An alternating current is given by the equation i = i _ 1 sin + i _ 2 cos . The rms current will be:

Correct answer: D. 1 2 ( i _ 1 ^ 2 + i _ 2 ^ 2 )(1)/(2) i = i _ 1 sin + i _ 2 ( + 90) i = i _ 1 ^ 2 + i _ 2 ^ 2 ( + ϕ) i _ rms = i _ 0 2 = i _ 1 ^ 2 + i _ 2 ^ 2 2

JEE Main 202126 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 26 February
Shift: Morning Shift
Subject: Physics
Chapter: Alternating Current

JEE Main 2021 · 26 February · Morning Shift · Physics · Alternating Current

An alternating current is given by the equation i = i _ 1 sin + i _...

An alternating current is given by the equation

i = {i}_{1}sin\omegat + {i}_{2}cos\omegat

. The rms current will be:

Options

A\dfrac{1}{\sqrt{2}}{ $\left({i}_{1} + {i}_{2}\right)}^{2}$
B $\dfrac{1}{\sqrt{2}}\left({i}_{1} + {i}_{2}\right)$
C\dfrac{1}{2} $\left({i}_{1}^{2} + {i}_{2}^{2}\right)\dfrac{1}{2}$
D\dfrac{1}{\sqrt{2}} $\left({i}_{1}^{2} + {i}_{2}^{2}\right)\dfrac{1}{2}$

Correct answer

D. \dfrac{1}{\sqrt{2}} $\left({i}_{1}^{2} + {i}_{2}^{2}\right)\dfrac{1}{2}$

Step-by-step solution

$i = {i}_{1}sin\omegat + {i}_{2}\sin \left(\omegat + 90\right)$

$i = \sqrt{{i}_{1}^{2} + {i}_{2}^{2}}\sin \left(\omegat + ϕ\right)$

${i}_{rms} = \dfrac{{i}_{0}}{\sqrt{2}} = \dfrac{\sqrt{{i}_{1}^{2} + {i}_{2}^{2}}}{\sqrt{2}}$

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