What is the average value of a sine wave Vm sinωt over a full cycle?
0
Calculating the Average Value of a Sine Wave
The question asks for the average value of a sine wave, given by \(V(t) = V_m \sin(\omega t)\), over a full cycle. Understanding the average value of periodic functions like sine waves is fundamental in electrical engineering and physics.
Definition of Average Value Over a Full Cycle
For a periodic function \(f(t)\) with period \(T\), the average value over one full cycle is defined as:
\(\text{Average Value} = \frac{1}{T} \int_{t_0}^{t_0+T} f(t) dt\)
For a sine wave \(V(t) = V_m \sin(\omega t)\), the angular frequency is \(\omega\). The period \(T\) is given by \(T = \frac{2\pi}{\omega}\). We can calculate the average value starting from \(t_0 = 0\) to \(t_0 + T = \frac{2\pi}{\omega}\).
Step-by-Step Calculation
We need to evaluate the integral of \(V_m \sin(\omega t)\) over the interval \([0, \frac{2\pi}{\omega}]\):
\(\text{Average Value} = \frac{1}{2\pi/\omega} \int_{0}^{2\pi/\omega} V_m \sin(\omega t) dt\)
Let's first evaluate the indefinite integral:
\(\int V_m \sin(\omega t) dt = V_m \int \sin(\omega t) dt\)
Using the standard integral \(\int \sin(ax) dx = -\frac{1}{a}\cos(ax)\), we get:
\(V_m \left(-\frac{\cos(\omega t)}{\omega}\right) + C = -\frac{V_m}{\omega} \cos(\omega t) + C\)
Now, we evaluate the definite integral from \(0\) to \(\frac{2\pi}{\omega}\):
\(\int_{0}^{2\pi/\omega} V_m \sin(\omega t) dt = \left[-\frac{V_m}{\omega} \cos(\omega t)\right]_{0}^{2\pi/\omega}\)
\( = \left(-\frac{V_m}{\omega} \cos\left(\omega \cdot \frac{2\pi}{\omega}\right)\right) - \left(-\frac{V_m}{\omega} \cos(\omega \cdot 0)\right)\)
\( = \left(-\frac{V_m}{\omega} \cos(2\pi)\right) - \left(-\frac{V_m}{\omega} \cos(0)\right)\)
We know that \(\cos(2\pi) = 1\) and \(\cos(0) = 1\).
\( = \left(-\frac{V_m}{\omega} \cdot 1\right) - \left(-\frac{V_m}{\omega} \cdot 1\right)\)
\( = -\frac{V_m}{\omega} + \frac{V_m}{\omega} = 0\)
Finally, we substitute this back into the average value formula:
\(\text{Average Value} = \frac{1}{2\pi/\omega} \cdot 0 = 0\)
Conceptual Understanding
A sine wave \(V(t) = V_m \sin(\omega t)\) is symmetric about the horizontal axis. Over a full cycle, the positive area above the axis is exactly equal in magnitude to the negative area below the axis. When calculating the average value using integration, these positive and negative areas cancel each other out, resulting in a net average of zero.
Consider the graph of a sine wave over one period:
- From \(t=0\) to \(t=\pi/\omega\), the function is positive.
- From \(t=\pi/\omega\) to \(t=2\pi/\omega\), the function is negative.
The integral represents the net area under the curve. Because the positive area cancels the negative area, the total integral over a full cycle is zero, leading to an average value of zero.
Summary of Average Value Calculation
The average value of a sine wave \(V_m \sin(\omega t)\) over a full cycle is found by integrating the function over one period and dividing by the period length.
\(\text{Average Value} = \frac{\omega}{2\pi} \int_{0}^{2\pi/\omega} V_m \sin(\omega t) dt\)
Evaluating the integral:
\(\int_{0}^{2\pi/\omega} V_m \sin(\omega t) dt = 0\)
Therefore,
\(\text{Average Value} = \frac{\omega}{2\pi} \cdot 0 = 0\)
Comparison with Other Values
It is important not to confuse the average value over a full cycle with other related values:
| Value Type | Description | Formula (for \(V_m \sin(\omega t)\)) |
|---|---|---|
| Peak Value (\(V_m\)) | Maximum instantaneous value | \(V_m\) |
| Peak-to-Peak Value | Difference between maximum and minimum values | \(2V_m\) |
| RMS Value (Root Mean Square) | Effective value, related to power dissipation | \(\frac{V_m}{\sqrt{2}}\) |
| Average Value (Full Cycle) | Mean value over one complete period | 0 |
| Average Value (Half Cycle) | Mean value over the positive (or negative) half cycle | \(\frac{2V_m}{\pi}\) |
Revision Table: Sine Wave Properties
| Property | Value (for \(V_m \sin(\omega t)\)) | Calculation Method |
|---|---|---|
| Period (T) | \(\frac{2\pi}{\omega}\) | Calculated from angular frequency \(\omega\) |
| Frequency (f) | \(\frac{\omega}{2\pi}\) | Reciprocal of Period |
| Peak Value | \(V_m\) | Maximum amplitude of the wave |
| RMS Value | \(\frac{V_m}{\sqrt{2}}\) | \(\sqrt{\frac{1}{T}\int_0^T (V_m \sin(\omega t))^2 dt}\) |
| Average Value (Full Cycle) | 0 | \(\frac{1}{T}\int_0^T V_m \sin(\omega t) dt\) |
| Average Value (Half Cycle) | \(\frac{2V_m}{\pi}\) | \(\frac{1}{T/2}\int_0^{T/2} V_m \sin(\omega t) dt\) |
Additional Information: Average vs. RMS
While the average value of a sine wave over a full cycle is zero, the RMS (Root Mean Square) value is often used to represent the effective value of AC voltage or current. The RMS value is particularly important because it relates directly to the power delivered to a resistive load. A DC voltage equal to the RMS value of an AC voltage will deliver the same amount of power to a resistor as the AC voltage.
The average value over a half cycle (e.g., the positive half) is non-zero (\(\frac{2V_m}{\pi}\)) and is used in some applications, such as rectifiers, where only the positive or negative portion of the waveform is considered.
For the given question, the average value of the sine wave over a full cycle is indeed 0.
Important Questions from Sinusoidal Steady State Analysis
The total opposition offered to the flow of current in AC circuit is called-
A quantity whose magnitude has a definite repeating time cycle is called a-
The current drawn by a tungsten filament lamp is measured by an ammeter. The ammeter reading under steady state condition will be ______ the ammeter reading when the supply is switched on.
The current flowing through a pure inductor in an AC circuit lags the applied voltage by:
The peak factor of a sinusoidal waveform is:
