Question

An alternating current is given by \(i=i_{1}\sin \omega t-i_{2}\cos \omega t\). The rms current is given by

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

Hint:

Combine the sine and cosine into a single sinusoid and divide its amplitude by the square root of 2.

Answer 1

The Correct Option is C

Step 1: Single sinusoid
The sum $i_1\sin\omega t-i_2\cos\omega t$ is one sinusoid with peak value $i_0=\sqrt{i_1^{2}+i_2^{2}}$.

Step 2: RMS rule
For a sinusoid, $i_{rms}=\dfrac{i_0}{\sqrt{2}}$.

Step 3: Result
$i_{rms}=\dfrac{\sqrt{i_1^{2}+i_2^{2}}}{\sqrt{2}}=\sqrt{\dfrac{i_1^{2}+i_2^{2}}{2}}$. Option (D) is the peak value, not rms.

Final Answer:
Dividing the peak $\sqrt{i_1^{2}+i_2^{2}}$ by $\sqrt{2}$ gives the rms current. \[ \boxed{\text{(C) }\sqrt{\dfrac{i_{1}^{2}+i_{2}^{2}}{2}}} \]