Question

The r.m.s. value of a current given by \( i = (i_1 \cos \omega t + i_2 \sin \omega t) \) is:

• A. \( \frac{1}{\sqrt{2}} (i_1 + i_2) \)
• B. \( \frac{1}{\sqrt{2}} (i_1 - i_2) \)
• C. \( \frac{1}{\sqrt{2}} \sqrt{i_1^2 + i_2^2} \)
• D. \( \frac{1}{\sqrt{2}} (i_1^2 + i_2^2) \)

Hint:

To calculate the r.m.s. value for a combined sinusoidal current, square each term, take the mean, and then the square root.

Answer 1

The Correct Option is C

Finding the R.M.S. Value of the Current

1: Formula for r.m.s. Value The r.m.s. (root mean square) value of a current \( i(t) \) is given by: \[ i_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T i^2(t) \, dt} \] where \( T \) is the period of the oscillation.
2: Squaring the Current Expression Given that: \[ i = i_1 \cos \omega t + i_2 \sin \omega t \] The square of the current is: \[ i^2 = (i_1 \cos \omega t + i_2 \sin \omega t)^2 = i_1^2 \cos^2 \omega t + i_2^2 \sin^2 \omega t + 2 i_1 i_2 \cos \omega t \sin \omega t \]
3: Averaging Over One Period The average values of \( \cos^2 \omega t \) and \( \sin^2 \omega t \) over a period \( T \) are \( \frac{1}{2} \), and the average of \( \cos \omega t \sin \omega t \) is zero. Thus: \[ \langle i^2 \rangle = \frac{1}{2} i_1^2 + \frac{1}{2} i_2^2 \]
4: Calculating the r.m.s. Value Now, the r.m.s. value of the current is: \[ i_{\text{rms}} = \sqrt{\frac{1}{2} i_1^2 + \frac{1}{2} i_2^2} = \frac{1}{\sqrt{2}} \sqrt{i_1^2 + i_2^2} \] Thus, the correct answer is: \[ \boxed{(C) \, \frac{1}{\sqrt{2}} \sqrt{i_1^2 + i_2^2}} \]