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(C) The rated power $P_R$ of the lamp is $100\, W$ at a rated voltage $V_R = 220\, V$. The resistance of the filament at rated conditions is $R = \fra

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$> 25\, W$

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(C) The rated power $P_R$ of the lamp is $100\, W$ at a rated voltage $V_R = 220\, V$. The resistance of the filament at rated conditions is $R = \frac{V_R^2}{P_R} = \frac{220^2}{100} = 484\, \Omega$.If the resistance were constant,the power consumed at $V = 110\, V$ would be $P = \frac{V^2}{R} = \frac{110^2}{484} = 25\, W$.However,the resistance of a filament increases with temperature. When the lamp is connected to $110\, V$,the power dissipated is lower than the rated $100\, W$,so the filament temperature will be lower than its operating temperature at $220\, V$.Since the temperature is lower,the resistance of the filament will be lower than the rated resistance $R = 484\, \Omega$.Using the formula $P = \frac{V^2}{R_{new}}$,since $R_{new} < R$,the power $P$ will be greater than $25\, W$.

If the resistance of the filament increases with temperature,what will be the power dissipated in a $220\, V- 100\, W$ lamp when connected to a $110\, V$ power supply?

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