First slide

Electrical power

Question

A filament bulb $500 W$ , $100 V$ 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 $500 W$ . The value of R is

Easy

A

$230 Ω$
B

$46 Ω$
C

$26 Ω$
D

$13 Ω$

Solution

$Resistance of bulb, R_{B} = \frac{V^{2}}{P} = \frac{(100)^{2}}{500} = 20 Ω$
Power of the bulb in the circuit,
$\begin{array}{l} P = V I \\ I = \frac{P}{V_{B}} = \frac{500}{100} = 5 A \end{array}$
$V_{R} = I R \Rightarrow (230 - 100) = 5 \times R$
$∴ R = 26 Ω$

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