A `12Omega` resitance and an inductance of 0.05/ `piH` with negligible resistance are connected in series. Across the end of this circuit is connected a 130 V alternating voltage of frequency 50Hz. Calculate the alternating current in the circuit and potential difference across the resistance and that across the inductance.

Explanation

The impedance of the circuit is given by
`Z=sqrt((R^(2)+omega^(2)L^(2)))=sqrt([R^(2)+(2pifL)^(2)])`
`" "=sqrt([(12)^(2)+{2 times 3.14 times 50 times 0.05//3.14}^(2)])`
Current in the circuit `i=E//Z=130/13=10amp`
Potential difference across resistance
`V_(R)=iR=10 times 12=120` volt
Inductive reactance of coil `X_(L)=omegaL=2pifL`
`therefore X_(L)=2pi times 50 times (0.05/pi)=5ohm`
Potential difference across inductance
`" "V_(L)=i times X_(L)=10 times 5=50` volt