A coil of inductance 0.50H and resistance 100 `Omega` is connected to a 240V. 50Hz ac supply.
(a) What is the maximum current in the coil? (b) What is the time lag between the voltage maximum and the current maximum?

Explanation

Here `L = 0.50 H, R = 100` ohm,
`E_(rms) = 240 V, v = 50 Hz`
Step I :
We know
`E_(0) = sqrt(2) E_(rms) = 1.414 xx 240 = 339.4 V`
Impedance of LR circuit is
`Z_(L) = sqrt(R^(2) + omega^(2)L^(2)) = sqrt(R^(2) + 4pi^(2)v^(2)L^(2))`
`= sqrt((100)^(2) + 4 xx (3.14)^(2) xx (50)^(2) xx (0.50)^(2)) = 186.1` ohm
Peak current, `I_(0) = (E_(0))/(Z_(L)) = (339.4)/(186.1) = 1.82 A`
In LR circuit the phase difference between current and voltage `phi` is given by
`tan phi = (0.50 xx 2 xx 3.14 xx 50)/(100) = 1.57`
`phi = tan^(-1)(1.57) = 57^(@)30. = 57.5^(@)`
`= (57.5 xx pi)/(180) = 0.3194 pi` radians
Here `E = E_(0) sin omega t , E_(0) = E_(rms) sqrt(2)`
`therefore` Time lag between `E_(0)` and `I_(0)` is given by
`t = (phi)/(omega) = (phi)/(2pi v) = (0.3194 pi)/(2pi xx 50)`
`= 0.003194 s = 3.194 xx 10^(-3)s`.