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(C) The power dissipated by a lamp,which determines its illumination,is given by $P = I^2 R$,where $I$ is the current and $R$ is the resistance.Let th

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$36$

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(C) The power dissipated by a lamp,which determines its illumination,is given by $P = I^2 R$,where $I$ is the current and $R$ is the resistance.Let the initial current be $I_0$. Then the initial power is $P_0 = I_0^2 R$.If the current drops by $20 \%$,the new current becomes $I_f = I_0 - 0.20 I_0 = 0.80 I_0$.The new power is $P_f = (0.80 I_0)^2 R = 0.64 I_0^2 R = 0.64 P_0$.The percentage change in power is $\frac{P_f - P_0}{P_0} 	imes 100 = (0.64 - 1) 	imes 100 = -36 \%$.The negative sign indicates a decrease. Therefore,the illumination decreases by $36 \%$.

By what percentage will the illumination of the lamp decrease if the current drops by $20 \%$ (in $\%$)?

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