enter image description here Here are two circuits diagram simulated in Falstad. The voltmeter on the left measures the potential difference across the $8 Ω$ resistor. $6V$ at $P_1$ and $2V$ at $P_2$, therefore the Potential difference is $4V$.

The voltmeter reading on the right is $0V$, indicating a balanced circuit. We can also arrive at the same conclusion with a galvanometer: no current flows as there is no potential difference. But in the case when we use a voltmeter to check whether a circuit is balanced, I can't see what is the p.d the voltmeter is measuring to give $0V$ and why it can be used for the purpose of finding unknown EMF, for example. My explanation: Since no current flows through the voltmeter, $P_1$' is $0V$ on the negative side of the battery, $P_2$' must be $0V$ as well assuming my $P_1$' is true. If both $P_1$' and $P_2$' are correct, there will be no voltage down the circuit for the $4 Ω$ resistor, with this argument, voltage at $P_1$' and $P_2$' will be all wrong.

For the sake of further analysis, I disconnected the $4V$ battery on the right with a $5V$ battery, now the $5V$ battery 'wins' the $4V$ $8Ω$ resistor by $1V$, so $0V$ at $P_1$' and $1V$ at $P_2$', $0-1=-1$. The same argument above can of course be mentioned here again.

  • $\begingroup$ "The voltmeter reading on the right is $0V$, indicating a balanced circuit". Not sure I follow you. To me, the voltmeter on the right simply indicates that the resistance of the "wire" between P1 and P2 is negligible. $\endgroup$
    – Bob D
    Sep 28, 2021 at 17:44
  • $\begingroup$ The voltmeter reading on the right is $0 \,\rm V$, indicating a balanced circuit should read The ammeter reading on the right is $0 \,\rm A$, indicating a balanced circuit as the voltmeter is in fact acting as an ammeter. $\endgroup$
    – Farcher
    Sep 28, 2021 at 22:38
  • $\begingroup$ @BobD I have simplified the circuit of potentiometer with a movable contact in order to, hopefully, makes it easier to see why does the voltmeter give that reading. $\endgroup$
    – radastro
    Sep 28, 2021 at 23:32
  • $\begingroup$ @Farcher Yep, that's what I see in exercises problems, ammeter and voltmeter can be used interchangeably in this purpose to test whether a circuit is balanced. I am not sure why the voltmeter can do so. If I swap out the 4V battery with a 5V battery, the voltmeter seems to 'know' the EMF difference with the 6V battery, giving -1 as the output. Whereas in the case when the circuit is balanced, the voltmeter doesn't care about the EMF difference, giving 0 as the output. $\endgroup$
    – radastro
    Sep 28, 2021 at 23:48
  • $\begingroup$ Alright I added a small piece of wire after the 4V battery, so the reading shown the will be what the voltmeter is reading, not showing the EMF of the battery. Now I can say for certain that $P_1$' is 2V, and $P_2$' is 2V. The confusion can now be broken down into: Why $P_1$' is 2V. I think there is some cancelling in potential difference here, 6V-4V=2V $\endgroup$
    – radastro
    Sep 29, 2021 at 0:19

1 Answer 1


You have connected the cell on the right in the opposite direction. In this configuration it is providing a -4 V p.d. (If we take the original cell to be providing +6 V). This means that 4 V are “dropped” across that cell, and then the voltmeter is comparing 2 V to 2 V (i.e. no potential difference).

Your next question: about how it can be used to find an unknown p.d. In this configuration, I can’t see this circuit providing any information about an unknown p.d. You can, however, make a potentiometer.

In that case, you would need a variable resistor on the right of P2’. (Although I’ve usually seen potentiometer questions which have one length of resistance wire with a moveable contact). In that case, by varying the resistance of the variable resistor, you also change the number of volts dropped across the 8 ohm resistor (because you are changing the total resistance of the circuit). The idea is that both the 8 ohm resistor and the unknown cell (which is connected in the opposite direction to the main cell) are lowering the potential. In order to find the voltage of the unknown cell, you vary the variable resistor until the galvanometer reads no current. At that point, you know that the 8 ohm resistor and the cell are dropping an equal number of volts. You can then use the value of the 8 and 4 ohm resistors and the variable resistor to calculate things like current in the circuit etc. and then you can work out the voltage dropped across the 8 ohm resistor, giving you the voltage of the unknown cell.

Look up some potentiometer circuits if you are still confused.

Hope this helps.


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