Question

A sinusoidal voltage of amplitude \( 25 \) V and frequency \( 50 \) Hz is applied to a half-wave rectifier using a P-N junction diode. No filter is used, and the load resistor is \( 1000\Omega \). The forward resistance \( R_f \) of the ideal diode is \( 10\Omega \). The percentage rectifier efficiency is:

• A. \( 40\% \)
• B. \( 20\% \)
• C. \( 30\% \)
• D. \( 15\% \)

Hint:

For rectifiers:
- A full-wave rectifier is more efficient than a half-wave rectifier.
- The efficiency of a half-wave rectifier is around \( 40.6\% \).
- The presence of filters improves DC output quality.

Answer 1

The Correct Option is A

To determine the percentage rectifier efficiency, we use the formula for the efficiency of a half-wave rectifier: \[ \eta = \frac{DC { Power Delivered to Load}}{{AC Power Supplied to Rectifier}} \times 100 \] Step 1: Calculate the RMS Value of the AC Input Voltage The given peak voltage is: \[ V_m = 25V \] The RMS value of the input voltage is: \[ V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{25}{\sqrt{2}} = 17.68V \] Step 2: Calculate DC Output Voltage For a half-wave rectifier, the DC output voltage is given by: \[ V_{DC} = \frac{V_m - I_{DC} R_f}{\pi} \] Since \( I_{DC} \) is unknown, we approximate \( V_{DC} \) as: \[ V_{DC} \approx \frac{V_m}{\pi} = \frac{25}{\pi} = 7.96V \] Step 3: Calculate the DC Power Delivered to Load The DC output current is: \[ I_{DC} = \frac{V_{DC}}{R_L} = \frac{7.96}{1000} = 7.96 { mA} \] Thus, the DC power delivered to the load is: \[ P_{DC} = V_{DC} I_{DC} = 7.96V \times 7.96 \times 10^{-3} A = 63.37 { mW} \] Step 4: Calculate the AC Power Supplied to Rectifier The AC power supplied to the rectifier is given by: \[ P_{AC} = \frac{V_{rms}^2}{R_{eq}} \] where \( R_{eq} \) is the equivalent resistance: \[ R_{eq} = R_L + R_f = 1000 + 10 = 1010\Omega \] \[ P_{AC} = \frac{(17.68)^2}{1010} = \frac{312.64}{1010} = 0.3098 W = 309.8 { mW} \] Step 5: Calculate Efficiency \[ \eta = \frac{P_{DC}}{P_{AC}} \times 100 = \frac{63.37}{309.8} \times 100 = 40\% \] Thus, the percentage rectifier efficiency is: \[ {40\%} \]