Alternating Current

JEE Main level, single correct, medium

42 of 190

Question 42 of 190

An alternating current is given by the equation i=i1sin⁡ωt+i2cos⁡ωti={i}_{1}\sin \omega t+{i}_{2}\cos \omega t. The rms current will be:

Show answer and solution

Answer: (D) 12(i12+i22)12\dfrac{1}{\sqrt{2}}{\left({i}_{1}^{2}+{i}_{2}^{2}\right)}^{\dfrac{1}{2}}

i=i1sin⁡ωt+i2sin⁡(ωt+90)i={i}_{1}\sin \omega t+{i}_{2}\sin \left(\omega t+90\right)
i=i12+i22sin⁡(ωt+ϕ)i=\sqrt{{i}_{1}^{2}+{i}_{2}^{2}}\sin \left(\omega t+\phi \right)
irms=i02=i12+i222{i}_{rms}=\dfrac{{i}_{0}}{\sqrt{2}}=\dfrac{\sqrt{{i}_{1}^{2}+{i}_{2}^{2}}}{\sqrt{2}}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: AC circuits, phasors and resonance simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Alternating Current (Alternating Current, Voltage and Power), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.