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 solve this, we need to consider how electric bulbs behave when connected in series across a mains supply.
1. Resistor Behavior in Series: When \( n \) identical electric bulbs are connected in series, the total resistance \( R_{\text{total}} \) is the sum of the individual resistances: \[ R_{\text{total}} = R_1 + R_2 + \dots + R_n \] Each bulb has the same resistance, so we have: \[ R_{\text{total}} = nR \] where \( R \) is the resistance of each bulb.
2. Total Power Consumption in Series: The total power drawn by the combination of bulbs can be determined using the power formula: \[ P_{\text{total}} = \frac{V^2}{R_{\text{total}}} \] where \( V \) is the voltage across the combination. Since the bulbs are identical, the total power drawn by the series combination of bulbs is: \[ P_{\text{total}} = \frac{V^2}{nR} \] Since the power drawn by each individual bulb is \( p = \frac{V^2}{R} \), we substitute this into the equation: \[ P_{\text{total}} = \frac{p}{n} \]
Thus, the total power drawn by the combination is \( \frac{p}{n} \), and the correct answer is (3).