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A filament bulb $600 W$ , $100 V$ is to be used in a $230 V$ main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes $600 W$ . The value of R is

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$21.66 Ω$

$230 Ω$

$13 Ω$

$46 Ω$

answer is C.

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Detailed Solution

$Resistance of bulb, R_{B} = \frac{V^{2}}{P} = \frac{(100)^{2}}{600} = \frac{50}{3} Ω$

Power of the bulb in the circuit,

$\begin{array}{l} P = V I \\ I = \frac{P}{V_{B}} = \frac{600}{100} = 6 A \end{array}$

$V_{R} = I R \Rightarrow (230 - 100) = 6 \times R$

$∴ R = 21.66 Ω$

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