A 100 watt bulb working on 200 volt and a200 watt bulb working on 100 volt have

A
resistances in the ratio of 4 : 1
B
maximum current ratings in the ratio of 1 : 4
C
resistances in the ratio of 2 : 1
D
maximum current ratings in the ratio of 1 : 2

Solution:

$\begin{array}{l} R = \frac{V^{2}}{P} \Rightarrow R_{1} = \frac{200 \times 200}{100} = 400 Ω a n d \\ R_{2} = \frac{100 \times 100}{200} = 50 Ω \end{array}$

Maximum current rating, $i = \frac{P}{V}$

$So i_{1} = \frac{100}{200} and i_{2} = \frac{200}{100} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{1}{4}$

Solution:

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