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.   

 

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