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JEE Main 202118 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Evening Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2021 · 18 March · Evening Shift · Physics · Alternating Current

In a series LCR circuit, the inductive reactance ( X _ L ) is 10Ω...

In a series LCR circuit, the inductive reactance (XL)\left({X}_{L}\right) is 10Ω10Ω and the capacitive reactance (XC)\left({X}_{C}\right) is 4Ω4Ω. The resistance (R)( R ) in the circuit is 6Ω6Ω. The power factor of the circuit is :In a series LCR circuit, the inductive reactance ( X _ L ) is 10Ω and the capacitive reactance ( X _ C ) is 4Ω . TheIn a series LCR circuit, the inductive reactance ( X _ L ) is 10Ω and the capacitive reactance ( X _ C ) is 4Ω . The

Options

  1. A12\dfrac{1}{2}
  2. B122\dfrac{1}{2\sqrt{2}}
  3. C12\dfrac{1}{\sqrt{2}}
  4. D32\dfrac{\sqrt{3}}{2}

Correct answer

C. 12\dfrac{1}{\sqrt{2}}

Step-by-step solution

In a series LCR circuit, the inductive reactance ( X _ L ) is 10Ω and the capacitive reactance ( X _ C ) is 4Ω . The

We know that power factor is cosϕcosϕ,

cosϕ=RZ...(1)cosϕ = \dfrac{R}{Z} . . . \left(1\right)

Z=R2+(XL−XC)2...(2)Z = \sqrt{{R}^{2} + {\left({X}_{L} - {X}_{C}\right)}^{2}} . . . \left(2\right)

In a series LCR circuit, the inductive reactance ( X _ L ) is 10Ω and the capacitive reactance ( X _ C ) is 4Ω . The

⇒Z=62+(10−4)2⇒ Z = \sqrt{{6}^{2} + ( 10 - 4{)}^{2}}

⇒Z=62∣cosϕ=662⇒ Z = 6\sqrt{2} ∣ cosϕ = \dfrac{6}{6\sqrt{2}}

cosϕ=12cosϕ = \dfrac{1}{\sqrt{2}}

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