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In a physics worksheet that I have, I stumbled upon a simple circuit with a battery (with it's own internal resistance), a voltmeter parallel to the battery, and a variable resistor parallel to the battery.

Apparently, varying the resistance of the variable resistor can change the reading on the voltmeter.

What I previously believed: Changing the resistance of the resistor doesn't change the potential difference detected by the voltmeter, but instead, it changes the current going through the resistor (less resistance, more current).

Can someone explain this concept to me? Although this may seem like a simple concept, I just can't seem to wrap my head around the basic ideas of voltage, resistance, and current, since the beginning of high school physics...

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  • $\begingroup$ "Changing the resistance of the resistor doesn't change the potential difference detected by the voltmeter..." - That's only when you assume the voltmeter to have $\infty$ resistance. If you try to write the equations for the circuit assuming that the voltmeter has a finite resistance, you will see for yourself that the potential difference between the two terminals of the voltmeter is no longer constant. If you still don't get it, then I think I will post an answer. $\endgroup$ – user243267 Dec 3 '19 at 17:29
  • $\begingroup$ Real-life voltmeters have quite huge resistances but not $\infty$. $\endgroup$ – user243267 Dec 3 '19 at 17:36
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Changing the resistance of the resistor doesn't change the potential difference detected by the voltmeter, but instead, it changes the current going through the resistor

What you say is true only in cases when the circuit has only one resistor.

But in your case here, there are two resistors connected in series, the external variable resistor and the internal resistance of the battery. When multiple resistors are connected in the circuit, voltage dropped across each one depends on their arrangement in the circuit. Moreover no voltmeter is truly ideal (i.e. having infinite resistance). So the resistance of the voltmeter also becomes a part of the circuit since current now flows through it.

So in such a case you may first find the current flowing in the circuit by Ohm's law assuming the connected voltmeter doesn't have any effect on the circuit.

$$I=\frac V {R+r}$$

Now that we've found the current, you can find the voltage dropped across the variable resistor easily by Ohm's law.

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What I previously believed: Changing the resistance of the resistor doesn't change the potential difference detected by the voltmeter, but instead, it changes the current going through the resistor (less resistance, more current).

See the circuit diagram below. It assumes a very high input impedance of the voltmeter.

Current delivered to the resistor $R_L$ connected in parallel with the battery causes a voltage drop across the internal battery resistance $R_b$. That means the voltage across the battery terminals read by the voltmeter will be less than the battery EMF.

$$V_{voltmeter}=EMF-IR_b$$

Varying the resistance across the battery changes the current $I$ which in turn changes the battery terminal voltage read by the voltmeter.

Hope this helps.

enter image description here

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