If the current in an electric bulb is changed | vedclass

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(B) The power $P$ dissipated in a resistor $R$ is given by $P = I^2 R$.Taking the logarithmic derivative,we get $\frac{\Delta P}{P} = 2 \frac{\Delta I

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$2$

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(B) The power $P$ dissipated in a resistor $R$ is given by $P = I^2 R$.Taking the logarithmic derivative,we get $\frac{\Delta P}{P} = 2 \frac{\Delta I}{I}$.Given that the percentage change in current $\frac{\Delta I}{I} 	imes 100 = 1\%$.Therefore,the percentage change in power is $\frac{\Delta P}{P} 	imes 100 = 2 	imes (\frac{\Delta I}{I} 	imes 100)$.Substituting the value,we get $2 	imes 1\% = 2\%$.Thus,the percentage change in power is $2\%$.

If the current in an electric bulb is changed by $1\%$,then the percentage change in power will be ............. $\%$.

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