Simple - Inductance Question

Discussion in 'Homework Help' started by ForgottenMemorie, Oct 6, 2008.

  1. ForgottenMemorie

    Thread Starter New Member

    Mar 26, 2008
    1. The problem statement, all variables and given/known data

    An AC generator produces a voltage of 230<45º volts. It is connected across an impedance of 17.4<-33º ohms.

    The supply frequency = 50Hz

    I got my assignment back and this question wrong, maybe you can guide me, please because I’m not sure entirety. Help is appreciated.

    Question: Calculate the value of inductance that will make the circuit become series resonant at 50Hz?

    2. Relevant equations

    1/2π fc = 2 π fl

    3. The attempt at a solution

    I= 17.4 sin33 = 9.47
    C= 1/2 π (50x9.47) = 3.36x10^-6
    Would it be: 2 π fl > 2 π(50) = 314 ?
  2. mik3

    Senior Member

    Feb 4, 2008
    From the angle of the impedance we can say it is composed of a capacitor and a resistor.

    The value of the resistor=17.4*cos(33)=14.59 ohms

    and the reactance (XC) of the capacitor=17.4*sin(33)=9.48 ohms

    To have a series resonance the reactance of the inductor (XL) must equal the reactance of the capacitor (XC), thus:


    L=XL/(2π*f)=9.48/(2π*50)=30.2 mH