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:
Show Hint
- \( np \)
- \( \frac{p}{n^2} \)
- \( \frac{p}{n} \)
- \( p \)
The Correct Option is C
Approach Solution - 1
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).
Approach Solution -2
Step 1: Understanding the problem.
We are given \( n \) identical electric bulbs, each drawing a power \( p \) independently from the mains supply. The bulbs are now connected in series across the main supply. We are tasked with finding the total power drawn by the combination.
Step 2: Power in a series combination.
In a series combination of bulbs, the same current flows through each bulb. The total power drawn from the supply can be determined using the formula for power in a series circuit. For a series connection, the total resistance increases, which reduces the total current drawn from the mains supply. Since power \( p \) drawn by each bulb is given, and the current through each bulb is the same, the total power \( P_{\text{total}} \) drawn by the entire combination is: \[ P_{\text{total}} = \frac{p}{n} \] This results from the fact that in series, the power is divided among the \( n \) bulbs, reducing the total power drawn by a factor of \( n \).
Step 3: Conclusion.
Thus, the total power drawn by the combination is \( \boxed{\frac{p}{n}} \).
Final Answer:
\[ \boxed{C}. \]
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