A coil having inductance of 5 H and resistance of 30 `Omeg` has a peak voltage as `9 sqrt(3)` volts 50 Hz connected across it . Find the current through the coil and the absorbed power.

Explanation

L = 5 H, R = 30 , `E_(0) = 9 sqrt(3) `V, f = 50 Hz
`X_(L) = 2 pi f L = 2 xx 3.14 xx 50 xx 5 = 1570 `H
Z = `sqrt(R^(2) + (X_(L))^(2))`
= `sqrt( (30)^(2) + (1570)^(2) ) = 1570.29 Omega`
`E_(v) = (E_(0))/(sqrt(2)) = (9 sqrt(3))/(2) = 7.7 ` V
`I_(v) = (E_(v))/(Z) = (7.7)/(1570) = 0.004 `A
power absorbed = `E_(V) I_(V) cos phi`
`= E_(V) I_(V) ((R)/(Z))`
= `7.7 xx 0.004 xx (30)/(1570) = 0.0005 ` W