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(D) The rated power is $P_1 = \frac{V_1^2}{R_1}$,where $V_1 = 220 \ V$ and $P_1 = 100 \ W$. Thus,$R_1 = \frac{220^2}{100}$.When connected to $V_2 = 22

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Between $100 	imes (0.8)^2 \ W$ and $100 	imes 0.8 \ W$

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(D) The rated power is $P_1 = \frac{V_1^2}{R_1}$,where $V_1 = 220 \ V$ and $P_1 = 100 \ W$. Thus,$R_1 = \frac{220^2}{100}$.When connected to $V_2 = 220 	imes 0.8 \ V$,the new power is $P_2 = \frac{V_2^2}{R_2}$.Since the voltage decreases,the temperature of the filament decreases,which causes the resistance to decrease $(R_2 If the resistance were constant $(R_2 = R_1)$,the power would be $P_{const} = \frac{(220 	imes 0.8)^2}{R_1} = (0.8)^2 	imes P_1 = 100 	imes (0.8)^2 \ W = 64 \ W$.Since $R_2 Also,if we consider the current $I = \frac{V}{R}$,since $R$ decreases as $V$ decreases,the current $I_2$ is less than $I_1$ but the ratio $\frac{I_2}{I_1} > 0.8$. Thus,$P_2 = V_2 I_2 = (0.8 V_1) I_2 Therefore,the actual power lies between $64 \ W$ and $80 \ W$,which is between $100 	imes (0.8)^2 \ W$ and $100 	imes 0.8 \ W$.

The resistance of a bulb filament changes with temperature. If a bulb rated at $220 \ V$ and $100 \ W$ is connected to a supply of $(220 \times 0.8) \ V$,the actual power consumed will be:

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