(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
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