# Voltmeter reading in a parallel circuit with two batteries

1. Sep 30, 2016

### moenste

1. The problem statement, all variables and given/known data
In the circuit below the batteries have negligible internal resistance and the voltmeter V has a very high resistance. What would be the reading of the voltmeter?

2. The attempt at a solution
I used the Kirchhoff's rule: the current that flows from 10 V is I1, current that flows to the voltmeter is (I1 - I3) and the current that flows into the 8 V is I3. So I got 10 - 8 = 4 I3 + 1 I1 and 10 = RV (I1 - I3) + 1 I1. But this doesn't look like it can be solved. We have R which I don't what to do with it. Any ideas please?

2. Sep 30, 2016

### Staff: Mentor

The voltmeter has a high resistance (ideally infinite resistance) so it passes no current. What does that tell you about the relationship between $I_1$ and $I_3$?

3. Sep 30, 2016

### moenste

That I1 - I3 = 0?

I also calculated: I = V / R = 10 / 1 = 10 A and so V1 = 10 * 1 = 10 V and I = 8 / 4 = 2 A, so V2 = 8 V.

4. Sep 30, 2016

### Staff: Mentor

Sure, or in other words, $I_1 = I_3$. So there's really only one current:

No, you need to take the entire loop into consideration in order to determine the current. Write KVL around the loop and solve for the current.

5. Sep 30, 2016

### moenste

Don't we get 10 - 8 = 4 * I + 1 * I → 2 = 5 * I → I = 0.4 A?

6. Sep 30, 2016

### Staff: Mentor

Looks good.

7. Sep 30, 2016

### moenste

And then we have V = I R + 0.4 * 1 = 0.4 V or 0.4 * 4 = 1.6 V. And then either 10 - 0.4 = 9.6 V or 8 + 1.6 = 9.6 V. Should be correct.

8. Sep 30, 2016

Yes.