Verified Solution • 2025–26 Session moderate

QJEEM23Ph9fcf: JEE Main 2023 Physics - Class 12 JEEM23Ph9fcf Solution: In a series LR circuit with , power factor P 1 . If a capacitor …

In a series LR circuit with XL=R\mathrm{X_L=R} , power factor P1. If a capacitor of capacitance C with XC=XL\mathrm{X_C=X_L} is added to the circuit the power factor becomes P2. The ratio of P1 to P2 will be :

Multiple Choice Options

  • A 1 : 2\sqrt2
  • B 1:03
  • C 1:02
  • D 1:01

Step-by-Step Verified Solution

XL=R{X_L} = R

⇒P1=RXL2+R2=12\Rightarrow {P_1} = {R \over {\sqrt {X_L^2 + {R^2}} }} = {1 \over {\sqrt 2 }}

Now, XL=XC=R{X_L} = {X_C} = R

⇒P2=RR2+(XL−XC)2=1\Rightarrow {P_2} = {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }} = 1

⇒P1P2=12\Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}

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