Question

There are ‘n’ number of identical electric bulbs, each is designed to draw a power $ p $ independently from the mains supply. They are now joined in series across the main supply. The total power drawn by the combination is:

• A. \( np \)
• B. \( \frac{p}{n^2} \)
• C. \( \frac{p}{n} \)
• D. \( p \)

Hint:

In a series circuit, the total power drawn is inversely proportional to the number of identical bulbs connected in series.

Answer 1

The Correct Option is C

To address this, we must examine the behavior of electric bulbs when connected in series to a mains supply.

1. Resistor Behavior in Series:

When \( n \) identical electric bulbs are connected in series, the aggregate resistance, denoted as \( R_{\text{total}} \), is the sum of their individual resistances:
\[
R_{\text{total}} = R_1 + R_2 + \dots + R_n
\]
As each bulb possesses identical resistance, this simplifies to:
\[
R_{\text{total}} = nR
\]
where \( R \) represents the resistance of a single bulb.


2. Total Power Consumption in Series:

The total power consumed by the bulb arrangement can be calculated using the power formula:
\[
P_{\text{total}} = \frac{V^2}{R_{\text{total}}}
\]
where \( V \) is the voltage applied across the combination. Given the bulbs are identical, the total power drawn by the series configuration is:
\[
P_{\text{total}} = \frac{V^2}{nR}
\]
Since the power consumed by each individual bulb is \( p = \frac{V^2}{R} \), we can substitute this into the equation:
\[
P_{\text{total}} = \frac{p}{n}
\]

Therefore, the total power consumed by the combination is \( \frac{p}{n} \), making option (3) the correct answer.