An AC source operating at 60 Hz with a maximum voltage of173 V is connected in s

Question

An AC source operating at 60 Hz with a maximum voltage of173 V is connected in series with aresistor (R = 1.5 k) and acapacitor (C = 2.5 µF). (a) What is the maximum value of the current inthe circuit?
A

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

(c) When the current is zero, what are the magnitudes of thepotential difference across the resistor, the capacitor, and the ACsource?
VR = V
VC = V
Vsource = V
How much charge is on the capacitor at this instant?
qc = µC

(d) When the current is at a maximum, what are the magnitudes ofthe potential differences across the resistor, the capacitor, andthe AC source?
VR = V
VC = V
Vsource = V
How much charge is on the capacitor at this instant?
qc = µC
(a) What is the maximum value of the current inthe circuit?
A

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

(c) When the current is zero, what are the magnitudes of thepotential difference across the resistor, the capacitor, and the ACsource?
VR = V
VC = V
Vsource = V
How much charge is on the capacitor at this instant?
qc = µC

(d) When the current is at a maximum, what are the magnitudes ofthe potential differences across the resistor, the capacitor, andthe AC source?
VR = V
VC = V
Vsource = V
How much charge is on the capacitor at this instant?
qc = µC

Explanation / Answer

   we are given with
   f = 60 Hz
   Vmax = 173 V
   R = 1.5 k
       = 1.5 x 103
   C = 2.5 C
       = 2.5 x 10-6 C
   the capacitive reactance is given by
   XC = 1 / 2 f C
         = ........
   the impedence Z is given by
   Z = [R2 + (XL -XC)2]
   as XL = 0
   Z = [R2 +(XC)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
   VCmax = ImaxXC
             = ........ V
(c)
   when the instantaneous current i is zero theinstantaneous voltage across the resistor is
   vR = i R
        = 0
   the instantaneous voltage across the capacitor isalways 90o or a quarter cycle out of phase with
   the instantaneous current
   so we get
   when i = 0
   vC = VCmax
        = ........ V
   charge is
   qC = C vC
        = ......... C
   the kirchoffs rule always applies to the instantaneousvoltage around a closed loop
   for the series circuit
   vsource = vR + vC
   vsource = 0 + VCmax
              = ....... V
(d)
   when the instantaneous current is maximum that is i =Imax the instantaneous voltage across the
   resistor will be
   vR = i R
        = ImaxR
        = VRmax
        = .......... V
   the instantaneous voltage across the capacitor isa quarter cycle out of phase with the current so
   when i = Imax we must get
   vC = 0
   qC = C vC
         = 0
   now applying the kirchoofs loop rule to theinstantaneous voltage around the series circuit at the
   instant when i = Imax gives
   vsource = vR + vC
               = VRmax + 0
               = ......... V

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