An AC source operating at 60 Hz with a maximum voltage of 170 V isconnected in s

Question

An AC source operating at 60 Hz with a maximum voltage of 170 V isconnected in series with a resistor ( R = 1.2 k Ohms.) and aninductor L = 2.8H.
a) What is the maximum value of the current in the circuit?
B) What are the maximum values of the potential difference acrossthe resistor and the inductor?
C) When the current is at a maximum, what are the magnitudes of thepotential differences across the resistor, the inductor, and the ACsource?
D) When the current is zero what are the magnitudes of thepotential difference across the resistor, te inductor and the ACsource?



I am stuck with this problem. I thought I would use thisequation to find the Vmax but it didn't work.

Vmax = Imax * R

I'm given the V max and the R, I don't understand why this doesn'twork. Thanks for any help.

Explanation / Answer

   we are given with
   f = 60 Hz
   Vmax = 170 V
   R = 1.2 k
       = 1.2 x 103
   L = 2.8 H
   the inductive reactance is given by
   XL = 2 f L
         = ........
   the impedence Z is given by
   Z = [R2 + (XL -XC)2]
   as XC = 0
   Z = [R2 +(XL)2]
       = .........
(a)
   the maximum current will be
   Imax = Vmax / Z
           =........ A
(b)
   the maximum values of the potential difference acrossthe resistor will be
   VRmax = Imax R
             = ........ V
   the maximum values of the potential difference acrossthe capacitor will be
   VLmax = ImaxXL
             = ........ V
(c)
   when the instantaneous current i is zero theinstantaneous voltage across the resistor is
   vR = i R
        = 0
   the instantaneous voltage across the inductor isalways 90o or a quarter cycle out of phase with
   the instantaneous current
   so we get
   when i = Imax
   vL = 0
   the kirchoffs rule always applies to the instantaneousvoltage around a closed loop
   for the series circuit
   vsource = vR + vL
   when i = Imax
   vsource = Imax R + 0
              = ....... V
(d)
   when the instantaneous current is zero
   the instantaneous voltage across the resistor willbe
   vR = 0
   the instantaneous voltage across the inductor isa quarter cycle out of phase with the current so
   when i = 0 we must get
   vL = VL,max
        = ......... V
   now applying the kirchoofs loop rule to theinstantaneous voltage around the series circuit at the
   instant when i = Imax gives
   vsource = vR + vL
               = 0 + VLmax
               = ......... V

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