potential

1. The problem statement, all variables and given/known data

The resistances in the figures A and B are all 7.0 ohms, and the batteries are ideal 13 V batteries.

When switch S in figure b is closed, what is the change in the electric potential VR1 across resistor 1?

2. Relevant equations
I know that in figure A if S is closed the change in electric potential across R1 is 0. How does the addition of R3 in figure B change this fact?

Ohm's Law: V=iR

3. The attempt at a solution
I'm not really sure where to begin. I thought the change in electric potential in figure B would be 0 as well. Anybody care to enlighten me?

Last edited: Mar 4, 2008
2. Mar 4, 2008

### dynamicsolo

What is the equivalent resistance of the network in figure (b) with the switch open, and then with it closed? In figure (a), the current will change when the switch is thrown, but the parallel pair are the only resistors the current passes through and the only resistors the voltage drops across. So the voltage across R1 remains 13 V.

The upstream resistor R3, however, will have a different voltage drop before and after the switch is thrown, so R1 and R2 will also.

Last edited: Mar 4, 2008
3. Mar 5, 2008

### daimoku

In figure b, the equivalent resistance is 14 ohms and 10.5 ohms with the switch opened and closed respectively.

So, the change in current across R1 with S open is 6.5V and with S closed it's 4.33V. Therefore, the change in potential difference is -2.17V. Does this sound right?

4. Mar 5, 2008

### dynamicsolo

I think you mean the change in voltage across the resistor...

Yes, I agree with these results.

5. Mar 5, 2008

### daimoku

Yeah, I did mean change in voltage! Thanks a lot for your help!