Verified Solution • 2025–26 Session moderate

QJEEM22Phde31: JEE Main 2022 Physics - Class 12 JEEM22Phde31 Solution: In a series circuit and power factor of the circuit is . When…

In a series LRL R circuit XL=RX_{L}=R and power factor of the circuit is P1P_{1} . When capacitor with capacitance CC such that XL=XCX_{L}=X_{C} is put in series, the power factor becomes P2P_{2} . The ratio P1P2\frac{P_{1}}{P_{2}} is:

Multiple Choice Options

  • A 12\frac{1}{2}
  • B 12\frac{1}{\sqrt{2}}
  • C 32\frac{\sqrt{3}}{\sqrt{2}}
  • D 2:01

Step-by-Step Verified Solution

P1=cos⁡ϕ=12(XL=R){P_1} = \cos \phi = {1 \over {\sqrt 2 }}({X_L} = R)

P2=cos⁡ϕ′=1{P_2} = \cos \phi ' = 1 (will become resonance circuit)

So, P1P2=12{{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}

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