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

When the coil is connected to a d.c. source, its
resistance R is given by
`R=V/i=12/4=3Omega`
When it is connected to a.c. source, the impedance
Z of the coil is given by
`Z=V_(rms)/i_(rms)=12/2.4=5Omega`
For a coil, `Z=sqrt([R^(2)+(omegaL)^(2)])`
`therefore " "5=sqrt([(3)^(2)+(50L)^(2)])`
or `" "25=[(3)^(2)+(50L)^(2)]`
Solving we get L = 0.08 henry
When the coil is connected with a condenser in
series, the impedance `Z^(')` is given by
`Z^(')=sqrt([R^(2)+(omegaL-1/(omegaC))^(2)])`
`=[(3)^(2)+(50 times 0.08-1/(50 times 2500 times 10^(-6)))^(2)]^(1//2)=5ohm`
Power developed `P=V_(rms) times i_(rms) times cosphi`
where `cosphi=R//Z^(')=3//5=0.6`
`therefore P=12 times 2.4 times 0.6=17.28` watt