An alternating current is given by I = IA t + IB t. The r.m.s. current will be:

Given: I = IA t + IB t Find: The r.m.s. current. The r.m.s. value is obtained from the mean square current: I rms = I^2 Square the given current: I^2 = (IA t + IB t)^2 = IA^2 ^2 t + IB^2 ^2 t + 2IA IB t t Now take time average over one complete period. Using ^2 t = 12, ^2 t = 12, t t = 0 we get I^2 = IA^2 12 + IB^2 12 = IA^2 + IB^22 For a sinusoidal current of amplitude Im, the r.m.s. value is I rms = Im 2 So, the two components have r.m.s. values IA 2 and IB 2 respectively. Since the sine and cosine terms are orthogonal over a complete cycle, their mean squares add: I rms = ( IA 2)^2 + ( IB 2)^2 = IA^22 + IB^22 = IA^2 + IB^22 Therefore, the correct option is C.

An alternating current is given by I = IA t + IB t. The r.m.s. cu… — Solution | JEE Physics

MCQMediumJEE 2025Alternating Current Basics

An alternating current is given by $I = I_A \sin \omega t + I_B \cos \omega t.$ The r.m.s. current will be:

A
$\sqrt{I_A^2 + I_B^2}$
B
$\frac{\sqrt{I_A^2 + I_B^2}}{2}$
C
$\sqrt{\frac{I_A^2 + I_B^2}{2}}$
D
$\frac{|I_A + I_B|}{\sqrt{2}}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

I = I_A \sin \omega t + I_B \cos \omega t

Find: The r.m.s. current.

The r.m.s. value is obtained from the mean square current:

I_{\text{rms}} = \sqrt{\langle I^2 \rangle}

Square the given current:

I^2 = \left(I_A \sin \omega t + I_B \cos \omega t\right)^2

= I_A^2 \sin^2 \omega t + I_B^2 \cos^2 \omega t + 2I_A I_B \sin \omega t \cos \omega t

Now take time average over one complete period. Using

\langle \sin^2 \omega t \rangle = \frac{1}{2}, \qquad \langle \cos^2 \omega t \rangle = \frac{1}{2}, \qquad \langle \sin \omega t \cos \omega t \rangle = 0

we get

\langle I^2 \rangle = I_A^2 \cdot \frac{1}{2} + I_B^2 \cdot \frac{1}{2} = \frac{I_A^2 + I_B^2}{2}

Using rms of orthogonal sinusoidal components

For a sinusoidal current of amplitude $I_m$ , the r.m.s. value is

I_{\text{rms}} = \frac{I_m}{\sqrt{2}}

So, the two components have r.m.s. values

\frac{I_A}{\sqrt{2}} \quad \text{and} \quad \frac{I_B}{\sqrt{2}}

respectively.

Since the sine and cosine terms are orthogonal over a complete cycle, their mean squares add:

I_{\text{rms}} = \sqrt{\left(\frac{I_A}{\sqrt{2}}\right)^2 + \left(\frac{I_B}{\sqrt{2}}\right)^2}

= \sqrt{\frac{I_A^2}{2} + \frac{I_B^2}{2}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}

Therefore, the correct option is C.

Common mistakes

Adding amplitudes first and writing $I_{\text{rms}} = \frac{I_A + I_B}{\sqrt{2}}$ is incorrect because sine and cosine components are phase-shifted by $90^\circ$ . Instead, square the expression, take time average, and then take the square root.
Using $\sqrt{I_A^2 + I_B^2}$ as the r.m.s. value ignores the factor of $\frac{1}{2}$ coming from the averages of $\sin^2 \omega t$ and $\cos^2 \omega t$ . Use $\langle \sin^2 \omega t \rangle = \langle \cos^2 \omega t \rangle = \frac{1}{2}$ .
Keeping the cross term in the average is a mistake. Over one full period, $\langle \sin \omega t \cos \omega t \rangle = 0$ , so that mixed term does not contribute to the r.m.s. current.

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Answer

Step-by-step solution

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