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:

• A. \( i_0 \)
• B. \( \dfrac{i_0}{\sqrt6} \)
• C. \( \dfrac{i_0}{\sqrt2} \)
• D. \( \dfrac{i_0}{\sqrt3} \)

Hint:

For linearly varying currents, always square the function first before integrating for r.m.s. values.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The Root Mean Square (r.m.s) value of a time-varying current \(i(t)\) over a time period \(T\) is defined as the square root of the average of the square of the current over that period.
Step 2: Key Formula or Approach:
The formula for r.m.s current is:
\[ I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} i^2 \, dt} \]
Step 3: Detailed Explanation:
Given current function: \(i = i_0 \frac{t}{T}\)
First, we find the average of the square of the current:
\[ i^2 = \left( i_0 \frac{t}{T} \right)^2 = \frac{i_0^2}{T^2} t^2 \]
Integrating over the period \(t = 0\) to \(T\):
\[ \int_{0}^{T} i^2 \, dt = \int_{0}^{T} \frac{i_0^2}{T^2} t^2 \, dt = \frac{i_0^2}{T^2} \left[ \frac{t^3}{3} \right]_{0}^{T} \]
\[ = \frac{i_0^2}{T^2} \left( \frac{T^3}{3} - 0 \right) = \frac{i_0^2 T}{3} \]
Now, divide by the time period \(T\):
\[ \text{Mean of } i^2 = \frac{1}{T} \left( \frac{i_0^2 T}{3} \right) = \frac{i_0^2}{3} \]
Taking the square root to find \(I_{rms}\):
\[ I_{rms} = \sqrt{\frac{i_0^2}{3}} = \frac{i_0}{\sqrt{3}} \]
Step 4: Final Answer:
The r.m.s current for the period \(t = 0\) to \(t = T\) is \(\frac{i_0}{\sqrt{3}}\).