JEE PYQ: Alternating Current Question 3
Question 3 - 2021 (17 Mar Shift 1)
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:
(a) $\sqrt{\frac{I_1^2 - I_2^2}{2}}$ (b) $\sqrt{\frac{I_1^2 + I_2^2}{2}}$ (c) $\frac{I_1 + I_2}{\sqrt{2}}$ (d) $\frac{I_1 + I_2}{2\sqrt{2}}$
Show Answer
Answer: (b)
Solution
$I = I_1 \sin \omega t + I_2 \cos \omega t$. $I_0 = \sqrt{I_1^2 + I_2^2}$. $I_{rms} = \frac{I_0}{\sqrt{2}} = \sqrt{\frac{I_1^2 + I_2^2}{2}}$.
BETA
