Question

Assertion $\left(A\right)$: The r.m.s. value of alternating current is defined as the square root of the average of $I^2$ during a complete cycle.

Reason $\left(R\right)$: For sinusoidal a.c.
$(I=I_0\sin \omega t)\text{and}{I}_{\text{rms}}=\frac{I_0}{\sqrt{2}}$.

• A. Both $\left(A\right)$ and $\left(R\right)$ are true, and $\left(R\right)$ is the correct explanation of $\left(A\right)$
• B. Both $\left(A\right)$ and $\left(R\right)$ are true, but $\left(R\right)$ is not the correct explanation of $\left(A\right)$
• C. $\left(A\right)$ is true but $\left(R\right)$ is false
• D. $\left(A\right)$ is false but $\left(R\right)$ is true

Answer 1

The correct option is A Both $\left(A\right)$ and $\left(R\right)$ are true, and $\left(R\right)$ is the correct explanation of $\left(A\right)$

The r.m.s value of a.c. current is,

$I_{\text{rms}}^2=\frac{{\int }_0^T{I}^{2}dt}{{\int }_0^{T}dt}$

$I=I_0\sin \left(\omega t\right)$

$I_{\text{rms}}^2=\frac{{\int }_0^T{I}_0^2{\sin }^2\left(\omega t\right)dt}{{\int }_0^{T}dt}=\frac{I_0^2}{T}{\int }_0^T\left[\frac{1-\cos 2\omega t}{2}\right]dt$

$=\frac{I_0^2}{2T}{\left[\frac{t-\sin 2\omega t}{2\omega }\right]}_0^T$

$\Rightarrow I_{\text{rms}}^2=\frac{I_0^2}{2}$

$\Rightarrow I_{\text{rms}}=\frac{I_0}{\sqrt{2}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.