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(B) The power dissipated in a resistor by an $AC$ current is given by $P = I_{rms}^2 R$.For a sinusoidal current,the root-mean-square $(RMS)$ current

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$\frac{1}{2}I_p^2R$

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(B) The power dissipated in a resistor by an $AC$ current is given by $P = I_{rms}^2 R$.For a sinusoidal current,the root-mean-square $(RMS)$ current $I_{rms}$ is related to the peak current $I_p$ by the formula $I_{rms} = \frac{I_p}{\sqrt{2}}$.Substituting this into the power formula,we get $P = \left( \frac{I_p}{\sqrt{2}} \right)^2 R$.Simplifying this expression,$P = \frac{I_p^2}{2} R = \frac{1}{2} I_p^2 R$.Therefore,the correct option is $B$.

$A$ sinusoidal $AC$ current flows through a resistor of resistance $R$. If the peak current is $I_p$,then the power dissipated is

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