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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:
A
26 $\Omega$
B
13 $\Omega$
C
230 $\Omega$
D
46 $\Omega$
Explanation
Same current, so voltage divides in the ratio of resistances.
Detailed Solution
$R_{bulb} = \frac{V^2}{P} = \frac{100^2}{500} = 20\ \Omega$
R drops the remaining 130 V, so $\frac{R}{R_{bulb}} = \frac{130}{100}$
$\frac{R}{20} = \frac{130}{100} \Rightarrow R = 26\ \Omega$