Question

$\text{DC}$ power output of half-wave rectifier is equal to

$(I_0$ is the peak current and $R_L$ is the load resistance$)$

• A. $\frac{2I_0^2}{{\pi }^2}{R}_L$
• B. $\frac{I_0^2}{2{\pi }^2}{R}_L$
• C. $\frac{I_0^2}{{\pi }^2}{R}_L$
• D. $\frac{4I_0^2}{{\pi }^2}{R}_L$

Answer 1

The correct option is C $\frac{I_0^2}{{\pi }^2}{R}_L$

Given:
$I_0=$ Peak current
So, Peak voltage, $V_0=I_0{R}_L$

And we know that,

$I_{avg}=\frac{I_0}{\pi }\&V_{avg}=\frac{V_0}{\pi }$

Now, the average $\text{DC}$ power of a half-wave rectifier is given by,

$P=V_{DC}\times I_{DC}=\frac{V_0}{\pi }\times \frac{I_0}{\pi }=\frac{V_0{I}_0}{{\pi }^2}$

$\therefore P=\frac{I_0^2}{{\pi }^2}{R}_L$

Hence, option $\left(\text{C}\right)$ is correct.