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

To determine the total power drawn by \( n \) identical light bulbs connected in series, let's consider the following:

• Each bulb has a power rating of \( P \), which means when it is connected individually to the supply voltage, it would draw power \( P \).
• When bulbs are connected in series, the total voltage is divided among them.
• If \( V \) is the total voltage of the source, each bulb will receive a voltage of \( \frac{V}{n} \) because the total voltage is shared equally among the bulbs in series. 
• The power drawn by each bulb when connected in series is given by the formula for power: \( P = \frac{V^2}{R} \), where \( R \) is the resistance of one bulb.
• In series, the voltage across each bulb is \( \frac{V}{n} \), so the power across each bulb is:

\[\text{Power per bulb} = \left(\frac{V}{n}\right)^2 \times R\]

• The total power drawn by all \( n \) bulbs is:

\[\text{Total power} = n \times \left(\frac{V}{n}\right)^2 \times R = \frac{nV^2}{n^2R} = \frac{V^2}{nR}\]

• Since each bulb alone draws power \( P = \frac{V^2}{R} \), the expression for total power becomes:

\[\text{Total power} = \frac{P}{n}\]

Therefore, the total power drawn by the bulbs when connected in series is \( \frac{P}{n} \).

Answer 2

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: \[ P_{\text{total}} = \frac{P}{n} \]