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Saturday, 31 October

JEE Main 2022 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 29 July 2022, Shift 1, Physics Q43

Question 43 of 90 in this shift, Physics question 13 of 30, Section A.

An alternating emf E=440sin⁡100πtE = 440\sin 100\pi t is applied to a circuit containing an inductance of 2π\frac{\sqrt{2}}{\pi} H. If an a.c. ammeter is connected in the circuit, its reading will be :
  1. (1)4.4 A
  2. (2)1.55 A
  3. (3)2.2 AOfficial answer
  4. (4)3.11 A

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
  1. 17 Mar 2021, Shift 1 · Q7An AC current is given by I=I1sin⁡ωt+I2cos⁡ωtI=I_1\sin\omega t+I_2\cos\omega t. A hot wire ammeter will give a reading :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.