Question

The rms value of a.c voltage with peak of 311V is
A. 220V
B. 311V
C. 180V
D. 320V

Answer 1

Hint:
One must know, how to find the RMS value of an alternating voltage, that is for alternating voltage $V={{V}_{0}}\sin \omega t$ the RMS value is \[\dfrac{V_{0}^{{}}}{\sqrt{2}}.\] The peak value of the a.c voltage can be found, by equating the term: $V={{V}_{0}}\sin \omega t$ for maximum value of voltage, which is the peak voltage.

Step by step solution:
Let’s first start by understanding how to find out the RMS value of any function f(x). The average value of the function f(x) having time period (T) is, Average value $=\dfrac{1}{T}\int\limits_{0}^{T}{f(x)dx}$.
Therefore, the average value of square of f(x) is, $\dfrac{1}{T}\int\limits_{0}^{T}{{{[f(x)]}^{2}}dx}$.
Hence, the RMS value of of the function f(x) will be, RMS value $=\sqrt{\dfrac{1}{T}\int\limits_{0}^{T}{f(x)dx}}$.
Now, let’s consider the case for the alternating voltage given by $V={{V}_{0}}\sin \omega t$. Since, this is a sinusoidal wave equation, hence, the time period is $2\pi .$ Hence, using these values, that is T=$2\pi $ and f(x)= $V={{V}_{0}}\sin \omega t$. Therefore, the RMS value becomes, $\sqrt{\dfrac{1}{2\pi }\int\limits_{0}^{2\pi }{{{({{V}_{0}}\sin \omega t)}^{2}}dt}}$.
That is, $\sqrt{\dfrac{V_{0}^{2}}{2\pi }\int\limits_{0}^{2\pi }{{{(\sin \omega t)}^{2}}dt}}$, here we will substitute the value of${{\sin }^{2}}(\omega t)$ as ${{\cos }^{2}}(\omega t)=\dfrac{1}{2}(1+\cos 2\omega t).$Further, we will also take out the constant values. This makes the RMS value to be, $\dfrac{V_{0}^{{}}}{\sqrt{2\pi }}\sqrt{\int\limits_{0}^{2\pi }{\dfrac{1}{2}(1+\cos 2\omega t)dt}}$. That is, $\dfrac{V_{0}^{{}}}{2\sqrt{\pi }}\sqrt{\int\limits_{0}^{2\pi }{(1+\cos 2\omega t)dt}}$.
We know that,$\omega =\dfrac{2\pi }{T}$. Hence the integral becomes, $\dfrac{V_{0}^{{}}}{2\sqrt{\pi }}\sqrt{[t]_{0}^{2\pi }+[\dfrac{\sin \omega t}{\omega }]_{0}^{2\pi }}=\dfrac{V_{0}^{{}}}{2\sqrt{\pi }}\sqrt{[2\pi -0]+[\dfrac{\sin 2\pi -\sin 0}{\omega }]}$.
Therefore, the RMS eventually becomes, \[\dfrac{V_{0}^{{}}}{2\sqrt{\pi }}\sqrt{2\pi }=\dfrac{V_{0}^{{}}}{\sqrt{2}}.\]
In the problem, we are given that the peak value of the ac voltage is 311V. That means, the maximum value of V given by: $V={{V}_{0}}\sin \omega t$ is 311V. Therefore; ${{V}_{\max }}=311={{V}_{0}}\sin \omega t$.
However, since ${{V}_{0}}$ is a constant, hence we need to find the maximum value of $\sin \omega t$. That is: $\sin \omega t=1$. Therefore; ${{V}_{\max }}=311={{V}_{0}}\sin \omega t\Rightarrow {{V}_{0}}=311$.
Substituting in the value of ${{V}_{0}}=311$ into the RMS value, we get \[{{V}_{RMS}}=\dfrac{V_{0}^{{}}}{\sqrt{2}}=\dfrac{311}{\sqrt{2}}=220\].
Therefore, the RMS value of ac voltage with peak value of 311V is 220V, given by Option A.

Note:
You may think that, we are always considering the value of alternating current as $V={{V}_{0}}\sin \omega t$ for most of the equations. The reason is, that this value of ${{V}_{0}}\sin \omega t$ is the only necessary part of the alternating voltage carrying the important information required for the calculations done above. The additional term is the phase angle.
When we consider, ${{V}_{1}}={{V}_{0}}\sin (\omega t\pm \delta )$, here $(\delta )$ refers to the phase angle. The phase angle is the amount or angle with which the alternating voltage $({{V}_{1}})$ will lead or lag against $V={{V}_{0}}\sin \omega t$.
$({{V}_{1}})$ will be leading against $V={{V}_{0}}\sin \omega t$, when the phase angle is positive. That is, ${{V}_{1}}={{V}_{0}}\sin (\omega t+\delta )$. Similarly, $({{V}_{1}})$ will be lagging behind $V={{V}_{0}}\sin \omega t$, when the phase angle is negative. That is, ${{V}_{1}}={{V}_{0}}\sin (\omega t-\delta )$.