The Maximum Efficiency of a Full-Wave Rectifier is

100%

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25.20%

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42.2%

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81.2%

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Solution

The correct option is D
81.2%

Step 1: Given
The efficiency of the rectifier is $η$ .
The maximum current is $I_{m}$ .
The DC current is $I_{d c}$ .
Step 2: Formula used
$η = \frac{P_{d c}}{P_{a c}}$ , where $P_{d c}$ is the power of DC power and $P_{a c}$ is the AC power.
$P_{a c} = {I^{2}}_{r m s} (r_{f} + R_{l})$ , where , $I_{r m s}$ is the root mean square current, $r_{f}$ is the forward resistance, and $R_{l}$ is the load resistance.
$P_{d c} = {I^{2}}_{d c} R_{l}$
$I_{d c} = \frac{2 I_{m}}{π}$
Step 3: Find $P_{d c}$ and $P_{a c}$
Find an expression for $P_{d c}$ by substituting $I_{d c} = \frac{2 I_{m}}{π}$ in $P_{d c} = {I^{2}}_{d c} R_{l}$ .
$\begin{array}{rcl} P_{d c} & = & {I^{2}}_{d c} R_{l} \\ = & {(\frac{2 I_{m}}{π})}^{2} R_{l} \end{array}$
Find an expression for $P_{a c}$ by substituting $I_{r m s} = \frac{I_{m}}{\sqrt{2}}$ in $P_{a c} = {I^{2}}_{r m s} (r_{f} + R_{l})$ .
$\begin{array}{rcl} P_{a c} & = & {I^{2}}_{r m s} (r_{f} + R_{l}) \\ = & {(\frac{I_{m}}{\sqrt{2}})}^{2} (r_{f} + R_{l}) \end{array}$
Step 3: Find Efficiency
Find the efficiency by substituting the obtained expressions in the formula. For maximum efficiency the $r_{f} < < < < R_{l}$ , so $r_{f} + R_{l} = R_{l}$ .
$\begin{array}{rcl} η & = & \frac{P_{d c}}{P_{a c}} \\ = & \frac{{(\frac{2 I_{m}}{π})}^{2} R_{l}}{{(\frac{I_{m}}{\sqrt{2}})}^{2} (r_{f} + R_{l})} \\ = & \frac{{(\frac{2 I_{m}}{π})}^{2} R_{l}}{{(\frac{I_{m}}{\sqrt{2}})}^{2} R_{l}} \\ = & \frac{8}{π^{2}} \end{array}$
Find the percentage by multiplying by 100.
$\frac{8}{π^{2}} \times 100 % = 81.2 %$
Hence, the answer is (D).

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