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.
....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
As we are considering inductance
(a)
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)
Circuit impedance
Z^2=R^2+X_L^2
whereX_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