A) \[100%\]
B) \[25.20%\]
C) \[40.6%\]
D) \[81.2%\]
Correct Answer: D
Solution :
Efficiency of a rectifier is given by \[\eta =\frac{DC\text{ }power\text{ }output}{AC\text{ }power\text{ }input}\] For full wave rectifier DC power output \[=I_{DC}^{2}\times {{R}_{L}}\] \[=\left( \frac{2{{I}_{0}}}{\pi } \right)\times {{R}_{L}}\] AC input power \[=I_{rms}^{2}({{r}_{f}}+{{R}_{L}})\] \[={{\left( \frac{{{I}_{0}}}{\sqrt{2}} \right)}^{2}}({{r}_{f}}+{{R}_{L}})\] \[\therefore \] Rectifier efficiency \[\eta =\frac{{{\left( \frac{2{{I}_{0}}}{\pi } \right)}^{2}}{{R}_{L}}}{{{\left( \frac{{{I}_{0}}}{\sqrt{2}} \right)}^{2}}({{r}_{f}}+{{R}_{L}})}=\frac{0.812{{R}_{L}}}{{{r}_{f}}+{{R}_{L}}}\] \[\eta \] will be maximum, if \[{{r}_{f}}\] is negligible as compared to \[{{R}_{L}}\]. \[\therefore \] Maximum rectified efficiency \[=81.2%\]
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