A long solenoid connected to a 12V DC source passes a steady current of 2 A. When the solenoid is connected to an AC source of 12V at 50 Hz, the current flowing is 1A. Calculate inductance of the solenoid.

Explanation

Remainder that the solenoid mentioned here is not just a solenoid but has the resistance also due to its wire material. Now, in a DC, the solenoid does'nt offer any reactance after a few milliseconds, however in the beginning it offers any reactance after a few milliseconds, however in the beginning it offers infinte resistance. So, the curreny by a DC into an inductor is, `I=V/R`
or `R=12/2=6Omega`
But when an AC is contained tothe same inductor, the reactance will also come into picture.
So, `I=V/(sqrt(R^(2)+omega^(2)L^(2)))`
Here,` I=1A, V=12V, R=6 Omega`
`omega=2pi50=100pirad^(-1)0,L=?`
or `10=12/(sqrt((6)^(2)+(100piL)^(2))rArr = L =33mH`