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Physics

A battery of e.m.f. 3.0 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 1.5 A, the voltmeter reads 2.7 V. Find the internal resistance of the battery.

Current Electricity

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Answer

Given,

e.m.f. (ε) = 3.0 V

Current (I) = 1.5 A

Potential difference (V) = 2.7 V

r = ?

From relation,

V = ε – Ir

Substituting the values in the formula above we get,

2.7=3−(1.5r)1.5r=3−2.71.5r=0.3r=0.31.5r=315⇒r=0.2Ω2.7 = 3 - (1.5 r) \\[0.5em] 1.5 r = 3 - 2.7 \\[0.5em] 1.5 r = 0.3 \\[0.5em] r = \dfrac{0.3}{1.5} \\[0.5em] r = \dfrac{3}{15} \\[0.5em] \Rightarrow r = 0.2 Ω

Hence, internal resistance of the battery = 0.2 Ω

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