Question

A capacitor of capacitance `100mu F`,a coil of resistence `120Omega` and inductance `100"mH"` are connected to an ac source of emf `e=30sin(100t)" V"`.Calculate the peak value of the current............?

Answer 1

`|I| = 0.2" A"`

Explanation:

The voltage in exponential form:

`V = |V|e^(i(omegat))`

Substitute `|V| = 30" V"` and `omega=100" rad/s"`

`V = (30" V")e^(i((100" rad/s")t))`

The formula for impedance in exponential form:

`Z = |Z|e^(itheta)`

`Z = sqrt((R)^2+ (omegaL-1/(omegaC))^2)e^(itan^-1((omegaL-1/(omegaC))/R))`

Substitute `R = 120Omega`, `omega = 100" rad/s"`, `L = 100xx10^-3" H"`, and `C = 100xx10^-6" F"`

`|Z | =sqrt((R)^2+ (omegaL-1/(omegaC))^2)`

#|Z| = sqrt((120)^2+(100(100xx10^-3)- 1/(100(100xx10^-6)))^2)#

`|Z| = 150Omega`

Because we want the peak value of the current, the phase angle does not really matter but, nevertheless, we shall compute it:

`theta = tan^-1((omegaL-1/(omegaC))/R)`

Substitute `R = 120Omega`, `omega = 100" rad/s"`, `L = 100xx10^-3" H"`, and `C = 100xx10^-6" F"`

`theta = tan^-1((100(100xx10^-3)-1/(100(100xx10^-6)))/120)`

`theta = -36.87^@`

The peak value of the current is the magnitude:

`|I| = |V|/|Z|`

`|I| = (30" V")/(150" "Omega)`

`|I| = 0.2" A"`