An AC current is given by $I=I_{1} \sin \omega t+I_{2} \cos \omega t$. A hot wire ammeter will give a reading:

(B) $\sqrt{\frac{I_{1}^{2}+I_{2}^{2}}{2}}$

(A) $\frac{I_{1}+I_{2}}{2 \sqrt{2}}$

(C) $\frac{I_{1}+I_{2}}{\sqrt{2}}$

(D) $\sqrt{\frac{I_{1}^{2}-I_{2}^{2}}{2}}$

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Alternating Current

An AC current is given by

I = I_{1} sin ω t + I_{2} cos ω t

. A hot wire ammeter will give a reading:

Answer: (B)

The two terms are a quarter of a cycle apart, so they combine into a single sinusoid of amplitude

I_{0} = I_{1}^{2} + I_{2}^{2}

A hot-wire instrument responds to the heating effect, which goes as

I^{2}

, so it reads the root mean square value

I_{rms} = \frac{I _{0}}{2} = \frac{I _{1}^{2} + I _{2}^{2}}{2}

The ammeter reads

\frac{I _{1}^{2} + I _{2}^{2}}{2}

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Single correct option