Voltmeter reading in a parallel circuit with two batteries

In summary, the voltmeter in the given circuit will have a reading of 9.6 V due to the negligible internal resistance of the batteries and the high resistance of the voltmeter. Using Kirchhoff's rule and taking the entire loop into consideration, it can be determined that the current in the loop is 0.4 A, leading to a voltage drop of 0.4 V or 1.6 V across the resistors. This results in a voltage reading of 9.6 V for the voltmeter.
  • #1
moenste
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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?

ccd53ec942ce.jpg


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 = 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?
 
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  • #2
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##?
 
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  • #3
gneill said:
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.
 
  • #4
moenste said:
That I1 - I3 = 0?
Sure, or in other words, ##I_1 = I_3##. So there's really only one current:
upload_2016-9-30_8-44-46.png

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.
 
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  • #5
gneill said:
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?
 
  • #6
moenste said:
Don't we get 10 - 8 = 4 * I + 1 * I → 2 = 5 * I → I = 0.4 A?
Looks good.
 
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  • #7
gneill said:
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.
 
  • #8
Yes.
 
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Related to Voltmeter reading in a parallel circuit with two batteries

1. How do I calculate the voltmeter reading in a parallel circuit with two batteries?

The voltmeter reading in a parallel circuit with two batteries can be calculated by adding the voltage of each battery and dividing by the total number of branches. This is because in a parallel circuit, the voltage is the same across each branch.

2. Why is the voltmeter reading different in a parallel circuit compared to a series circuit with two batteries?

In a series circuit, the voltage of each battery adds up to create a larger total voltage, resulting in a higher voltmeter reading. In a parallel circuit, the voltage across each branch remains constant, resulting in a lower voltmeter reading.

3. How does the resistance of the voltmeter affect the reading in a parallel circuit with two batteries?

The resistance of the voltmeter affects the reading in a parallel circuit by creating a voltage drop. This means that the voltmeter reading will be slightly lower than the actual voltage of the batteries.

4. Can the voltmeter reading in a parallel circuit be higher than the voltage of one battery?

No, the voltmeter reading in a parallel circuit cannot be higher than the voltage of one battery. This is because the voltmeter is measuring the voltage across each branch, and in a parallel circuit, the voltage across each branch is the same as the voltage of one battery.

5. How does the placement of the voltmeter affect the reading in a parallel circuit?

The placement of the voltmeter does not affect the reading in a parallel circuit as long as it is connected in parallel to the circuit. This is because the voltmeter is measuring the voltage across each branch, and in a parallel circuit, the voltage across each branch remains constant regardless of the placement of the voltmeter.

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