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?
Best Answer
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)