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Q. A battery supplies 150 W and 196 W power to two resistors of $6\,\Omega$ and $4\,\Omega$ when they are connected separately to it. The internal resistance of the battery is

KCETKCET 2001Current Electricity

Solution:

P $\propto \, \frac{R}{(R + r)^2}$
$\therefore $ $\frac{P_1}{P_2} = \frac{R_1 }{R_2} \frac{(R_2 + r)^2}{(R_1 + r)^2}$
$i.e.$ $\frac{150}{196} = \frac{6}{4} \frac{(4 + r)^2}{4 ( 6 + r)^2} \, i.e. \, r = 1 \Omega$