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A battery of e.m.f. 6.0 V supplies current through a circuit in which 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.

Current Electricity

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Answer

Given,

e.m.f. (ε) = 6.0 V

Current (I) = 3 A

Potential difference (V) = 5.4 V

r = ?

From relation,

V = ε – Ir

Substituting the values in the formula above we get,

5.4=6(3r)3r=65.43r=0.6r=0.63r=630r=210r=0.2Ω5.4 = 6 - (3 r) \\[0.5em] 3 r = 6 - 5.4 \\[0.5em] 3 r = 0.6 \\[0.5em] r = \dfrac{0.6}{3} \\[0.5em] r = \dfrac{6}{30} \\[0.5em] r = \dfrac{2}{10} \\[0.5em] \Rightarrow r = 0.2 Ω

Hence, internal resistance of the battery = 0.2 Ω

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