A battery of 24 V and internal resistance 2 is connected to a load (external resistance) R. The value of current so that power delivered to the load will be maximum, is

6 A

The correct answer is Option (3) → 6 A

Maximum power is delivered to the load when $R = r$ (internal resistance of the battery).

Given: $V = 24\ \text{V}$, $r = 2\ \Omega$, $R = r = 2\ \Omega$

Current through the load: $I = \frac{V}{R + r} = \frac{24}{2 + 2} = \frac{24}{4} = 6\ \text{A}$

Current for maximum power = 6 A