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Saturday, 31 October

JEE Main 2025 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 3 April 2025, Shift 2, Physics Q37

Question 37 of 75 in this shift, Physics question 12 of 25, Section A.

An electric bulb rated as 100W-220V is connected to an ac source of rms voltage 220V. The peak value of current through the bulb is :
  1. (1)0.45 A
  2. (2)0.32 A
  3. (3)0.64 AOfficial answer
  4. (4)2.2 A

Official answer

Option 3

NTA final key.

Same topic in other shifts

All RMS and Peak Values questions
  1. 7 Apr 2025, Shift 1 · Q35An ac current is represented as i=52+10cos⁡(650πt+π6)i = 5\sqrt{2} + 10\cos\left(650\pi t + \frac{\pi}{6}\right) Amp The r.m.s value of the current isEasySingle correct
  2. 7 Apr 2025, Shift 2 · Q48An inductor of reactance 100 Ω100\ \Omega, a capacitor of reactance 50 Ω50\ \Omega, and a resistor of resistance 50 Ω50\ \Omega are connected in…EasyNumerical value
  3. 4 Apr 2024, Shift 1 · Q58A alternating current at any instant is given by i=[6+56sin⁡(100πt+π/3)]i = \left[6 + \sqrt{56}\sin(100\pi t + \pi/3)\right] A. The rms value of the current is…MediumNumerical value
  4. 30 Jan 2024, Shift 2 · Q43An alternating voltage V(t)=220sin⁡100πtV(t) = 220\sin 100\pi t volt is applied to a purely resistive load of 50 Ω50\ \Omega. The time taken for the…MediumSingle correct
  5. 27 Jul 2022, Shift 1 · Q44A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of 3 Ω\Omega and 2 Ω\Omega respectively.…EasySingle correct
  6. 29 Jul 2022, Shift 1 · Q43An 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…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.