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#
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#