Question

The average value of a half wave rectified sinusoidal voltage with peak amplitude 10 Volts is

• A. $\frac{20}{\pi}$ Volts
• B. $\frac{10}{\pi}$ Volts
• C. $0$ Volts
• D. $\frac{12}{\pi}$ Volts

Hint:

Always remember:
Half-Wave Rectifier: Average = $\frac{V_m}{\pi}$, RMS = $\frac{V_m}{2}$.
Full-Wave Rectifier: Average = $\frac{2V_m}{\pi}$, RMS = $\frac{V_m}{\sqrt{2}}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question asks for the average (DC) value of a half-wave rectified sine wave when the peak amplitude of the original sinusoidal voltage is $V_m = 10\text{ V}$.

Step 2: Key Formula or Approach:
A half-wave rectified signal has a non-zero value during one half-cycle and is zero during the other.
The average value $V_{avg}$ over a full period $T = 2\pi$ is given by:
\[ V_{avg} = \frac{1}{2\pi} \int_{0}^{\pi} V_m \sin(\theta) d\theta \]
Integrating this yields the standard relation:
\[ V_{avg} = \frac{V_m}{\pi} \]

Step 3: Detailed Explanation:

• Given peak amplitude: $V_m = 10\text{ Volts}$.

• Substituting this into the average value formula:
\[ V_{avg} = \frac{10}{\pi}\text{ Volts} \]

• (Note: For a full-wave rectifier, the average value would be double, i.e., $\frac{2V_m}{\pi} = \frac{20}{\pi}$ Volts).

Step 4: Final Answer:
The average value is $\frac{10}{\pi}$ Volts, which corresponds to Option (B).