The resistance of the filament of an electric | vedclass

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(D) The rated power is $P_1 = 100\,W$ at $V_1 = 220\,V$. The resistance at this temperature is $R_1 = \frac{V_1^2}{P_1}$.When connected to $V_2 = 220

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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 $P_1 = 100\,W$ at $V_1 = 220\,V$. The resistance at this temperature is $R_1 = \frac{V_1^2}{P_1}$.When connected to $V_2 = 220 	imes 0.8\,V$,the new power is $P_2 = \frac{V_2^2}{R_2}$.Taking the ratio: $\frac{P_2}{P_1} = \frac{V_2^2}{V_1^2} 	imes \frac{R_1}{R_2} = (0.8)^2 	imes \frac{R_1}{R_2}$.Since the voltage decreases,the temperature of the filament decreases,which causes the resistance to decrease $(R_2  1$.This implies $P_2 > (0.8)^2 	imes P_1 = 100 	imes (0.8)^2\,W$.Also,since $P = VI$,and both $V$ and $I$ decrease compared to the rated values,the power $P_2$ must be less than the power calculated if resistance remained constant at $R_1$ (which would be $100 	imes 0.8\,W$).Thus,the actual power $P_2$ lies between $100 	imes (0.8)^2\,W$ and $100 	imes 0.8\,W$.

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

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