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?

Answer

`1.82A; 3.2 xx10^(-3)s`

Explanation

Impedance of the coil:
`Z=sqrt(R^(2)+(2 pi f l)^(2)) =sqrt(100^(2)+4 pi^(2) (50)^(2)(0.5)^(2))=186.2 Omega`
Maximum current: `I_(0) =(E_0)/(Z) =(240 sqrt(2))/(186.2) = 1.82A`
`tan phi - (X_(C )-X_(L))/(R ) =(0-2 pi (50)(0.5))/(100) = -(pi)/(2)`
`implies phi = - tan^(-1) ((pi)/(2)) =-27.51^(@) =-1.003 rad `
Let emf be maximum at `(t_1)`, then `(omega t_(1))=(pi)/(2)`
[ we have `E=E_(0) sin omega t and I=(I_0) sin (omega t + phi]`
Let current be maximum at `(t_2)`, then` (omega t_(2))+phi =(pi)/(2)`
form above we have `omega(t_(2)-t_(1)) =-phi`
Time lag between voltage maximum and current maximum :
`(t_2)-(t_1) =-(phi)/(omega) = (1.003)/(2 pi f ) =3.2 xx10^(-3)s`.