A cell supplies a current of 0.6 A through a 3 Ω resistor and a current of 0.2 A through a 10 Ω resistor. The internal resistance of the cell is

0.5 Ω

The correct answer is Option (1) → 0.5 Ω

Given:

When external resistance $R_1 = 3\ \Omega$, current $I_1 = 0.6\ A$

When external resistance $R_2 = 10\ \Omega$, current $I_2 = 0.2\ A$

Let emf of the cell be $E$ and internal resistance be $r$.

From Ohm’s law:

$E = I_1 (R_1 + r) = I_2 (R_2 + r)$

$0.6(3 + r) = 0.2(10 + r)$

$1.8 + 0.6r = 2 + 0.2r$

$0.4r = 0.2$

$r = \frac{0.2}{0.4} = 0.5\ \Omega$

Hence, the internal resistance of the cell is $0.5\ \Omega$.