Get Answers to all your Questions

header-bg qa

An ideal inductor of \frac{5}{\pi} H inductance is connected to a 200 V, 50 Hz ac supply.

(a) Calculate the rms and peak value of current in the inductor.

(b) What is the phase difference between current through the inductor and the applied voltage? How will it change if a small resistance is connected in series with this inductor in the circuit?

 

 

 
 
 
 
 

Answers (1)

a)

I_{rms}=\frac{V_{rms}}{2\pi fL}=\frac{200}{2\pi \times50\times\frac{5}{\pi}}=\frac{2}{5}A=0.4A

I_p=\sqrt{2}I_{rms}=\sqrt{2}\times0.4=0.56A

b) Phase difference=90 degree

If a small resistance is connected in series with this inductor the phase difference decreases.   

 

Posted by

Safeer PP

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads