A current of `4A` flows in a coil when connected to a `12V DC` source. If the same coil is connected to a `12V, 50 rad//s AC` source, a current of `2.4A` flows in the circuit. Determine the inductance of the coil. Also, find the power developed in the circuit if a`2500muF` capacitor is connected in series with the coil.

Explanation

A coil consists of an inductance (L) and a resistance (R).
In la DC, only resistance is effective, Hence,
`R=V/I = 12/4 = 3Omega`
In AC, `I_(rms) = V_(rms)/Z = V_(rms)/sqrt(R^(2)+omega^(2)L^(2))`
`therefore L^(2) = 1/omega^(2)[(V_(rms)/I_(rms))^(2)-R^(2)]`
`therefore L = 1/omegasqrt((V_(rms)/I_(rms))^(2)-R^(2)))`
Substituting the values, we have
`L=1/50sqrt((12/2.4)^(2)-(3)^(2))=0.08`H
ii) When capacitor is connected to the circuit, the impedance is,
`Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))`
Here, `R=3Omega`
`X_(L) = omegaL=(50)(0.08)=4Omega`
and `X_(C) = 1/(omegaC)= 1/(50)(2500 xx 10^(-6))=8Omega`
`Z=sqrt((3)^(2)+(4-8)^(2))=5Omega`
Now, `P=V_(rms)I_(rms)cosphi`
`=V_(rms)xxV_(rms)/Z xx R/Z = (V_(rms)/Z)^(2)xxR`
Substituting the values, we have `P=(12/3)^(2) xx 3 = 17.28`W