Question

An alternating current is given by the equation \[ i=i_1\sin\omega t+i_2\cos\omega t \] The \(i_{\mathrm{rms}}\) will be:

• A. \[ \frac{1}{\sqrt{2}}(i_1+i_2) \]
• B. \[ \frac{1}{\sqrt{2}}(i_1^2+i_2^2) \]
• C. \[ \frac{1}{\sqrt{2}}(i_1+i_2)^2 \]
• D. \[ \frac{1}{\sqrt{2}}\left(i_1^2+i_2^2\right)^{1/2} \]

Hint:

For AC signals, \[ \langle \sin^2\theta\rangle = \langle \cos^2\theta\rangle = \frac12, \qquad \langle \sin\theta\cos\theta\rangle=0 \] These identities are frequently used in RMS-value calculations.

Answer 1

The Correct Option is D

Concept: The rms value of an alternating current is given by \[ i_{\mathrm{rms}} = \sqrt{\left\langle i^2 \right\rangle} \] where \(\langle \cdot \rangle\) denotes time average over one complete cycle.

Step 1: Write the given current equation. \[ i=i_1\sin\omega t+i_2\cos\omega t \] Squaring both sides, \[ i^2 = i_1^2\sin^2\omega t + i_2^2\cos^2\omega t + 2i_1i_2\sin\omega t\cos\omega t \]

Step 2: Take the time average. Using \[ \left\langle \sin^2\omega t \right\rangle = \frac12 \] \[ \left\langle \cos^2\omega t \right\rangle = \frac12 \] \[ \left\langle \sin\omega t\cos\omega t \right\rangle = 0 \] we get \[ \left\langle i^2 \right\rangle = \frac{i_1^2}{2} + \frac{i_2^2}{2} \] \[ \left\langle i^2 \right\rangle = \frac{i_1^2+i_2^2}{2} \]

Step 3: Calculate the rms value. \[ i_{\mathrm{rms}} = \sqrt{\frac{i_1^2+i_2^2}{2}} \] \[ i_{\mathrm{rms}} = \frac{1}{\sqrt{2}} \sqrt{i_1^2+i_2^2} \]

Step 4: State the answer. \[ \boxed{ i_{\mathrm{rms}} = \frac{1}{\sqrt{2}} \left(i_1^2+i_2^2\right)^{1/2} } \]