Question

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

• A. 0.45

Answer 1

The correct option is A 0.45

$V_0=\frac{2V_m}{\pi }$
$V_{0n}=\frac{4V_m}{\pi }\sum _{\pi =2,4,6,...}^{\infty }\left(\frac{\cos n\omega t}{n^2-1}\right)$

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

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

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

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


Only considering second harmonic current as ripple current

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

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

$I_o=\frac{2V_m}{\pi R}=6.902$

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

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

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

$2\omega L=281.26$

$L=0.447\approx 0.45H$