Question

Efficiency of a half-wave rectifier is nearly:
$\begin{align}
  & \left( A \right)80\% \\
 & \left( B \right)60\% \\
 & \left( C \right)40\% \\
 & \left( D \right)20\% \\
\end{align}$

Answer 1

Hint:A half wave rectifier (also known as a filter) is a rectifier which only allows one half cycle of an AC voltage supply to pass while stopping the other half cycle. The rectifier efficiency of a half wave rectifier can be calculated by calculating the ratio of output DC power and the input AC power. The formula that we are going to use for this efficiency is given by: $\eta \%=\dfrac{{{P}_{DC}}}{{{P}_{AC}}}\times 100$.


Complete step-by-step solution:
Let us first assign some terms that we are going to use later in our solution.
Let the maximum current be denoted by ‘${{I}_{m}}$’.
Let us denote the root mean square current by ‘${{I}_{rms}}$ ’ and the output DC current by ‘${{I}_{DC}}$’.
And let the load resistance be ‘R’.
Then, we know that half of maximum current is equal to the root mean square current. This could be written as:
$\Rightarrow {{I}_{rms}}=\dfrac{{{I}_{m}}}{2}$
So, the input AC power can be calculated as follows:
$\begin{align}
  & \Rightarrow {{P}_{AC}}=\dfrac{{{I}_{rms}}^{2}}{R} \\
 & \Rightarrow {{P}_{AC}}=\dfrac{{{\left( \dfrac{{{I}_{m}}}{2} \right)}^{2}}}{R} \\
\end{align}$
Now, we know that the DC current value is equal to:
$\Rightarrow {{I}_{DC}}=\dfrac{{{I}_{m}}}{\pi }$
Therefore, the DC output power can be calculated as follows:
$\begin{align}
  & \Rightarrow {{P}_{DC}}=\dfrac{{{I}_{DC}}^{2}}{R} \\
 & \Rightarrow {{P}_{DC}}=\dfrac{{{\left( \dfrac{{{I}_{m}}}{\pi } \right)}^{2}}}{R} \\
\end{align}$
Now that we have calculated the input AC power and output DC power, the efficiency of the rectifier can be calculated as follows:
$\Rightarrow \eta \%=\dfrac{{{P}_{DC}}}{{{P}_{AC}}}\times 100$
Putting the values of both the power terms in the above equation, we get:
$\begin{align}
  & \Rightarrow \eta \%=\dfrac{\left( \dfrac{{{\left( \dfrac{{{I}_{m}}}{\pi } \right)}^{2}}}{R} \right)}{\left( \dfrac{{{\left( \dfrac{{{I}_{m}}}{2} \right)}^{2}}}{R} \right)}\times 100 \\
 & \Rightarrow \eta \%=\dfrac{4}{{{\pi }^{2}}}\times 100 \\
\end{align}$
Here, we can estimate the value of pi squared as 10. This will give us the efficiency as:
$\begin{align}
  & \Rightarrow \eta \%\approx \dfrac{4}{10}\times 100 \\
 & \Rightarrow \eta \%\approx 40 \\
 & \therefore \eta \approx 40\% \\
\end{align}$
Hence, the efficiency of a half-wave rectifier is nearly 40%.
Hence, option (C) is the correct option.

Note: The values of terms like rms current, DC current, efficiency, etc. should always be remembered thoroughly as it helps save time during a time bound exam. Some applications of half-wave rectifiers are, they are used in pulse generator circuits, soldering iron circuits, signal demodulation circuits, etc.