If the electric current in a lamp decreases b | vedclass

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(D) The power $P$ dissipated in a lamp is given by the formula $P = I^2 R$,where $I$ is the current and $R$ is the resistance.Taking the logarithmic d

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$10$

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(D) The power $P$ dissipated in a lamp is given by the formula $P = I^2 R$,where $I$ is the current and $R$ is the resistance.Taking the logarithmic differentiation of the equation,we get $\frac{dP}{P} = 2 \frac{dI}{I}$.Given that the current decreases by $5 \ \%$,we have $\frac{dI}{I} = -0.05$.Substituting this value into the equation,we get $\frac{dP}{P} = 2 	imes (-0.05) = -0.10$.This indicates that the power output decreases by $10 \ \%$.

If the electric current in a lamp decreases by $5 \ \%$,then the power output decreases by . . . . . . . (in $\%$)

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