A coil when connected across a 10 V d.c. supply draws a current of 2 A. When it is connected across 10 V - 50 hz a.c. supply the same coil draws a current of 1 A. Explain why? Hence determine self inductance of the coil.

Explanation

In case of DC supply
V= 10 , I =2 A
Resistance of the coil `R= (V)/(I) = (10)/(2) = 5 Omega`
In case of AC supply the coil will also offer reactance.
in this case `epsilon_("eff") = 10 V , I_("eff") = 1 A `
`:. " " Z = (epsilon_("eff"))/ (I_("eff")) = (10)/(1) = 10 Omega`
Inductive reactance is given by
`X_(L) =\sqrt(Z^(2)-R^(2))=sqrt(10^(2)-5^(2))=5sqrt(3)`
Also if L is the self inductance of the coil
`2pi f L = X_(L)= 5sqrt(3)`
`implies L = (5sqrt(3))/(2pif) = (5sqrt(3))/(2xx3.14xx50)= 0.028 H `