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(D) The rated power $P_0 = 100\, W$ at voltage $V_0 = 220\, V$. The resistance $R$ of the filament at this temperature is $R = \frac{V_0^2}{P_0} = \fr

Vedclass · Accepted Answer

Between $81\, W$ and $90\, W$

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(D) The rated power $P_0 = 100\, W$ at voltage $V_0 = 220\, V$. The resistance $R$ of the filament at this temperature is $R = \frac{V_0^2}{P_0} = \frac{220^2}{100} = 484\, \Omega$.When the voltage drops by $10\%$,the new voltage is $V' = 220 	imes 0.9 = 198\, V$.If the resistance $R$ remained constant,the power would be $P' = \frac{V'^2}{R} = \frac{(0.9 V_0)^2}{R} = 0.81 	imes P_0 = 81\, W$.However,since the voltage decreases,the power dissipated decreases,leading to a lower filament temperature. Since the resistance of a metallic filament decreases with a decrease in temperature,the new resistance $R' Since $P = \frac{V'^2}{R'}$,and $R' Since the drop in temperature is relatively small,the power will remain less than the original $100\, W$ and specifically less than the linear approximation of $90\, W$ (which would occur if $P \propto V$). Thus,the power lies between $81\, W$ and $90\, W$.

The resistance of the filament of a lamp increases with the increase in temperature. $A$ lamp rated $100\, W, 220\, V$ is connected across a $220\, V$ power supply. If the voltage drops by $10\%$,then the power of the lamp will be:

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