When switch S is open, the voltmeter across the battery reads 1.52V. When theswitch is closed , the voltmeter reading drops to 1.37V and the ammeter reads 1.5A. Find the internal resistance of the battery.

r = 0.1 Ω

Explanation:

We will use Ohm's Law in this question: V = IR, where I is the current and R is the resistance.

When the switch is open, the voltmeter reads 1.52 V, and there is no current in the circuit. We can deduce that the internal resistor in the battery causes 0.15 V to dissipate into heat, since the voltmeter reads 1.37 V when the switch is closed.

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