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Question
A battery of e.m.f. 6.0 V supplies current through a circuit in which the resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 3 A, the voltmeter reads 5.4 V. Find the internal resistance of the battery.
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Solution
Since a battery's terminal potential difference is less than its e.m.f., an increase in circuit current causes the voltmeter reading to fall.
Now E = 6.0 V,
V = 5.4 V,
I = 3.0 A
Internal resistance r = `("E" - "V")/"I"`
= `(6 - 5.4)/3.0`
= `0.3/1.5`
r = 0.2 Ω
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