A `9//(100 pi) H` inductor and a `12 Omega` resistance are connected in series to a `225 V`, `50 Hz` ac source. Calculate the current in the circuit and the phase angle between the current and the source voltage.

Explanation

Here, `R = 12 Omega, L = (0.05)/(pi) H, E_(v) = 130 V`, `v = 50 Hz`
`X_(L) = omega L = 2 pi v L = 2 pi xx 50 xx (0.05)/(pi) = 5 Omega`
`Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(12^(2) + 5^(2)) = 13 ohm`
`I_(v) = (E_(v))/(Z) = (130)/(13) = 10 A`
In RL circuit, current lags behind the voltage by phase angle`phi`, where
`tan phi = (X_(L))/(R ) = (5)/(12) = 0.4167`
`phi = tan^(-1) (0.4167) = 22.6^(@)`