Question

An electric bulb whose filament is of resistance 4 Ω operates at 12 V for 20 s. The energy released by the bulb is:

पर्याय

• 700 J

• 720 J

• 680 J

• 740 J

Answer 1

720 J

Explanation:

$$W = \frac{V^2}{R}t$$

$$= \frac{12^2}{4} \times 20$$

$$= 720\ \mathrm{J}$$