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JEE Main 202229 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
29 June
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2022 · 29 June · Morning Shift · Physics · Alternating Current

For a series LCR circuit, I vs ω curve is shown (a) To the left of...

For a series LCRLCR circuit, II vs ω\omega curve is shown

For a series LCR circuit, I vs ω curve is shown (a) To the left of ω _ r , the circuit is mainly capacitive. (b) To the

(a) To the left of ωr{\omega}_{r}, the circuit is mainly capacitive.
(b) To the left of ωr{\omega}_{r}, the circuit is mainly inductive.
(c) At ωr{\omega}_{r}, impedance of the circuit is equal to the resistance of the circuit.
(d) At ωr{\omega}_{r}, impedance of the circuit is 00.

Choose the most appropriate answer from the options given below.

Options

  1. A(a) and (d) only
  2. B(b) and (d) only
  3. C(a) and (c) only
  4. D(b) and (c) only

Correct answer

C. (a) and (c) only

Step-by-step solution

The impedance of an AC circuit is given by, Z=R2+(XC−XL)2Z = \sqrt{{R}^{2} + {\left({X}_{C} - {X}_{L}\right)}^{2}}.

For current to be maximum ZZ should be minimum. The value of ZZ will be minimum when XC−XL=0{X}_{C} - {X}_{L} = 0. Therefore, the value of Z=RZ = R. Hence, statement C is correct and statement D is wrong.

To the left of ωr{\omega}_{r} i.e. when ω<ωr\omega < {\omega}_{r}, XC>XL{X}_{C} > {X}_{L}. Therefore, the circuit is mainly capacitive. Hence, statement A is correct and statement B is wrong.

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