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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

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:$V_{L,av} = \frac{V_{L,peak}}{\pi} = \frac{V_m}{\pi} \cdot \frac{R_L}{R_S+R_L} = \frac{R_L}{R_S+R_L} \cdot \frac{V_m}{\pi}$.

List $I$	List $II$
$(a)$ Rectifier	$(i)$ Used either for stepping up or stepping down the $a.c.$ voltage
$(b)$ Stabilizer	$(ii)$ Used to convert $a.c.$ voltage into $d.c.$ voltage
$(c)$ Transformer	$(iii)$ Used to remove any ripple in the rectified output voltage
$(d)$ Filter	$(iv)$ Used for constant output voltage even when the input voltage or load current change

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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