Voltmeter reading in a parallel circuit with two batteries

[moenste]

Homework Statement

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?

Answer: 9.6 V.

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 = R_V (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?

 

[gneill]

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 $I_3$?

 

[moenste]

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 $I_3$?

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.

 

[gneill]

That I₁ - I₃ = 0?

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

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.

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.

 

[moenste]

Sure, or in other words, $I_1 = I_3$. So there's really only one current:
View attachment 106742

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.

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

 

[gneill]

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

Looks good.

 

[moenste]

Looks good.

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.

 

[gneill]

Yes.