Question

n identical light bulbs, each designed to draw power \( P \) from a certain voltage supply are joined in series across that supply. The total power which they will draw is:

• A. \( nP \)
• B. \( P \)
• C. \( \frac{P}{n} \)
• D. \( Pn^{-2} \)

Hint:

In a series circuit, the total power is the sum of the powers dissipated across all resistors (light bulbs in this case).

Answer 1

The Correct Option is C

In a series circuit, the current through each light bulb is the same. The total resistance \( R_{\text{total}} \) is the sum of the individual resistances \( R_{\text{individual}} \) of each bulb. The power drawn by each bulb is \( P = \frac{V^2}{R_{\text{individual}}} \), and the total power is: \[ P_{\text{total}} = \frac{V^2}{R_{\text{total}}} \] Since the total resistance in series is \( R_{\text{total}} = nR_{\text{individual}} \), the total power becomes: \[ P_{\text{total}} = \frac{P}{n} \]