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Q.

A battery with e.m.f E and internal resistance r is connected across an external resistance R. The maximum power in the external circuit is 9 W. The current flowing through the circuit in these conditions is 3A. Then which of the following is correct

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a

$r = \frac{R}{2}$

b

$r = 1 .5 Ω$

c

E = 6V

d

$r = 3 Ω$

answer is A.

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Detailed Solution

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Current through the circuit $I = (\frac{E}{r + R})$
Power dissipated in External resistance ${P=I}^{2} R = \frac{E^{2} R}{(R + r)^{2}}$
Power dissipated is maximum when $(\frac{δP}{δR}) = 0, i . e, r = R$
$\begin{array}{l} P_{max} = \frac{E^{2} r}{(2 r)^{2}} = \frac{E^{2}}{4 r} = 9 w \\ i = \frac{E}{2 r} = 3 A \Rightarrow E = 6 V & r = 2 Ω \end{array}$

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