For an R-L series circuit, the power factor is 3 2 , for R-L frequency f Hz. If the frequency doubles, the new power factor will be

Correct answer: D. (3)/(7) The initial power factor is given by φ = R R^2 + X_L^2 = 3 2 Squaring both sides, we get: (R^2)/(R^2 + X_L^2) = (3)/(4) 4R^2 = 3R^2 + 3X_L^2 R^2 = 3X_L^2 X_L = R 3 The inductive reactance is X_L = 2π f L . When the frequency is doubled ( f' = 2f ), the new inductive reactance becomes: X_L' = 2X_L = 2R 3 The new power factor is: φ' = R R^2 + (X_L')^2 φ' = R R^2 + ( 2R 3 )^2 φ' = R R^2 + (4R^2)/(3) φ' = R (7R^2)/(3) = (3)/(7) Answer: (3)/(7)

MHT CET 202625 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 25 April
Shift: Evening Shift
Subject: Physics
Chapter: Alternating Current

MHT CET 2026 · 25 April · Evening Shift · Physics · Alternating Current

For an R-L series circuit, the power factor is 3 2 , for R-L...

For an R-L series circuit, the power factor is

\dfrac{\sqrt{3}}{2}

, for R-L frequency

f

Hz. If the frequency doubles, the new power factor will be

Options

A $\dfrac{3}{\sqrt{7}}$
B $\dfrac{\sqrt{3}}{7}$
C $\dfrac{3}{7}$
D $\sqrt{\dfrac{3}{7}}$

Correct answer

D. $\sqrt{\dfrac{3}{7}}$

Step-by-step solution

The initial power factor is given by

\cos \phi = \dfrac{R}{\sqrt{R^2 + X_L^2}} = \dfrac{\sqrt{3}}{2}

Squaring both sides, we get:

\dfrac{R^2}{R^2 + X_L^2} = \dfrac{3}{4}

4R^2 = 3R^2 + 3X_L^2

R^2 = 3X_L^2 \Rightarrow X_L = \dfrac{R}{\sqrt{3}}

The inductive reactance is

X_L = 2\pi f L

. When the frequency is doubled (

f' = 2f

), the new inductive reactance becomes:

X_L' = 2X_L = \dfrac{2R}{\sqrt{3}}

The new power factor is:

\cos \phi' = \dfrac{R}{\sqrt{R^2 + (X_L')^2}}

\cos \phi' = \dfrac{R}{\sqrt{R^2 + \left(\dfrac{2R}{\sqrt{3}}\right)^2}}

\cos \phi' = \dfrac{R}{\sqrt{R^2 + \dfrac{4R^2}{3}}}

\cos \phi' = \dfrac{R}{\sqrt{\dfrac{7R^2}{3}}} = \sqrt{\dfrac{3}{7}}

Answer:

\sqrt{\dfrac{3}{7}}

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