MCQ
$n$ identical bulbs, each designed to draw a power $p$ from a certain voltage supply, are joined in series across that supply. The total power which they will draw is
- A$p/{n^2}$
- ✓$p/n$
- C$p$
- D$np$
✓
Answer
Correct option: B.
$p/n$
b
(b) Let resistance each bulb $=R$ and voltage of source $=V$
(b) Let resistance each bulb $=R$ and voltage of source $=V$
Power of each bulb is $P=V^{2} / R$ or $R=V^{2} / P$
As $n$ bulbs are in series so total resistance $R_{t}=n R$ and current $i=V / R_{t}=V / n R$
Total power drawn is $P_{t}=i^{2} R_{t}=\frac{V^{2}}{n^{2} R^{2}}(n R)=\frac{V^{2}}{n R}=P / n$
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