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 the alternating current expression:

\[ I = I_A \sin \omega t + I_B \cos \omega t, \] where \( I_A \) and \( I_B \) are constants, and \( \omega \) is the angular frequency of the current.

Step 1: Understanding the RMS Current

The root mean square (RMS) value of an alternating current is given by: \[ I_{\text{rms}} = \sqrt{\langle I^2 \rangle}, \] where \( \langle I^2 \rangle \) is the time average of the square of the current.

Step 2: Calculate \( \langle I^2 \rangle \)

The square of the current \( I \) is: \[ I^2 = \left( I_A \sin \omega t + I_B \cos \omega t \right)^2. \] Expanding the square: \[ I^2 = I_A^2 \sin^2 \omega t + I_B^2 \cos^2 \omega t + 2 I_A I_B \sin \omega t \cos \omega t. \] Taking the time average, we use the following identities: - \( \langle \sin^2 \omega t \rangle = \frac{1}{2} \), - \( \langle \cos^2 \omega t \rangle = \frac{1}{2} \), - \( \langle \sin \omega t \cos \omega t \rangle = 0 \). So, the time average of \( I^2 \) is: \[ \langle I^2 \rangle = I_A^2 \cdot \frac{1}{2} + I_B^2 \cdot \frac{1}{2} = \frac{I_A^2 + I_B^2}{2}. \]

Step 3: Calculate the RMS Current

The RMS current is the square root of the time average: \[ I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}. \]

Final Answer:

The RMS current is \( \boxed{\sqrt{\frac{I_A^2 + I_B^2}{2}}} \).