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(A) Given,$AC$ voltage,$V_s=V_m \sin \omega t$,where $V_m$ is the maximum value of the voltage.In the given half-wave rectifier circuit,the diode cond

Vedclass · Accepted Answer

$\frac{R_L}{\left(R_L+R_S\right)} \frac{V_m}{\pi}$

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(A) Given,$AC$ voltage,$V_s=V_m \sin \omega t$,where $V_m$ is the maximum value of the voltage.In the given half-wave rectifier circuit,the diode conducts only during the positive half-cycle of the input $AC$ signal.For an ideal diode,the voltage across the load $R_L$ during the conduction period is given by the voltage divider rule: $V_{L,peak} = V_m \cdot \frac{R_L}{R_S+R_L}$.The average value of a half-wave rectified sine wave is given by $V_{av} = \frac{V_{peak}}{\pi}$.Substituting the peak voltage across the load,we get:$V_{L,av} = \frac{V_{L,peak}}{\pi} = \frac{V_m}{\pi} \cdot \frac{R_L}{R_S+R_L} = \frac{R_L}{R_S+R_L} \cdot \frac{V_m}{\pi}$.

In the diode-based rectifier circuit given below,if $V_s=V_m \sin \omega t$ and the diode is ideal,then the average value of $V_L$ is

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