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Q.
The maximum efficiency of full wave rectifier is
J & K CETJ & K CET 2004
Solution:
Efficiency of a rectifier is given by
$\eta=\frac{D C \text { power output }}{A C \text { power input }}$
For full wave rectifier $DC$ power output
$=I_{D C}^{2} \times R_{L}=\left(\frac{2 I_{0}}{\pi}\right) \times R_{L}$
AC input power $=I_{rms}^{2}\left(r_{f}+R_{L}\right)$
$=\left(\frac{I_{0}}{\sqrt{2}}\right)^{2}\left(r_{f}+R_{L}\right)$
$\therefore $ Rectifier efficiency
$\eta=\frac{\left(\frac{2 I_{0}}{\pi}\right)^{2} R_{L}}{\left(\frac{I_{0}}{\sqrt{2}}\right)^{2}\left(r_{f}+R_{L}\right)}$
$=\frac{0.812 R_{L}}{r_{f}+R_{L}} \eta$
will be maximum, if $r_{f}$ is negligible as compared to $R_{L} $
$\therefore $ Maximum rectified efficiency $=81.2\%$