Question

A full wave rectifier has a 120 V, 60 Hz source and load including, R = 500 $\Omega$ and C = 100 $\mu F$ as shown below. The value of steady state dc current drawn from the rectifier circuit will be (Assume charging time of the capacitor is negligible).

Answer 1

For full wave rectifier,

The load voltage, $V_0=V_m-\frac{V_r}{2}$

Where Vr is ripple voltage (peak-to-peak)

Average current through capacitor is zero

$V_0=I_{DC}{R}_L$ ...$\left(i\right)$

$V_r=I_{DC}\times \frac{T}{2C}$
(where, T = total time period)

$V_0=V_m-I_{DC}\frac{T}{2C}$ ...$\left(ii\right)$

Using equations (i) and (ii)

$I_{DC}{R}_L=V_m-I_{DC}\frac{1}{4fC}$

$I_{DC}=\frac{V_m}{R_L+\frac{1}{4fC}}$

$=\frac{120\sqrt{2}}{500+\frac{1}{4\times 60\times 100\times {10}^{-6}}}=0.3133A$