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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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Verified Answer
The correct answer is:
$\frac{R_L}{\left(R_L+R_S\right)} \frac{V_m}{\pi}$
Given, AC voItage, $V_s=V_m \sin \omega t$ and $V_m=$ maximum value of voltage In the given circuit diode will only conduct current in forward bias, so output across $R_L$ will be
Average voltage, $V_{\mathrm{av}}=\frac{V_m}{\pi}$ average current $I_{\mathrm{av}}=\frac{V_{\mathrm{av}}}{R}$ Total resistance, $R=R_S+R_L$
$$
I_{\mathrm{av}}=\frac{V_m}{\left(R_S+R_L\right) \pi}
$$
Now, voltage across $R_L$,
$$
\begin{aligned}
& V_L=I_{\mathrm{av}} \cdot R_L \\
& V_L=\frac{R_L}{R_S+R_L} \cdot \frac{V_m}{\pi}
\end{aligned}
$$
Average voltage, $V_{\mathrm{av}}=\frac{V_m}{\pi}$ average current $I_{\mathrm{av}}=\frac{V_{\mathrm{av}}}{R}$ Total resistance, $R=R_S+R_L$
$$
I_{\mathrm{av}}=\frac{V_m}{\left(R_S+R_L\right) \pi}
$$
Now, voltage across $R_L$,
$$
\begin{aligned}
& V_L=I_{\mathrm{av}} \cdot R_L \\
& V_L=\frac{R_L}{R_S+R_L} \cdot \frac{V_m}{\pi}
\end{aligned}
$$
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