An alternating emf given by $E=300\sin \left[ \left( 100\pi \right)t \right]$ volt is applied to a resistance of 100 ohms. The rms current through the circuit is (in amperes):
\[A.\quad \dfrac{3}{\sqrt{2}}\]
\[B.\quad \dfrac{9}{\sqrt{2}}\]
\[C.\quad 3\]
\[D.\quad \dfrac{6}{\sqrt{2}}\]

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 \[\dfrac{V_{0}^{{}}}{\sqrt{2}}.\] Using the Ohm’s law: $V=IR\Rightarrow I=\dfrac{V}{R}$, we can find correspondingly the RMS value of current in the circuit as; ${{I}_{rms}}=\dfrac{{{V}_{rms}}}{R}$.

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 standard 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, \[\dfrac{V_{0}^{{}}}{2\sqrt{\pi }}\sqrt{2\pi }=\dfrac{V_{0}^{{}}}{\sqrt{2}}.\]
We will now use Ohm's law, which is given by; $V=IR\Rightarrow I=\dfrac{V}{R}$. The given value of the alternating emf is: $E=300\sin \left[ \left( 100\pi \right)t \right]$. Comparing this value against the standard alternating voltage value of: $V={{V}_{0}}\sin \omega t$. Therefore;${{V}_{0}}=300$Volts. Hence, the RMS value of the emf is: \[\dfrac{V_{0}^{{}}}{\sqrt{2}}=\dfrac{300}{\sqrt{2}}\Rightarrow {{V}_{rms}}=\dfrac{300}{\sqrt{2}}\].
The given value of resistance is $R=100\Omega $.
Hence, using the Ohm’s law relation, we will get; ${{I}_{rms}}=\dfrac{{{V}_{rms}}}{R}$. That is; ${{I}_{rms}}=\dfrac{{{V}_{rms}}}{R}=\dfrac{300}{\sqrt{2}}\times \dfrac{1}{100}\Rightarrow {{I}_{rms}}=\dfrac{3}{\sqrt{2}}$Ampere. That is, Option A.

Note:
The original value of an alternating voltage is, \[V={{V}_{0}}\sin (\omega t\pm \delta )\], here $(\delta )$refers to the phase angle. For the current problem, the phase angle is equal to zero. The phase angle is the amount or angle with which the alternating voltage $({{V}_{1}}={{V}_{0}}\sin (\omega t\pm \delta ))$ 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, ${{V}_{1}}={{V}_{0}}\sin (\omega t-\delta )$.