Question:

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:

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For linearly varying currents, always square the function first before integrating for r.m.s. values.
Updated On: Feb 4, 2026
  • \( i_0 \)
  • \( \dfrac{i_0}{\sqrt6} \)
  • \( \dfrac{i_0}{\sqrt2} \)
  • \( \dfrac{i_0}{\sqrt3} \)
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The Correct Option is D

Solution and Explanation

Concept: The root mean square (r.m.s.) value of a time-varying current \( i(t) \) over a period \( T \) is defined as: \[ i_{\text{rms}}=\sqrt{\frac{1}{T}\int_0^T i^2(t)\,dt} \]
Step 1: Substitute the given expression for current \[ i(t)=i_0\frac{t}{T} \] \[ i^2(t)=i_0^2\frac{t^2}{T^2} \]
Step 2: Apply the r.m.s. formula \[ i_{\text{rms}}=\sqrt{\frac{1}{T}\int_0^T i_0^2\frac{t^2}{T^2}\,dt} \] \[ =i_0\sqrt{\frac{1}{T^3}\int_0^T t^2\,dt} \]
Step 3: Evaluate the integral \[ \int_0^T t^2\,dt=\left[\frac{t^3}{3}\right]_0^T=\frac{T^3}{3} \]
Step 4: Compute the r.m.s. value \[ i_{\text{rms}}=i_0\sqrt{\frac{T^3}{3T^3}}=\frac{i_0}{\sqrt3} \]
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