Question

The maximum efficiency of full-wave rectifier is

• A. $100\%$
• B. $25.20\%$
• C. $42.2\%$
• D. $81.2\%$

Answer 1

The correct option is D

$81.2\%$

Step 1: Given

A full-wave rectifier is given. Its efficiency is $\eta$.

Step 2: Formula Used

A full wave rectifier converts AC (alternating current) voltage into DC voltage (direct current). The efficiency of the full wave rectifier can be calculated by finding the percentage of the ratio of DC power $P_{DC}$ to AC power $P_{AC}$. Thus, the efficiency is given by $\eta =\frac{P_{DC}}{P_{AC}}\times 100\%$.

The power of the DC circuit is $P_{DC}=\frac{{V_{DC}}^2}{R_L}$, where $V_{DC}=\frac{2V_M}{\pi }$ is voltage in the DC circuit, $V_M$ is peak voltage and $R_L$ is load resistance.

The power of the AC circuit is $P_{AC}=\frac{{V_{RMS}}^2}{R_L}$, where $V_{RMS}=\frac{V_M}{\sqrt{2}}$ is root mean square voltage in the AC circuit, $V_M$ is peak voltage and $R_L$ is load resistance.

Step 3: Calculate the Efficiency

Substitute all the expressions in the efficiency formula.

$\begin{array}{rcl}\eta & =& \frac{P_{DC}}{P_{AC}}\times 100\%\\ & =& \frac{\frac{{V_{DC}}^2}{R_L}}{\frac{{V_{RMS}}^2}{R_L}}\times 100\%\\ & =& \frac{{V_{DC}}^2}{{V_{RMS}}^2}\times 100\%\end{array}$

Further substitute $V_{DC}=\frac{2V_M}{\pi }$ and $V_{RMS}=\frac{V_M}{\sqrt{2}}$.

$\begin{array}{rcl}\eta & =& \frac{{V_{DC}}^2}{{V_{RMS}}^2}\times 100\%\\ & =& \frac{{\left({\displaystyle \frac{2V_M}{\pi }}\right)}^2}{{\left({\displaystyle \frac{V_M}{\sqrt{2}}}\right)}^2}\times 100\%\\ & =& {\left(\frac{2\sqrt{2}}{\pi }\right)}^2\times 100\%\\ & =& \frac{8}{{\pi }^2}\times 100\%\\ & =& 81.2\%\end{array}$

Hence, the maximum efficiency of a full-wave rectifier is $81.2\%$.