An electric bulb is marked 100 W, 230 V. If the supply voltage drops to 115 V, what is the heat and light energy produced by the bulb in 20 min?

30,000 J

Here, P = 100 W; V = 230 V

Let R be resistance of the filament of the bulb.

Now, electric power, $\mathrm{P}=\mathrm{VI}=\frac{\mathrm{V}^2}{\mathrm{R}}$

Therefore, $\mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(230)^2}{100}=529 \Omega$

When the voltage drops to 115 V, heat and light energy produced by the bulb in 20 min is given by

$\mathrm{V}=\mathrm{VIt}=\frac{\mathrm{V}^2}{\mathrm{R}} \mathrm{t}=\frac{(115)^2}{529} \times 20 \times 60=30,000 \mathrm{~J}$