An electric motor is driven by d.c. supply of emf $V_0$ and internal resistance $r$. The current when output power is maximum will be

$V_0/2r$

The correct answer is Option (4) → $V_0/2r$

Let load resistance = $R$

Current in the circuit: $I = \frac{V_0}{R + r}$

Power delivered to load: $P = I^2 R = \frac{V_0^2 R}{(R + r)^2}$

Maximum power occurs when $R = r$

Thus, current at this condition:

$I = \frac{V_0}{R + r} = \frac{V_0}{2r}$

Current when output power is maximum = $\frac{V_0}{2r}$