Question

Calculate the heat energy given out by the filament of an electric bulb in 20 s, when its resistance is 4 Ω and the p.d. across its ends is 12 V.

Answer 1

R = 4 Ω; V = 12 V; t = 20 s

\[\text{Heat energy} = \frac{\mathrm{V}^{2} \times t}{\mathrm{R}}\]

\[= \frac{12 \times 12 \times 20}{4}\]

\[= \frac{2880}{4}\]

= 720 J