Expression for the RMS value of an alternating EMF

Knowledge Increases By Sharing...

In this short piece of the article, we are going to derive an expression for the RMS value of an alternating voltage, so let’s get started…

Root mean square value of an alternating emf

It is defined as that value of a steady voltage that produces the same amount of heat in a given resistance as is produced by the given alternating emf when applied to the same resistance for the same time. It is also called the virtual or effective value of the alternating emf. It is denoted by $\varepsilon_{r m s}$ or $\varepsilon_{e f f}$ or $\varepsilon_{v}$.

Read Also

Alternating current class 12 – amplitude, time period, frequency

Relation between the RMS value and the peak value of an alternating emf

RMS Voltage of a Sinusoidal AC Waveform
Relation between the RMS value and the peak value of an alternating emf

Suppose an alternating emf $\varepsilon$ applied to a resistance $R$ is given by $$ \varepsilon=\varepsilon_{0} \sin \omega t $$ Heat produced in a small time $d t$ will be $$ d H=\frac{\varepsilon^{2}}{R} d t=\frac{\varepsilon_{0}^{2}}{R} \sin ^{2} \omega t d t $$ Let $T$ be the time period of the alternating emf. Then heat produced in time $T$ will be

$$ \begin{aligned} H &=\int d H=\int_{0}^{T} \frac{\varepsilon_{0}^{2}}{R} \sin ^{2} \omega t d t \\ &=\frac{\varepsilon_{0}^{2}}{R} \int_{0}^{T} \frac{(1-\cos 2 \omega t)}{2} d t=\frac{E_{0}^{2}}{2 R}\left[t-\frac{\sin 2 \omega t}{2 \omega}\right]_{0}^{T} \\ &=\frac{\varepsilon_{0}^{2}}{2 R}\left[(T-0)-\frac{1}{2 \omega}\left|\sin \frac{4 \pi}{T} t\right|_{0}^{T}\right] \\ &=\frac{\varepsilon_{0}^{2}}{2 R}\left[T-\frac{1}{2 \omega} \sin (4 \pi-\sin 0)\right] \\ \text { or } \quad H &=\frac{\varepsilon_{0}^{2}}{2 R}[T-0]=\frac{\varepsilon_{0}^{2} T}{2 R} \end{aligned} $$

If $\varepsilon_{rms}$ is the root mean square value of the alternating emf, then the amount of heat produced by it in the same resistance $R$ in the time $T$ will be $$ H=\frac{\varepsilon_{rms}^{2} T}{R} $$

Read Also

From the above two equations, we get $$\frac{\varepsilon_{rms}^2 T}{R}=\frac{\varepsilon_{0}^2 T}{2R}$$ $$\text{Or}\qquad \varepsilon_{rms}=\frac{\varepsilon_0}{\sqrt{2}}=0.707\varepsilon_0$$

Stay tuned with Laws Of Nature for more useful and interesting content.

Was this article helpful?
YesNo
Knowledge Increases By Sharing...

Newsletter Updates

Enter your email address below and subscribe to our newsletter

1 1 vote
Rate this Article
Subscribe
Notify of
guest

0 Comments
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x
| Reply