JEE Mains · Physics · STD 12 - 3. current electricity
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 \(\frac{\mathrm{p}}{\mathrm{n}^2}\)
- C \(\frac{\mathrm{p}}{\mathrm{n}}\)
- D p
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{p}}{\mathrm{n}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & R_s=R_1+R_2+R_3+\ldots \ldots..+R_n \\ & \frac{V^2}{P_s}=\frac{V^2}{P}+\frac{V^2}{P}+\ldots \ldots.+\frac{V^2}{P_n} \\ & P_s=\frac{P}{n}\end{aligned}\)
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