Question

A battery has a short-circuit current of 30 A and an open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2 ohm^-1, the power dissipated by the bulb is

(a) 80 W

(b) 1800 W

(c) 112.5 W

(d) 228 W

The question was asked in an international level competition.

This is a very interesting question from Network Theroms topic in division EDC Overview of Electronic Devices & Circuits

Answer 1

Right option is (c) 112.5 W

Easy explanation: r = Voc/Isc = 1.2 ohm

Power used by bulb = (24 x 24) x 2/(1.2 + 2) x (1.2 + 2) or 11.5 Watt.