The following figure shows an arrangement to measure the emf ε and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistance. The voltmeter reads 1.52 V when the switch S is open. When the switch is closed, the voltmeter reading drops to 1.45 V and the ammeter reads 1.0 A. Find the emf and the internal resistance of the battery.
When the switch S is opened, loop 2 will be open and no current will pass through the ammeter. Loop 1 will be closed; so, a current will flow through it. But since the voltmeter has very high resistance, compared to the internal resistance of the battery, the voltage-drop across the internal resistance can be ignored, compared to the voltage drop across the voltmeter. So, the voltage appearing across the voltmeter will be almost equal to the emf of the battery.
∴ ε = 1.52 V
(b) When the switch is closed, current will pass through the circuit in loop 2. In this case, there will be a voltage drop across r due to current i flowing through it.
Applying the loop rule, we get:-
ε – ir = 1.45
⇒ 1.52 – ir = 1.45
⇒ ir = 0.07
⇒ 1.r = 0.07
r = 0.07 Ω
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