In an electrical circuit R, L, C and an a.c. voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is pi/3 . If instead, C is removed from the circuit, the phase difference is again pi/3 . The power factor of the circuit is:

Answers (1)

As we learnt in

Impedence -

Z= sqrt{R^{2}+left ( omega L -frac{1}{omega c}
ight )^{2}}

-

 

 

Power factor -

cos phi = frac{R}{Z}
 

- wherein

R
ightarrow resistance

Z
ightarrow impedence

 

 

 

     

\frac{Xc}{R}= \tan \frac{\pi }{3}\Rightarrow X_{c}=R \tan \frac{\pi }{3}

 when C is removed from the Circuit

\frac{X_{L}}{R}= \tan \frac{\pi }{3}\Rightarrow X_{L}=R \tan \frac{\pi }{3}

Z= \sqrt {R^{2}+\left ( X_{L}-X_{C} \right )^{2}} =Z=R

Power factor \cos \phi = \frac{R}{z}=\frac{z}{z}=1

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Test Series JEE Main 2024

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout JEE Main (One Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout JEE Main (Twelve Months Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 19999/- ₹ 14499/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions