The electric current in the circuit is given as i=i0 ( tT). The r.m.s. current for the period t=0 to t=T is:

Given: The current varies with time as i(t)=i0 tT over the interval from t=0 to t=T. Find: The r.m.s. current over one period. Concept: The root mean square value is defined by i rms= 1T 0^T i^2(t) dt Substitute the given current function: i(t)=i0 tT So, i^2(t)=i0^2 t^2T^2 Apply the r.m.s. formula: i rms= 1T 0^T i0^2 t^2T^2 dt =i0 1T^3 0^T t^2 dt Evaluate the integral: 0^T t^2 dt= [ t^33 ]0^T= T^33 Therefore, i rms=i0 T^33T^3= i0 3 Therefore, the correct option is D.

The electric current in the circuit is given as i=i0 ( tT). The r… — Solution | JEE Physics

MCQEasyJEE 2026Alternating Current Basics

The electric current in the circuit is given as $i=i_0\left(\frac{t}{T}\right).$ The r.m.s. current for the period $t=0$ to $t=T$ is:

A
$i_0$
B
$\dfrac{i_0}{\sqrt6}$
C
$\dfrac{i_0}{\sqrt2}$
D
$\dfrac{i_0}{\sqrt3}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The current varies with time as $i(t)=i_0\frac{t}{T}$ over the interval from $t=0$ to $t=T$ .

Find: The r.m.s. current over one period.

Concept: The root mean square value is defined by

i_{\text{rms}}=\sqrt{\frac{1}{T}\int_0^T i^2(t)\,dt}

Substitute the given current function:

i(t)=i_0\frac{t}{T}

So,

i^2(t)=i_0^2\frac{t^2}{T^2}

Apply the r.m.s. formula:

i_{\text{rms}}=\sqrt{\frac{1}{T}\int_0^T i_0^2\frac{t^2}{T^2}\,dt}

=i_0\sqrt{\frac{1}{T^3}\int_0^T t^2\,dt}

Evaluate the integral:

\int_0^T t^2\,dt=\left[\frac{t^3}{3}\right]_0^T=\frac{T^3}{3}

Therefore,

i_{\text{rms}}=i_0\sqrt{\frac{T^3}{3T^3}}=\frac{i_0}{\sqrt3}

Therefore, the correct option is D.

Common mistakes

Using the average current instead of the r.m.s. current is incorrect because r.m.s. requires squaring the function before integration. Use $i_{\text{rms}}=\sqrt{\frac{1}{T}\int_0^T i^2(t)\,dt}$ , not the simple average of $i(t)$ .
Forgetting to square $\frac{t}{T}$ gives the wrong integrand. Since $i(t)=i_0\frac{t}{T}$ , the square is $i^2(t)=i_0^2\frac{t^2}{T^2}$ , not $i_0^2\frac{t}{T}$ .
Making an error in the integral of $t^2$ leads to a wrong factor. The correct result is $\int_0^T t^2\,dt=\frac{T^3}{3}$ , so keep the power and limits carefully while evaluating.

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