Question

A single phase two pulse diode rectifier has input supply of 230 V, 50Hz and the load resistance $R=30\Omega$. The value of inductance to be connected in series with R so as to limit the current ripple factor to 5% is (Assume second harmonic component as the most dominating component of $I_r$ (ripple current) and neglect higher order even harmonics)

• A. 0.11 H
• B. 0.33 H
• C. 0.22 H
• D. 0.44 H

Answer 1

The correct option is D 0.44 H

$V_0=\frac{2V_m}{\pi }-\frac{4V_m}{\pi }\sum _{n=2,4,6,...}^{\infty }\left(\frac{\cos n\omega t}{n^2-1}\right)$

$\therefore \left(V_{02}\right)=\frac{4V_m}{3\pi }\cos 2\omega t$

$(V_{02}{)}_{rms}=\frac{4V_m}{3\pi \sqrt{2}}\cos 2\omega t$

$\therefore (I_2{)}_{rms}=\frac{(V_{02}{)}_{rms}}{|Z_2|}$

$\text{Current ripple factor}=CRF=\frac{I_r}{I_0}$

$\therefore (I_2{)}_{rms}=\frac{(V_{02}{)}_{rms}}{|Z_2|}$

Current ripple factor $=CRF=\frac{I_r}{I_0}$

Only considering second harmonic current as ripple current

$I_r=I_2=\frac{4V_m}{3\pi \sqrt{2}\sqrt{R^2+(2\omega L{)}^2}}$

$I_r=\frac{97.615}{\sqrt{900+(2\omega L{)}^2}}$

$I_0=\frac{2V_m}{\pi R}=6.902A$

$CRF=\frac{97.615}{\sqrt{900+(2\omega L{)}^2}\times 6.902}$

$0.05=\frac{97.615}{\sqrt{900+(2\omega L{)}^2}\times 6.902}$ [given CRF = 0.05]

$\sqrt{900+(2\omega L{)}^2}=282.86$

$2\omega L=281.26$

$L=0.447H$