An AC source operating at 65 Hz with amaximum voltage of 170 V is connected in s

Question

An AC source operating at 65 Hz with amaximum voltage of 170 V is connected in series with a resistor(R = 1.2 k) and an inductor (L =2.4 H). (a) What is the maximum value of the current inthe circuit?
Imax = A

(b) What are the maximum values of the potential difference acrossthe resistor and the inductor?
VR,max = V
VL,max = V

(c) When the current is at a maximum, what are the magnitudes ofthe potential differences across the resistor, the inductor, andthe AC source?
VR =    V
VL =     V
Vsource =    V

(d) When the current is zero, what are the magnitudes of thepotential difference across the resistor, the inductor, and the ACsource?
VR =   V
VL =    V
Vsource =   V (a) What is the maximum value of the current inthe circuit?
Imax = A

(b) What are the maximum values of the potential difference acrossthe resistor and the inductor?
VR,max = V
VL,max = V

(c) When the current is at a maximum, what are the magnitudes ofthe potential differences across the resistor, the inductor, andthe AC source?
VR =    V
VL =     V
Vsource =    V

(d) When the current is zero, what are the magnitudes of thepotential difference across the resistor, the inductor, and the ACsource?
VR =   V
VL =    V
Vsource =   V

Explanation / Answer

   we are given with
   f = 65 Hz
   Vmax = 170 V
   R = 1.2 k
       = 1.2 x 103
   L = 2.4 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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