# Potential divider with voltmeter

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.

• "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. Sep 28, 2021 at 17:44
• 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. Sep 28, 2021 at 22:38
• @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. Sep 28, 2021 at 23:32
• @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. Sep 28, 2021 at 23:48
• 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 Sep 29, 2021 at 0:19