Question

A half wave rectifier has an inductor and dc source connected as load as shown in figure below. The value of power absorbed by dc source at the load side will be ___ W. (Assume initial current in inductor as zero and ideal diode used)

Answer 1

At $\omega t=0^o$, the diode remains reverse biased until $V_{ac}=V_{dc}$

Let us assume at angle $\alpha$, diode starts conducting

$\therefore \alpha =si{n}^{-1}\left(\frac{E}{V_m}\right)=si{n}^{-1}\left(\frac{72}{120\sqrt{2}}\right)={25.1}^o$

$\alpha$ (in radian) = 0.438 radian

By KVL we can write,

$V_m\sin \omega t=L\frac{di\left(t\right)}{dt}+V_{dc}$

$V_m\sin \omega t=\omega L\frac{di\left(\omega t\right)}{dt}+V_{dc}$

Rearranging above equation,

$\frac{di\left(\omega t\right)}{dt}=\frac{V_{m}sin\omega t-V_{dc}}{\omega L}$

$i\left(\omega t\right)=\frac{1}{\omega L}{\int }_{\alpha }^{\omega t}{V}_{m}sin\lambda d\lambda -\frac{1}{\omega L}{\int }_{\alpha }^{\omega t}{V}_{dc}d\lambda$

$i\left(\omega t\right)=\frac{V_{in}}{\omega L}[-cos\lambda {]}_{\alpha }^{\omega t}-\frac{1}{\omega L}{V}_{dc}[\lambda {]}_{\alpha }^{\omega t}$

$=\frac{V_m}{\omega L}[cos\alpha -cos\omega t]+\frac{V_{dc}}{\omega L}(\alpha -\omega t)$

The time instant at which $i\left(\omega t\right)$ = 0

$0=\frac{120\sqrt{2}}{2\pi \times 60\times 50\times {10}^{-3}}[0.9055-cos\beta ]+\frac{72(0.438-\beta )}{2\pi \times 60\times 50\times {10}^{-3}}$

$0=8.152-9.003cos\beta +1.6730-3.819\beta$

$0=9.825-9.003cos\beta -3.819\beta$

$\beta =4.04rad$

Power absorbed by dc source

$I_0=\frac{1}{2\pi }{\int }_{0.488}^{4.04}[9.525-9.003cos(\omega t)-3.819\omega t]d\omega t$

= 2.46 A

$P_{dc}=V_{dc}{I}_0=2.46\times 72=177.12W$