How to Calculate RMS Voltage

The RMS value is only calculated for the time-varying waveforms where the magnitude of quantity varies with respect to time.

We cannot find the RMS value for the DC waveform as the DC waveform has a constant value for every instant of time.

There are two methods to calculate RMS value.

  • Graphical Method
  • Analytical Method

Graphical Method

In this method, we use a waveform to find the RMS value. The graphical method is more useful when the signal is not symmetrical or sinusoidal.

The accuracy of this method depends on the number of points taken from the waveform. Few points result in low accuracy, and a more significant number of points result in high accuracy. 

The RMS value is a square root of the average value of the squared function. For example, let’s take a sinusoidal waveform of voltage as shown below figure.

Follow these steps to calculate the RMS voltage by graphical method.

Step-1: Divide waveform into equal parts. Here, we consider the half cycle of the waveform. You can consider full-cycle also.

The first half cycle divides into ten equal parts; V1, V2, …, V10.

Step-2: Find square of each value.

\[ V_1^2, V_2^2, V_3^2, …, V_{10}^2 \]

Step-3: Take the average of these squared values. Find the total of these values and divide by the total number of points.

\[ \frac{V_1^2+V_2^2+V_3^2+V_4^2+V_5^2+V_6^2+V_7^2+V_8^2+V_9^2+V_{10}^2}{10} \]

Step-4 Now, take square root of this value.

\[ V_{RMS} = \sqrt{\frac{V_1^2+V_2^2+V_3^2+V_4^2+V_5^2+V_6^2+V_7^2+V_8^2+V_9^2+V_{10}^2}{10}} \]

These steps are same for all type of continuous waveforms.

Analytical Method

In this method, the RMS voltage can be calculated by a mathematical procedure. This method is more accurate for the pure sinusoidal waveform.

Consider a pure sinusoidal voltage waveform defined as VmCos(ωt) with a period of T.


Vm = Maximum value or Peak value of voltage waveform

ω = Angular frequency = 2π/T

Now, we calculate the RMS value of voltage.

\[ V_{RMS} = \sqrt{\frac{1}{T} \int_{0}^{T} V_m^2 cos^2(\omega t) dt} \]
\[ V_{RMS} = \sqrt{\frac{V_m^2}{T} \int_{0}^{T} cos^2(\omega t) dt} \]
\[ V_{RMS} = \sqrt{\frac{V_m^2}{T} \int_{0}^{T} \frac{1+cos(2 \omega t)}{2} dt} \]
\[ V_{RMS} = \sqrt{\frac{V_m^2}{2T} \int_{0}^{T} 1+cos(2 \omega t) dt} \]
\[ V_{RMS} = \sqrt{\frac{ V_m^2}{2T} \left[ t + \frac{sin(2 \omega t)}{2 \omega} \right ]_0^T \]
\[ V_{RMS} = \sqrt{\frac{ V_m^2}{2T} \left[ (T-0) + (\frac{sin(2 \omega T)}{2 \omega} - \frac{sin 0}{2 \omega} ) \right ] \]
\[ V_{RMS} = \sqrt{\frac{ V_m^2}{2T} \left[ T + \frac{sin(2 \omega T)}{2 \omega}  \right ] \]
\[ V_{RMS} = \sqrt{\frac{ V_m^2}{2T} \left[ T + \frac{sin(2 \frac{2 \pi}{T} T)}{2 \frac{2 \pi}{T} }  \right ] \]
\[ V_{RMS} = \sqrt{\frac{ V_m^2}{2T} \left[ T +\frac{sin(4 \pi)}{2 \frac{2 \pi}{T}} \right ] \]
\[ V_{RMS} = \sqrt{\frac{ V_m^2}{2T} [T+0]} \]
\[ V_{RMS} = \sqrt{\frac{ V_m^2}{2} \]
\[ V_{RMS} = V_m \frac{1}{\sqrt{2}} \]
\[ V_{RMS} = V_m 0.7071 \]

So, RMS value of pure sinusoidal waveform can derive from the peak (maximum) value.


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