Power of Electrical Circuit: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Power of Electrical Circuit. Understanding the underlying formula is key to arriving at the correct answer below:

Resistance of one bulb is $R = \frac{V^2}{P} = \frac{200^2}{60}$. In series, total resistance $R_{\text{eq}} = 3R$. Power drawn is $P' = \frac{V^2}{3R} = \frac{1}{3} \frac{V^2}{R} = \frac{60}{3} = 20\text{ watt}$.