Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

(a) What is the maximum value of the current inthe circuit? I max = 1 A (b) What

ID: 1745640 • Letter: #

Question

(a) What is the maximum value of the current inthe circuit?
Imax = 1 A

(b) What are the maximum values of the potential difference acrossthe resistor and the inductor?
VR, max = 2 V
VL, max = 3 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 = 4 V
vL = 5 V
vsource = 6 V

(d) When the current is zero, what are the magnitudes of thepotential difference across the resistor, the inductor, and the ACsource?
vR = 7 V
vL = 8 V
vsource = 9 V

Explanation / Answer


   we are given with
   f = 57 Hz
   Vmax = 170 V
   R = 1.2 k
       = 1.2 x 103
   L = 2.2 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