The Question is from neet-physics alternating-current , and it was created by the author VENKATESWARARAO

Question

A full-wave rectified sinusoidal voltage has a peak value of $V_m$. What is its average value?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Average value of a periodic waveform
 **Related Topics**: Rectification, sinusoidal waveforms, integration
 **Prerequisites**: Understanding of integration, trigonometric functions, and rectification techniques.
## SOLUTION PATHWAY
**Given Information**:
  Peak voltage = $V_m$
  Full-wave rectification
**Method & Steps**:
 1. The equation for a full-wave rectified sinusoidal voltage is given by $|V_m \sin(\omega t)|$.  To find the average value, we need to integrate this over one complete cycle and divide by the period.
  Reasoning: The average value represents the DC component of the rectified waveform.
  Formula/Rule: The average value of a periodic function $f(t)$ with period $T$ is given by: $\frac{1}{T} \int_0^T f(t) dt$
 2. The period of a full-wave rectified sine wave is $\frac{2\pi}{\omega}$.  Therefore, the average value is:
  Working: $\frac{\omega}{2\pi} \int_0^{\frac{2\pi}{\omega}} |V_m \sin(\omega t)| dt = \frac{\omega}{2\pi} \int_0^{\frac{2\pi}{\omega}} V_m \sin(\omega t) dt$ (since the integral over one cycle is positive)
  This simplifies to: $\frac{V_m}{2\pi} \int_0^{2\pi} \sin(x) dx$, where $x = \omega t$. The integral of $\sin(x)$ from 0 to $2\pi$ is 0, but since it is a full wave rectifier, we must consider the absolute value.  The integral becomes $\frac{V_m}{\pi} \int_0^{\pi} \sin(x) dx = \frac{V_m}{\pi} [ -\cos(x) ]_0^{\pi} = \frac{V_m}{\pi} [ -(-1) - (-1) ] = \frac{2V_m}{\pi}$
  Key Point:  The absolute value is crucial; ignoring it leads to an incorrect answer of 0.
 3. 
  Result: $\frac{2V_m}{\pi}$
  Verification: The result is consistent with the known average value of a full-wave rectified sine wave.
 ## OPTION ANALYSIS
A :x: This is the average value of a half-wave rectified sine wave.
B ✓: This is the correct average value of a full-wave rectified sine wave.
C :x: This would be the average value if the waveform were simply a square wave with amplitude $V_m$.
D :x: This is the peak value, not the average value.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly integrating without considering the absolute value of the rectified waveform: Remember that full-wave rectification involves taking the absolute value of the sine wave before calculating the average.
  2.Using the wrong period for integration: Use the period of the full-wave rectified waveform, which is half the period of the original sine wave.
 **Quick Tips**:
  1.Remember the formula for average value: $\frac{1}{T} \int_0^T f(t) dt$
  2.For a full-wave rectified sine wave, the average value is always $\frac{2V_m}{\pi}$

VENKATESWARARAO · Answer

$V_m/\pi$

VENKATESWARARAO · Answer

The correct Option is "$2V_m/\pi$"

VENKATESWARARAO · Answer

$V_m/2$

VENKATESWARARAO · Answer

$V_m$

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