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Q. A cell supplies a current of 0.9 A through a $2\,\Omega$ resistor and a current of 0.3 A through a $7\,\Omega$ resistor. The internal resistance of the cell is

KCETKCET 2002Current Electricity

Solution:

E = $I_1 \, (R_1 + r)$ and E = $I_2 (R_2 + r)$ $i.e.$ E = 0.9 (2 + r) and E = 0.3 (7 + r) $i.e.$ 1.8 + 0.9 r = 2.1 + 0.3$r$ $i.e.$ $x = \frac{0.3}{0.6} = 0.5 \, \Omega$