PHY 1160C
Homework
Chapter 22: Alternating Current
Ch 21: 7, 11, 21, 31, 37, 46, 53
22.7 An electric light, connected to an ordinary 120 V AC wall
outlet, uses 60 W of power. What are
a) the rms current through the light,
b) the resistance of the light, and
c) the maximum and minimum instantaneous values of
the power and of the current?
P = IV
60 W = (I) (120 V)
I = 60 W = 0.5 A
120 V
I rms = 0.5 A
I =V
R = V = 120 V = 240 W
,
or
R
I
0.5 A
i = I(t) = I max cos 2 f t = I max cos 120
I rms = 0.5 A
t
Irms = 1 Imax
2
I max = (1.414)(0.50 A)
I max = 0.707 A
I min = 0
Likewise,
V = Vrms = 120 V
V max = (1.414)(120 V) = 170 V
V min = 0
P=IV
Ch 22, p 1
P max = I max Vmax = (0.707 A)(170 V)
P max = 84.8 W
P min = 0
22.11 A laboratory power supply provides AC electricity with
a peak voltage of 27.0 volts. What is the rms voltage?
Vrms = 1 Vmax
2
Vrms = 1 (27 V)
2
V rms = 19.1 V
22.21
If 0.8 ampere flows through a capacitor when it is
connected across a 120-volt, 60 Hz AC voltage source, what is
the capacitance of the capacitor?
I = V
XC
XC = V
I
X c = 120 V / 0.8 A = 150 W
XC =
C=
C=
1
2 fC
1
2 f XC
1
2 (60
s –1 )(150
W )
C = 1.77 x 10–5 F
C = 17.7 x 10–6 F = 17.7 µF
Ch 22, p 2
22.31 If 0.20 ampere flows through a 200 mH inductor when
connected across a 30-volt AC voltage source, what is the
frequency of the source?
I = V
XL
XL = V
I
X L = 120 V = 600 W
0.20 A
XL = 2 f L
600 W
=2
f (200 mH)
f = 600 W / [2
f = 478 Hz
(200 x 10 –3 H)]
22.37 [Note the change in the values of the capacitor
and the frequency—and the change in the order of the
calculations or questions asked!]
Consider a series RCL circuit with R = 80 W , C = 15 µF,
L = 250 mH connected to a 12 V, 120 Hz AC source.
a) Calculate the reactances X C and XL.
b) What is the total impedance?
c) What current flows through the circuit?
d)
What are the three potential differences across
the three elements (V R , VC, and VL)?
XC =
1
2 fC
XC =
1
–6
2 (120 s –1 )(15 x 10 F)
X C = 88
XL = 2
fL
XL = 2
(120 s –1)(0.250 H) = 188 W
X L = 188
Ch 22, p 3
2
2
Z=
R + (XL – X C )
Z=
(80 W ) + (188 W
Z=
(80 W ) + (100 W )
2
– 88 W )
2
2
2
Z = 128
I = V = 12 V
Z
128 W
I = 0.094 A
The voltages we calculate across the circuit elements,
like the 12 volts of the power supply itself, are all rms-values.
V R = I R = (0.094 A)(80 W ) = 7.52 V
V C = I XC = (0.094 A)(88 W ) = 8.27 V
V L = I XL = (0.094 A)(188 W ) = 17.67 V
Some explanation or discussion of VC and, especially, VL seems
like it might be useful. The voltages across the capacitor and
the inductor are out of phase with either so the net voltage
across the two of them is
V net = | VL – VC | = | 17.67 V – 8.27 V | = 9.4 V
22.46 [Note the change in the value of the inductance!]
An 80 W resistor in connected in series with a 50 µF capacitor
and a 30 mH inductor across a 120 V, 60 Hz AC voltage source.
What is the impedance of the circuit?
How much current flows in the circuit?
What is the power factor (cos f ) of the circuit?
How much power is used by the circuit?
1
XC =
2 fC
XC =
1
–6
2 (60 s –1 )(50 x 10 F)
X C = 53
Ch 22, p 4
XL = 2
fL
XL = 2
(60 s –1)(0.030 H) = 11.3 W
X L = 11.3
2
2
Z=
R + (XL – X C )
Z=
(80 W ) + (53 W
Z=
(80 W ) + (41.7 W )
2
2
2
– 11.3 W )
2
Z = 90.2
I = V = 120 V
Z
90.2 W
I = 1.33 A
P = I 2 Z cos f = I V cos f
cos f = power factor
cos f = R = 80 W
= 0.887
Z 90.2 W
cos = 0.887
P = (1.33 A)(120 V)(0.887) = 141.5 W
P = 141.5 W
22.53 What is the resonance frequency of a series RCL circuit
with a 150 W resistor, a 25 µF capacitor, and a 80 mH inductor?
What current will flow in the circuit if it is connected to a 20 V
supply at this resonance frequency? What current will flow if
the frequency is increased or decreased by 20%?
At resonance Z = minimum or Z = R, so
XC = X L
1
=2 fL
2 fC
Ch 22, p 5
2
1
f =
2
4
f=
f=
2
LC
1
LC
1
(80 mH)(25 µF)
2
1
f=
–3
2
–6
(80 x 10
H)(25 x 10
F)
f = 112 Hz
At this resonance frequency of 112 Hz, the total
impedance is at its minimum, Zmin = R = 150 W so the current
through the circuit is
I = V = 20 V = 0.133 A
Z
150 W
At resonance, the current is 0.133 A
I res = 0.133 A
Now increase the frequency by 20%
f + = (1.20) fres = (1.20)(112 Hz) = 134 Hz
XC =
1
1
=
–6
2 fC
2 (134 Hz) (25 x 10 F)
X C = 47.5 W
X L = 2 f L = 2 (134 Hz)(80 x 10 –3 H)
X L = 67.4 W
2
2
Z=
R + (XL – X C )
Z=
150 + (67.4 – 47.5)
2
Z = 151.3 W
I = V = 20 V
Z
151.3 W
2
W
= 0.132 A
Ch 22, p 6
Now, decrease the frequency by 20%,
f – = (0.80) fres = (0.80)(112 Hz) = 89.6 Hz
XC =
1
1
=
–6
2 fC
2 (89.6 Hz) (25 x 10 F)
X C = 71.5 W
X L = 2 f L = 2 (89.6 Hz)(80 x 10 –3 H)
X L = 45.0 W
2
2
Z=
R + (XL – X C )
Z=
150 + (45.0 – 71.5)
2
Z = 152.3 W
I = V = 20 V
Z
152.3 W
2
W
= 0.131 A
And we see there is very little change in the current
when the frequency is changed by 20%.
It is interesting and fun to re-do this set of
calculations for R = 15 W
The resonance frequency remains the same, of course.
The values of the reactances, X L and XL, remain the same. Only R
has changed. At resonance, the new resonance current is now
I = V = 20 V = 1.33 A
Z
15 W
I res = 1.33 A
For the higher frequency, f + = 134 Hz, we still have
X C = 47.5 W
and X L = 67.4 W
2
2
Z=
R + (XL – X C )
Z=
15 + (67.4 – 47.5)
2
2
W
Z = 24.9 W
Ch 22, p 7
I = V = 20 V = 0.80 A
Z
24.9 W
Now, decrease the frequency by 20% again, so that
f – = 89.6 Hz, and we still have
X C = 71.5 W
and X L = 45.0 W
2
2
Z=
R + (XL – X C )
Z=
15 + (45.0 – 71.5)
2
2
W
Z = 30.4 W
I = V = 20 V = 0.657 A
Z
30.4 W
Now—with a smaller value of the resistance R—a
variation in the frequency of 20% does, indeed, make a very
large difference in the current.
Ch 22, p 8