An alternating current is given by i = i_1 ω t - i_2 ω t . The rms current is given by

Correct answer: C. (i_1^2 + i_2^2)/(2) The given alternating current is i = i_1 ω t - i_2 ω t The mean square value of the current over a complete cycle is: = = i_1^2 + i_2^2 - 2i_1 i_2 Using the standard average values over a full cycle: = (1)/(2) , = (1)/(2) , and = 0 = i_1^2 ((1)/(2)) + i_2^2 ((1)/(2)) - 0 = (i_1^2 + i_2^2)/(2) The rms current is the square root of the mean square current: i_ rms = = (i_1^2 + i_2^2)/(2) Answer: (i_1^2 + i_2^2)/(2)

MHT CET 202625 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 25 April
Shift: Morning Shift
Subject: Physics
Chapter: Alternating Current

MHT CET 2026 · 25 April · Morning Shift · Physics · Alternating Current

An alternating current is given by i = i_1 ω t - i_2 ω t . The rms...

An alternating current is given by

i = i_1\sin\omega t - i_2\cos\omega t

. The rms current is given by

Options

A $\dfrac{i_1 + i_2}{\sqrt{2}}$
B $\dfrac{\sqrt{i_1^2 + i_2^2}}{2}$
C $\sqrt{\dfrac{i_1^2 + i_2^2}{2}}$
D $\sqrt{i_1^2 + i_2^2}$

Correct answer

C. $\sqrt{\dfrac{i_1^2 + i_2^2}{2}}$

Step-by-step solution

The given alternating current is

i = i_1\sin\omega t - i_2\cos\omega t

The mean square value of the current over a complete cycle is:

<i^2> = <(i_1\sin\omega t - i_2\cos\omega t)^2>

<i^2> = i_1^2 <\sin^2\omega t> + i_2^2 <\cos^2\omega t> - 2i_1 i_2 <\sin\omega t \cos\omega t>

Using the standard average values over a full cycle:

< \sin^2\omega t > = \dfrac{1}{2}

< \cos^2\omega t > = \dfrac{1}{2}

, and

< \sin\omega t \cos\omega t > = 0

<i^2> = i_1^2 \left(\dfrac{1}{2}\right) + i_2^2 \left(\dfrac{1}{2}\right) - 0 = \dfrac{i_1^2 + i_2^2}{2}

The rms current is the square root of the mean square current:

i_{rms} = \sqrt{<i^2>} = \sqrt{\dfrac{i_1^2 + i_2^2}{2}}

Answer:

\sqrt{\dfrac{i_1^2 + i_2^2}{2}}

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