The electric current in the circuit is given as i=i_ o (t / T) . The r.m.s current for the period t=0 to t=T is \_\_\_\_ .

Correct answer: C. i_ o 3 The RMS current is calculated using I_ rms = (1)/(T)∫_0^T i^2 dt . Given i = i_0(t/T) : I_ rms = (1)/(T)∫_0^T i_0^2((t)/(T))^2 dt = (i_0^2)/(T^3)∫_0^T t^2 dt = (i_0^2)/(T^3) · (T^3)/(3) = (i_0^2)/(3) = i_0 3 .

JEE Main 202628 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 28 January
Shift: Morning Shift
Subject: Physics
Chapter: Alternating Current

JEE Main 2026 · 28 January · Morning Shift · Physics · Alternating Current

The electric current in the circuit is given as i=i_ o (t / T) ....

The electric current in the circuit is given as

i=i_{\mathrm{o}}(t / T)

. The r.m.s current for the period

t=0

t=T

\_\_\_\_

Options

A $i_{\mathrm{o}}$
B $\frac{i_{o}}{\sqrt{6}}$
C $\frac{i_{o}}{\sqrt{3}}$
D $\frac{i_{\mathrm{o}}}{\sqrt{2}}$

Correct answer

C. $\frac{i_{o}}{\sqrt{3}}$

Step-by-step solution

The RMS current is calculated using

I_{rms} = \sqrt{\frac{1}{T}\int_0^T i^2 dt}

.
Given

i = i_0(t/T)

I_{rms} = \sqrt{\frac{1}{T}\int_0^T i_0^2\left(\frac{t}{T}\right)^2 dt} = \sqrt{\frac{i_0^2}{T^3}\int_0^T t^2 dt} = \sqrt{\frac{i_0^2}{T^3} \cdot \frac{T^3}{3}}

= \sqrt{\frac{i_0^2}{3}} = \frac{i_0}{\sqrt{3}}

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