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Alternating Current · Question 15

JEE Main 2025 (Online) 4th April Evening ShiftNumericalEasy📈 Topic in 13% papers+4
1

An inductor of self inductance 1 H is connected in series with a resistor of 100 ohm and an ac supply of 100 volt, 50 Hz. Maximum current flowing in the circuit is A.

Answer

1

Solution

Inductive reactance:

XL = 2 f L = 2 50 1 = 100

Impedance:

Z = R2 + XL2 = (100 )2 + (100 )2 = 100 2

The supply voltage is the rms value, Vrms = 100 V:

Irms = VrmsZ = 100 100 2 = 1 2 A

Imax = 2 Irms = 1 A

Answer: 1

JEE Main 2025 (Online) 4th April Evening Shift

Keep practising — Alternating Current

2

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
JEE Main 2025 (Online) 3rd April Evening Shift
3

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
JEE Main 2025 (Online) 29th January Morning Shift

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