[Physics] Voltmeter forming a closed circuit

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A battery is connected to a 10Ω resistor as shown in Figure 2. The emf (electromotive force) of the battery is 6.0 V.

Circuit diagram

When the switch is open the voltmeter reads 6.0 V and when it is closed it reads 5.8V.
Explain why the readings are different.

The answer is:

when switch is closed a current flows (through the battery)
hence a pd/lost volts develops across the internal resistance

But why does the voltmeter not form a closed loop with the circuit and hence cause energy to be dissipated by internal resistance, without having the other loop's switch closed?

Best Answer

why does the voltmeter not form a closed loop with the circuit and hence cause energy to be dissipated by internal resistance

It does, but the voltmeter probably has an impedance of 10,000,000 ohms - as most multimeters do - so the current flowing is negligibly small (0.6 microAmps) and therefore the voltage drop due to battery internal resistance is also negligible.

4 x Alkaline AA has resistance maybe 4 x 300 milliohms. At 0.6 microAmps that's a voltage drop of 0.72 microvolts (if I did my sums right - please check)

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