Question

An alternating current is given by \[ I = I_A \sin \omega t + I_B \cos \omega t. \] The r.m.s. current will be:

• A. \( \sqrt{I_A^2 + I_B^2} \)
• B. \( \frac{\sqrt{I_A^2 + I_B^2}}{2} \)
• C. \( \sqrt{\frac{I_A^2 + I_B^2}{2}} \)
• D. \( \frac{|I_A + I_B|}{\sqrt{2}} \)

Hint:

The r.m.s. current is found by squaring the current expression, averaging over one period, and then taking the square root. When the current is a sum of sine and cosine functions, this simplifies to the formula given above.

Answer 1

The Correct Option is C

Given that the alternating current is expressed as: \[ I = I_A \sin \omega t + I_B \cos \omega t, \] where \( I_A \) and \( I_B \) are the amplitudes of the sine and cosine components, respectively. The r.m.s (root mean square) value of an alternating current is given by: \[ I_{{rms}} = \sqrt{\langle I^2 \rangle}, \] where \( \langle I^2 \rangle \) is the time average of \( I^2 \). To find \( \langle I^2 \rangle \), we square the expression for \( I \) and take the time average: \[ I^2 = (I_A \sin \omega t + I_B \cos \omega t)^2 = I_A^2 \sin^2 \omega t + I_B^2 \cos^2 \omega t + 2I_A I_B \sin \omega t \cos \omega t. \] Taking the time average \( \langle \cdot \rangle \) of each term: 
- \( \langle \sin^2 \omega t \rangle = \langle \cos^2 \omega t \rangle = \frac{1}{2} \), 
- \( \langle \sin \omega t \cos \omega t \rangle = 0 \) (because \( \sin \omega t \cos \omega t \) is an odd function with zero mean over one period).
Thus: \[ \langle I^2 \rangle = \frac{1}{2} I_A^2 + \frac{1}{2} I_B^2. \] Therefore, the r.m.s. current is: \[ I_{{rms}} = \sqrt{\langle I^2 \rangle} = \sqrt{\frac{I_A^2 + I_B^2}{2}}. \] Thus, the r.m.s. current is \( \boxed{\sqrt{\frac{I_A^2 + I_B^2}{2}}} \).