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JEE Main 2021 · PhysicsMultiple choiceSingle correctEasyConceptual

JEE Main 17 March 2021, Shift 1, Physics Q7

Question 7 of 90 in this shift, Physics question 7 of 30, Section A.

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 :

(1) $\frac{I_1+I_2}{\sqrt2}$
(2) $\frac{I_1+I_2}{2\sqrt2}$
(3) $\sqrt{\frac{I_1^2+I_2^2}{2}}$ Official answer
(4) $\sqrt{\frac{I_1^2-I_2^2}{2}}$

Official answer

Option 3

NTA final key.

Chapter: Alternating Current
Topic: RMS and Peak Values
Idea tested: AC Measuring Instruments

Same idea in other shifts

Asked 2× in all

29 Jul 2022, Shift 1 · Q43An alternating emf $E = 440\sin 100\pi t$ is applied to a circuit containing an inductance of $\frac{\sqrt{2}}{\pi}$ H. If an a.c. ammeter…EasySingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.