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

Step 1: A single bulb has resistance \(R=\dfrac{V^2}{P}\), fixed by its rating.

Step 2: Wiring \(n\) identical bulbs in series multiplies the resistance the supply sees by \(n\), so the new resistance is \(nR\), while the supply voltage \(V\) stays the same.

Step 3: Power delivered by a fixed voltage source scales as \(\dfrac{1}{\text{resistance}}\), so increasing the resistance by a factor of \(n\) scales the power down by the same factor of \(n\).\[ \boxed{P_{\text{total}} = \frac{P}{n}} \]

Answer 2

Power dissipated in a resistor scales with the square of the voltage across it, for a fixed resistance. In series, the supply voltage \(V\) splits equally across the \(n\) identical bulbs, so each bulb only gets \(\dfrac{V}{n}\) instead of its full rated \(V\).

Since power scales as voltage squared, each bulb's individual power drops to \(\left(\dfrac{1}{n}\right)^2\) of its rated value \(P\), which is \(\dfrac{P}{n^2}\) per bulb.

There are \(n\) such bulbs in the loop, so the total power drawn by the whole string is \(n\) times that per-bulb figure:\[ P_{\text{total}} = n\cdot\frac{P}{n^2} = \frac{P}{n} \]