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Alternating Current

JEE Main 2025 (Online) 3rd April Evening ShiftMCQEasy📈 Topic in 13% papers+4−1
Ques. 1JEE Main 2025, 3 April (Evening Shift)

An electric bulb rated as 100 W - 220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is:

  1. 0.32 A
  2. 0.64 A
  3. 0.45 A
  4. 2.2 A

Answer

(2) 0.64 A

Solution

At rated voltage the bulb consumes its rated power: P = Vrms Irms

Irms = 100220 = 0.4545 A

I0 = 2 Irms = 1.414 0.4545

I0 0.64 A

Answer: (2)

Keep practising — Alternating Current

Ques. 2JEE Main 2025, 29 January (Morning Shift)

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). In the light of the above statements, choose the most appropriate answer from the options given below :

  1. (A) is true but (R) is false
  2. Both (A) and (R) are true and (R) is the correct explanation of (A).
  3. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  4. (A) is false but (R) is true
Ques. 3JEE Main 2025, 24 January (Morning Shift)

An alternating current is given by I=IA t+IB t. The r.m.s current will be

  1. IA2+IB22
  2. IA2+IB22
  3. |IA+IB| 2
  4. IA2+IB2
Ques. 4JEE Main 2025, 23 January (Evening Shift)

In a series LCR circuit, a resistor of 300 , a capacitor of 25 nF and an inductor of 100 mH are used. For maximum current in the circuit, the angular frequency of the ac source is 104 radians s-1

Ques. 5JEE Main 2025, 22 January (Evening Shift)

A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is I0. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

  1. I02
  2. I0 2
  3. 2I0
  4. I0

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