In the circuit shown,the average power dissip | vedclass

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(B) The circuit represents a half-wave rectifier where the diode conducts only during the positive half-cycle of the input $AC$ voltage $E = E_0 \sin

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$\frac{E_0^2}{4 R}$

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(B) The circuit represents a half-wave rectifier where the diode conducts only during the positive half-cycle of the input $AC$ voltage $E = E_0 \sin \omega t$.For a half-wave rectified sine wave,the root mean square $(RMS)$ voltage is given by $E_{rms} = \frac{E_0}{2}$.The average power dissipated in the resistor $R$ is calculated using the formula $P_{av} = \frac{E_{rms}^2}{R}$.Substituting the value of $E_{rms}$,we get $P_{av} = \frac{(\frac{E_0}{2})^2}{R} = \frac{E_0^2}{4R}$.

In the circuit shown,the average power dissipated in the resistor is (assume diode to be ideal).

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