A series combination of a coil of inductance L and a resistor of resistance `12 Omega` is connected across a 12 V, 50 Hz supply. Calculate L if the circuit current is 0.5 A.

Explanation

Impedance, `Z = (E)/(1) = (12)/(0.5) = 24 Omega` and `Z = sqrt(R^(2) + omega^(2)L^(2))`
or `Z^(2) = R^(2) + omega^(2)L^(2)` or `L^(2) = (Z^(2) - R^(2))/(omega^(2))`
Here `omega = 2pi v = 2 xx(22)/(7) xx 50`
`= 31.4 "rad s"^(-1)` and `R = 12 Omega`
`L^(2) = ((24^(2))-(12)^(2))/((314)^(2))` or `L = (12sqrt(3))/(314) = 0.066H`