A coil with inductive resistance `X_(L)=30 Omega` and impedance `Z=50 Omega` is connected to the mains with effective oltage value `V=100V`. Find the phase difference between the current and the voltage, as well as the heat power generated in the coil.

Explanation

`cancel(Z)=sqrt(R^(2)+ X_(L)^(2))` or ` R_(0)=sqrt(cancel(Z^(2))-X_(L)^(2))`
The `tan theta=(X_(L))/( sqrt(cancel(Z^(2))-X_(L)^(2)))`
So ` cos varphi=(sqrt(cancel(Z^(2))-X_(L)^(2)))/(cancel(Z))=sqrt(1-((X_(L))/(cancel(Z)))^(2))`
`varphi=cos ^(-1) sqrt(1- ((X_(L))/( cancel(Z)))^(2))=37^(@)`.
The current lags by `varphi` behind the voltage.
also `P=VI cos varphi=(V^(2))/( cancel(Z^(2)))sqrt(cancel(Z^(2))-X_(L)^(2))=.160kW.`