Alternating Current

# RMS Value of AC

We know that the average or mean value of ac over one cycle is zero. So, the mean value of ac can not be used to specify ac current or voltage, and hence power. Thus, it was necessary to come up with another method to measure the effectiveness of an ac current or voltage. The obvious method would be to measure it in terms of dc that would do work (or produce heat) at the same average rate as ac under similar conditions. This equivalent dc is known as the rms value of ac.

When a dc flows through a resistance for a certain time, then heat is produced in the resistance. Now, if an ac produces same amount of heat in the same resistance in the same time, then that dc is as effective as the ac and this dc is called rms value of ac. RMS value is also known as virtual value or effective value of ac.

Hence, Root Mean Square (RMS) value of AC is defined as that value of steady current which when passed through a given resistance for a given time will produce the same amount of heat as the alternating current does in the same resistance in the same time. It is denoted by $I_{\text{rms}}$.

## Measurement of RMS Value of AC

Let the time period $(T)$ of AC (i.e. time required to complete one cycle) be divided into $n$ small equal intervals. Then each interval is equal to $\frac{T}{n}$. Also, let $\,i_1,i_2,…,i_n\,$ be the instantaneous currents during these intervals. If $\,H_1,H_2,…,H_n\,$ are the amounts of heat produced during these intervals when the currents pass through a resistance $R$, then according to the Joule’s laws of heating, $\begin{array}{c}H_1=i_1^2R\frac{T}{n}, & H_2=i_2^2R\frac{T}{n}, & … & H_n=i_n^2R\frac{T}{n} \end{array}$

Let $H$ be the total amount of heat produced over one cycle of ac. Then, $H=H_1+H_2+…+H_n$ $=i_1^2R\frac{T}{n}+i_2^2R\frac{T}{n}+…+i_n^2R\frac{T}{n}$ $\therefore H=(i_1^2+i_2^2+…+i_n^2)R\frac{T}{n}\text{ __(1)}$

Let $I_{\text{rms}}$ be the rms value of ac, then we can write $H=I_{\text{rms}}^2RT\text{ __(2)}$

From equations $\text{(1)}$ and $\text{(2)}$, $I_{\text{rms}}^2RT=(i_1^2+i_2^2+…+i_n^2)R\frac{T}{n}$ $I_{\text{rms}}^2=\frac{i_1^2+i_2^2+…+i_n^2}{n}$ $I_{\text{rms}}=\sqrt{\frac{i_1^2+i_2^2+…+i_n^2}{n}}$

Hence, rms value of ac is equal to the square root of mean of square of instantaneous currents.

## Relation between Peak Value and RMS Value of AC

The instantaneous ac at an instant of time $t$ is given by $I=I_0\sin\omega t$

If this alternating current flows through a resistance $R$ for small time $dt$, then according to the Joule’s laws of heating, the small amount of heat produced is given by, $dH=I^2R\:dt$ $dH=I_0^2\sin^2\omega t\:R\:dt$

The total amount of heat produced over one complete cycle can be obtained by integrating above equation from $t=0$ to $t=T$.

$\therefore H=\int_0^T I_0^2R\sin^2\omega t\:dt$ $=I_0^2R\int_0^T\frac{1-\cos 2\omega t}{2}\:dt$ $=\frac{I_0^2R}{2}\left[\int_0^T dt-\int_0^T\cos 2\omega t\:dt\right]$ $=\frac{I_0^2R}{2}\left[[t]_0^T-\left[\frac{\sin 2\omega t}{2\omega}\right]_0^T\right]$ $=\frac{I_0^2R}{2}\left[T-\frac{1}{2\omega}\left[\sin 2\omega T-\sin 0°\right]\right]$ $=\frac{I_0^2R}{2}\left[T-\frac{1}{2\omega}\left[\sin 2\frac{2π}{T} T-0\right]\right]$ $=\frac{I_0^2R}{2}[T-0]$ $\therefore H=\frac{I_0^2RT}{2}\text{ __(i)}$

If $I_{\text{rms}}$ is the rms value of ac, then we have $H=I_{\text{rms}}^2RT\text{ __(ii)}$ From $\text{(i)}$ and $\text{(ii)}$, $I_{\text{rms}}^2RT=\frac{I_0^2RT}{2}$ $\therefore I_{\text{rms}}=\frac{I_0}{\sqrt{2}}=0.707I_0$

Hence, rms value of ac is $0.707$ times the peak value of ac i.e. $70.7\%$ of the peak value.

Similarly, the rms value of alternating emf can be obtained as $E_{\text{rms}}=\frac{E_0}{\sqrt{2}}=0.707E_0$