Q62 / 75 · JEE Main 2026 (28 January Shift 1) · Physics
The electric current in the circuit is given as $i=i_{\mathrm{o}}(t / T)$. The r.m.s current for the period $t=0$ to $t=T$ is $\_\_\_\_$.
- A$i_{\mathrm{o}}$
- B$\frac{i_{o}}{\sqrt{6}}$
- Correct answerC$\frac{i_{o}}{\sqrt{3}}$
- D$\frac{i_{\mathrm{o}}}{\sqrt{2}}$
Solution
The RMS current is calculated using $I_{rms} = \sqrt{\frac{1}{T}\int_0^T i^2 dt}$.
Given $i = i_0(t/T)$:
$I_{rms} = \sqrt{\frac{1}{T}\int_0^T i_0^2\left(\frac{t}{T}\right)^2 dt} = \sqrt{\frac{i_0^2}{T^3}\int_0^T t^2 dt} = \sqrt{\frac{i_0^2}{T^3} \cdot \frac{T^3}{3}}$
$= \sqrt{\frac{i_0^2}{3}} = \frac{i_0}{\sqrt{3}}$.
Given $i = i_0(t/T)$:
$I_{rms} = \sqrt{\frac{1}{T}\int_0^T i_0^2\left(\frac{t}{T}\right)^2 dt} = \sqrt{\frac{i_0^2}{T^3}\int_0^T t^2 dt} = \sqrt{\frac{i_0^2}{T^3} \cdot \frac{T^3}{3}}$
$= \sqrt{\frac{i_0^2}{3}} = \frac{i_0}{\sqrt{3}}$.
Hence, the correct answer is option: 'C'.