Finding the average voltage of an alternating waveform is similar to finding its RMS value, but without squaring the instantaneous values or taking the square root of the final average.
The average voltage (or current) of any waveform for example like a sine wave, square wave, or triangular wave, is like a DC value for an AC waveform.
We calculate this by finding the average area under the waveform’s curve over a specific time period.
In simpler terms, it’s like taking the average of all the voltage values at different points in time (T).
With a balanced waveform, the area above the horizontal line (positive half cycle) is equal to the area below (negative half cycle).
This means, the average value over a complete cycle (360 degrees) is zero.
The positive and negative areas cancel each other out when calculated mathematically.
Therefore for symmetrical AC waveforms like sine waves, we calculate the average value only over half a cycle.
Since the average over a full cycle is always zero, regardless of the peak voltage.
The terms “average voltage,” “mean voltage,” and “average current” apply to both AC waveforms and calculations involving converting AC to DC (rectification).
These average values are commonly represented by symbols like Vavg or Iavg.
Finding Average Voltage Graphically
Just like with RMS voltage, we are able to estimate the average voltage of a waveform graphically. This method focuses on the positive half cycle of the waveform.
We can divide the positive half cycle into several equal sections called mid-ordinates. The number of sections is denoted by “n”.
Each mid-ordinate will have a specific width (in degrees or seconds), and a height representing the voltage value at that point on the waveform.
Calculating Average Voltage from Mid-Ordinates
Once we have the voltage values for each mid-ordinate, we simply add them all together (V1 + V2 + … + V12). This sum represents the total voltage across all the sections.
Finally we divide this total voltage by the number of mid-ordinates used (n). As shown below, this gives us the average voltage (VAV) which actually represents the average of all the voltage readings across the positive half cycle.
VAV = Total of all the mid-ordinates / number of mid-ordinates
Using the above basic formula, the average voltage could be thus calculated as:
VAV = V1 + V2 + V3 …..+ V11 + V12 / 12
Just like before, let’s revisit our example of a sine wave with a peak voltage of 20 volts. We’ll focus on analyzing only the positive half cycle of this waveform.
| Voltage | 6.2V | 11.8V | 16.2V | 16.8V | 19.0V | 16.8V | 16.2V | 11.8V | 6.2V | 0V |
| Angle | 18o | 36o | 54o | 72o | 90o | 108o | 126o | 144o | 162o | 180o |
Therefore to calculate the average voltage, we use the following formula:
VAV = 6.2 + 11.8 + 16.2 + 16.8 + 19 + 16.8 + 16.2 + 11.8 + 6.2 + 0 / 10
= 121 / 10 = 12.10 Volts
Using the graphical method with our example waveform (20V peak sine wave, positive half cycle only), we obtain an average voltage of approximately 12.64 volts.
Calculating Average Voltage through Analytic Method
We mentioned earlier, that for perfectly symmetrical waveforms (sine waves in this case), the average voltage over a full cycle is zero. This is because, the positive and negative halves cancel each other out.
However for non-symmetrical or complex waveforms, we need a different approach. Here the average voltage (or current) is calculated mathematically, by considering the entire periodic cycle.
This method involves approximating the area under the waveform’s curve. We achieve this, by dividing the area into smaller shapes, like triangles or rectangles, and then relating their areas to the total time period of the waveform.
Estimating the Area Under the Curve
Estimating the Area with Rectangles
By dividing the area under the waveform’s curve, into rectangles, we can get a rough estimate of the actual area for each rectangle. Adding the areas of all these rectangles, provides an approximation of the total “voltage activity” over the time period.
The more rectangles we use (ideally an infinite number of very thin rectangles), the closer our approximation becomes to the true value, which approaches 2/π for a sine wave.
There are various mathematical methods, to achieve this area approximation, such as the trapezoidal rule, the mid-ordinate rule, or Simpson’s rule.
For a specific case of a sine wave with peak voltage (Vp) and angular frequency (ω), defined by the equation V(t) = Vp.cos(ωt) and with a period of T, the mathematical integral representing the area under the positive half cycle can be expressed as:
Area = ∫π0 Vp sin(ωt)dt
Limits of Integration and Area Calculation
Since we’re interested in the average voltage over one half cycle, the integration limits are set from 0 to π. This captures the area under the curve for the positive half cycle.
By integrating the equation for the voltage waveform (V(t) = Vp.cos(ωt)) within these limits, we arrive at a final area of 2Vp. This represents the area under the positive (or negative) half cycle of the sine wave.
Average Voltage from Area
Knowing the area under the positive (or negative) half cycle allows us to calculate the average voltage for that specific region of the waveform. We achieve this, by dividing the integrated area by half the period (T/2). This takes into account that only one half-cycle is being considered.
Example: Average Voltage of a Sine Wave
Suppose, let’s say the instantaneous voltage of a sine wave is given by v = Vp.sin(θ) and its period is 2π. To find the average voltage:
- We integrate the voltage equation (v = Vp.sin(θ)) over half the period (θ = 0 to π).
- We divide the resulting integral value by half the period (T/2).
This calculation provides the average voltage for the positive half cycle of the sine wave.
- VAVE = 1/π(∫π0 Vp sinθdθ
- VAVE = VP/π(-cosθ)π0
- = 2VP/π = 2/π(VP) = 0.637 VP
Standard Equation for Average Voltage of a Sine Wave
Following the steps outlined above, we arrive at the standard equation for the average voltage (Vavg) of a sine wave over one half cycle:
VAVE = 2VP/π = 0.637 VP
Finding Average Voltage in Sine Waves
- The average voltage (Vavg) of a sine wave is calculated by multiplying its peak voltage (Vp) by a constant value of 0.637. This constant is simply 2 divided by pi (π).
- This average voltage is also referred to as the mean value and represents the equivalent DC value when considering both the area under the waveform and the time period.
- It’s important to note that the average voltage depends on the magnitude (peak voltage) of the waveform but is independent of its frequency or phase angle.
Why Average Voltage is Zero in a Full Cycle
Over a complete cycle of a sine wave the average voltage is zero. This happens because the positive and negative halves of the waveform mirror each other.
Suppose, we are calculating the average area under the curve for each half cycle (positive and negative). These areas would have opposite signs (one positive, one negative) due to the voltage being above and below zero in each half.
When we add these areas together (Vavg – (-Vavg)), the positive and negative areas cancel each other out resulting in a total average voltage of zero for the entire cycle.
Example: Average Voltage Calculation
Let’s revisit our graphical example, where the peak voltage (Vpk) was 19 volts. Now we’ll use the analytical method to calculate the average voltage.
VAV = Vpk x 0.637 = 19 x 0.637 = 12.10 volts
Equivalence to Graphical Method
This calculation using the constant 0.637 (Vavg x 0.637), gives the same result we would obtain from the graphical method with a reasonable level of accuracy.
We can also use this formula, to find the peak voltage (Vpk) from a known average voltage (Vavg). All you need to do is rearrange the formula, and divide the average voltage by the constant 0.637.
Suppose, let’s say the average voltage is 66 volts and we want to find the corresponding peak voltage (Vpk) of a sine wave.
Finding Peak Voltage from Average Voltage
Vpk = Vavg / 0.637 = 66 volts / 0.637 ≈ 103.6 volts
But remember, multiplying the peak value by 0.637, to find the average voltage only works for sinusoidal waveforms. This quick trick won’t give accurate results for other wave shapes, for example like square waves or triangular waves.
Conclusions
In AC voltages, or currents the term “average value” typically refers to the average over a full cycle. However the term “mean value” is sometimes used specifically for the average over one half cycle of a periodic waveform.
Key Points:
- The average voltage of a complete sine wave cycle is zero, because the positive and negative halves cancel each other out.
- Therefore for sine waves, the average voltage is calculated over one half cycle and equals 0.637 times the peak voltage (Vp). This applies to both AC voltage and current.
- This average voltage concept is useful for calculating the equivalent DC value of rectified AC outputs (like from rectifiers, or pulse-width modulated motor drivers) in which the voltage or current no longer reverses direction. In this cases RMS value becomes less important.
Average vs. RMS Voltage:
The key difference between average, and RMS voltage lies in how these account for the waveform’s shape.
- Average voltage considers the average area under the waveform’s curve… typically over one half cycle for sinusoids.
- RMS voltage on the other hand, represents the equivalent heating value of the AC waveform compared to a steady DC voltage. It’s calculated by taking the square root of the average of the squared instantaneous values over a complete cycle.
Simple Calculations for Sine Waves (Only):
In pure sine waves, both average and RMS voltage (or current) can be easily calculated, using these formulas:
- Average value = 0.637 × peak voltage (Vpk)
- RMS value = 0.707 × peak voltage (Vpk)
Reference: Voltage
https://www.vedantu.com/jee-main/physics-average-and-rms-value