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