An alternating current is given by the equation i = i1 sin \omega t + i2 cos \omega t. The rms current will be :

1 \over \sqrt 2 ( i_1^2 + i_2^2 )^1 \over 2

1 \over 2( i_1^2 + i_2^2 )^1 \over 2

Alternating Current PYQ — JEE Main 2021 · 26 Feb · Shift 1 · Q45 with Solution

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Correct answer: A

$i(t)=i_1\sin \omega t+i_2\cos \omega t$

We need the rms value, defined as

$i_{\text{rms}}=\sqrt{\langle i^2(t)\rangle}$

where $\langle \cdot \rangle$ denotes average over one complete cycle.

$i^2=(i_1\sin \omega t+i_2\cos \omega t)^2$

$i^2=i_1^2\sin^2\omega t+i_2^2\cos^2\omega t+2i_1i_2\sin\omega t\cos\omega t$

\qquad \langle \cos^2\omega t\rangle=\frac12, \qquad \langle \sin\omega t\cos\omega t\rangle=0$$ So, $$\langle i^2\rangle=i_1^2\cdot \frac12+i_2^2\cdot \frac12+2i_1i_2\cdot 0$$ $$\langle i^2\rangle=\frac12(i_1^2+i_2^2)$$ 4. Therefore, $$i_{\text{rms}}=\sqrt{\frac12(i_1^2+i_2^2)}$$ $$i_{\text{rms}}=\frac{1}{\sqrt2}\sqrt{i_1^2+i_2^2}$$ 5. Compare with options: - **A:** $$\frac{1}{\sqrt2}\sqrt{i_1^2+i_2^2}$$ ✅ - **B:** $$\frac{1}{\sqrt2}(i_1+i_2)$$ ❌ - **C:** $$\frac{1}{\sqrt2}(i_1+i_2)^2$$ ❌ - **D:** $$\frac12\sqrt{i_1^2+i_2^2}$$ ❌ Hence the correct option is **A**.