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Alternating Current question

2021 · 18 Mar · Shift 2 · Q47
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  5. /2021 · 18 Mar · Shift 2 · Q47

Alternating Current question

2021 · 18 Mar · Shift 2 · Q47

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
In a series LCR circuit, the inductive reactance (XL) is 10 Ω\OmegaΩ and the capacitive reactance (XC) is 4 Ω\OmegaΩ. The resistance (R) in the circuit is 6 Ω\OmegaΩ. The power factor of the circuit is :
  1. A
    12{1 \over 2}21​
  2. B
    32{{\sqrt 3 } \over 2}23​​
  3. C
    12{1 \over {\sqrt 2 }}2​1​
  4. D
    122{1 \over {2\sqrt 2 }}22​1​
View written solutionFree

Correct answer: C

  1. Given data
  • Inductive reactance: XL=10 ΩX_L = 10\,\OmegaXL​=10Ω
  • Capacitive reactance: XC=4 ΩX_C = 4\,\OmegaXC​=4Ω
  • Resistance: R=6 ΩR = 6\,\OmegaR=6Ω
  1. Net reactance

For a series LCR circuit,

X=XL−XC=10−4=6 ΩX = X_L - X_C = 10 - 4 = 6\,\OmegaX=XL​−XC​=10−4=6Ω

  1. Impedance of the circuit

The impedance is

Z=R2+X2Z = \sqrt{R^2 + X^2}Z=R2+X2​

Substitute the values:

Z=62+62=36+36=72=62 ΩZ = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\,\OmegaZ=62+62​=36+36​=72​=62​Ω

  1. Power factor

Power factor in an AC circuit is

cos⁡ϕ=RZ\cos\phi = \frac{R}{Z}cosϕ=ZR​

So,

cos⁡ϕ=662=12\cos\phi = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}cosϕ=62​6​=2​1​

  1. Match with options

12\frac{1}{\sqrt{2}}2​1​ corresponds to Option C.

Final Answer: 12\boxed{\frac{1}{\sqrt{2}}}2​1​​

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