A coil possessing both inductance L and resistance R is connected to a 24V dc supply having negligible internal resistance. The dc current in this circuit is found to be 3A. When the coil is connected to a 24V, 50Hz ac supply with negligible internal....?

....impedance, the circuit current is found to be 0.8 A. Find:

(a) the resistance of the coil.
(b) the inductance of the coil.

1 Answer
Jun 2, 2018

As we are considering inductance LL and resistance RR of the same coil, it becomes a series LRLR circuit.

(a) 24\ V dc supply connected.
Inductance does not play any role. The current in the coil is given by

I_(DC)=V_(DC)/R

Inserting given values we get

3=24/R
=>R=24/3=8\ Omega

(b) 24\ V, 50\ Hz ac supply connected.
Circuit impedance Z is given by the expression

Z^2=R^2+X_L^2
where X_L=omegaL, is inductive reactance of the coil. Now the current in the coil becomes
I_(AC)=V_(AC)/Z

Inserting various values we get

0.8=24/sqrt(8^2+(2pixx50xxL)^2)

Rearranging and squaring both sides

8^2+(2pixx50xxL)^2=(24/0.8)^2
=>(2pixx50xxL)^2=(24/0.8)^2-8^2
=>L=sqrt(836)/(100pi)
=>L=0.9\ H