Chapter 28 Alternating Current Circuits
“History teaches us that the searching spirit of man
required thousands of years for the discovery of the
fundamental principles of the sciences, on which the
superstructure was then raised in a comparatively
short time. But these very fundamental propositions
are nevertheless so clear and simple, that the
discovery of them reminds us, in more than one
respect, of Columbus’s egg…” Julius Robert Mayer
28.1 Introduction
Up to now the only electric circuits considered were DC circuits. It is only natural
that this historical sequence should be followed because the original power source
available for a circuit was the DC battery. However, with the knowledge gained
through electromagnetic induction, it soon became apparent that by mechanically
rotating a loop of wire in an external magnetic field, a sinusoidally varying
electromagnetic force could be obtained in the rotating loop of wire. This varying
emf gives rise to a varying alternating current. This new AC generator was
described in section 27.4. Recall that the alternating emf was given by
or
where
E = ωAB sin ωt
(27.30)
E = Emax sin ωt
(27.32
Emax = ωAB
(27.31)
Here ω is the angular speed of the rotating coil, A is the cross-sectional area of the
coil, and B is the magnetic field that the coil is rotated through.
The resulting alternating current in the coil was found to be
i = Emax sin ωt = imax sin ωt
R
(27.35)
The angular speed ω of the current loop is related to the frequency f of the
alternating current by
ω = 2πf
(28.1)
28.2 The Effective Current and Voltage in an AC Circuit
In contrast to a direct current, in which charge flows in one direction, in an
alternating current the charge flows first in one direction and then in the opposite
direction. The alternating current i varies sinusoidally with time, as shown in
figure 28.1(a).
Chapter 28 Alternating Current Circuits
Figure 28.1 An alternating current
The current i starts from zero and increases in one direction until it reaches
its maximum value imax. It then decreases to zero. It starts to increase again, but
now the charge is flowing in the opposite direction. This is shown as the negative
portion of the sine wave. The current increases negatively to −imax and then
decreases to zero, where the cycle begins again. We can express the alternating
current i as
(28.2)
i = imax sin(2πft)
where f is the frequency of the alternating current and t is the time.
With a DC current, the current is specified as having a certain value, let us
say 5 A. It always has that value because it is a constant current. We would also
like to describe an AC current by a single number, such as a current of 5 A.
However, with an AC current, we cannot say that the current has only one value; it
has many values because the current is constantly changing with time. It would
certainly be desirable to be able to refer to a single value for the alternating
current. We therefore need to find a different way to describe the alternating
current in order to consider it as having a single value, like a DC current. We
cannot simply take the average of an AC current, because the average current is
zero; that is, half the time the current is positive, half the time the current is
negative. We need to look at some other way to describe the effects of the
alternating current.
Consider the heating effect of the alternating current. The power P
dissipated in a resistor R in a DC circuit carrying a current I was found in chapter
24 to be
P = I2R
Using the same procedure, the power dissipated in an AC circuit is
P = i2R = [imax sin(2πft)]2R
28-2
(24.18)
Chapter 28 Alternating Current Circuits
= i2max sin2(2πft) R
(28.3)
Equation 28.3 shows that the power dissipated across the resistor is a function of
the time t. That is, the power dissipated is not a constant as in the DC circuit but
varies with time. Let us, instead, consider the average value of the power
generated, that is,
Pavg = i2max[sin2(2πft)]avgR
Although the average value of sin(2πft) is zero, the average value of its square is
not, as we can see from figure 28.1(b). We find the average value of sin2(2πft) by
using the trigonometric identity
sin2θ = 1 (1 − cos 2θ)
2
Applying that identity to the problem
Its average value is
sin2(2πft) = 1 [1 − cos(4πft)]
2
[sin2(2πft)]avg = 1 {1 − [cos(4πft)]avg}
2
However, the average value of the cosine is zero. Hence,
[sin2(2πft)]avg = 1
2
We can also see this in figure 28.1(b) by noting that the straight line labeled i2 = 0.5
cuts the sin2θ curve in half. Thus the shaded area above 0.5 is seen to be equal to
the shaded area below 0.5. Using this value of 1/2 for the average value of the
sin2(2πft), the average power expended in an AC circuit is found from equation 28.3
as
Pavg = 1 i2maxR
(28.4)
2
This average power expended in an AC circuit, with a current varying from −imax to
+imax, is not equivalent to the power expended in a DC circuit carrying the constant
current imax. To find the equivalence between the power expended in an AC circuit
and the power expended in a DC circuit, we introduce an effective current. An
effective current ieff is defined as a constant current that generates heat in a
resistor R at the same rate as an alternating current. Therefore, the average power
generated in an AC circuit is equal to the power generated in a DC circuit carrying
an effective current ieff, that is,
(P)DC = (Pavg)AC
28-3
Chapter 28 Alternating Current Circuits
i2effR = 1 i2maxR
2
Hence,
=
ieff
1
=
i
0.707 imax
2 max
(28.5)
Equation 28.5 gives the effective value of an alternating current. The effective value
is equivalent to a constant current of this value. It is 70.7% of the maximum or peak
value of the AC current. Even though the current is constantly changing with time,
the current can be referred to in terms of a single value, its effective or equivalent
value. The effective value of an AC current is equivalent to a constant DC current of
this value.
If an alternating current is applied to a resistor, the voltage drop across the
resistance R is
V = iR
Using the definition of the alternating current in equation 28.2, i = imax sin(2πft), we
can write this as
V = imaxR sin(2πft)
But imaxR = Vmax. Hence, the alternating voltage across the resistor is given by
V = Vmax sin(2πft)
(28.6)
Using the same analogy as for current, we obtain
=
Veff
1
=
Vmax 0.707 Vmax
2
(28.7)
Equation 28.7 gives the effective value of the voltage in an AC circuit. The
effective voltage is a constant value of the voltage that produces the same effect as
the alternating voltage. Thus, the constantly changing alternating voltage is
described in terms of the single valued, constant effective voltage. Whenever we
discuss current or voltage in an AC circuit, we mean its effective value. AC
ammeters and voltmeters measure the effective value of current and voltage,
respectively. Since this effective voltage or current is the square root of the mean or
average value of the voltage or current squared, it is often called the root-meansquared voltage or current, or simply the rms voltage or current. Hence, an
alternative notation is occasionally used with Ieff replaced by Irms, and Veff replaced
by Vrms.
28-4
Chapter 28 Alternating Current Circuits
Example 28.1
The effective value of an AC current. An AC current in a circuit varies from −3.50 A
to + 3.50 A. Find the effective value of this current.
Solution
The effective value of the current, found from equation 28.5, is
ieff = 0.707imax
= (0.707)(3.50 A)
= 2.47 A
Thus, even though the current is varying with time, if an AC ammeter were placed
in the circuit it would read the single value of 2.47 A.
To go to this Interactive Example click on this sentence.
Example 28.2
Finding the maximum value when the effective value is known. What is the
maximum voltage in a 60.0-Hz, 120-V line?
Solution
The effective voltage is 120 V and the maximum voltage, found from equation 28.7,
is
Veff = 0.707 Vmax
Vmax = Veff
0.707
= 120 V
0.707
= 170 V
Hence, even though an AC voltmeter would read the single value of 120 V, the
actual voltage would be varying between −170 V and +170 V.
To go to this Interactive Example click on this sentence.
In the analysis of an AC circuit, we will depart from our usual custom of
studying the simplest cases first and then building up to the most general case.
28-5
Chapter 28 Alternating Current Circuits
Instead, we will start with the general case, an RLC series circuit (i.e., a circuit
with a resistance R, an inductor L, and a capacitor C, connected in series) and take
as special cases the simpler examples of an RC circuit and an RL circuit. By now,
the student should be advanced enough to treat the general case first, and to be
able to see the beauty in the general case, in that it contains all the other examples
as special cases. The LC circuit will be left as a problem for the student to do.
28.3 An RLC Series Circuit
Consider a circuit containing a resistor R, an inductor L, and a capacitor C,
connected in series. This combination, shown in figure 28.2, is called an RLC
series circuit. Because R, L, and C are in series, we would expect the total applied
Figure 28.2 An RLC series circuit.
voltage to be equal to the sum of the voltages across R, L, and C. But if we placed a
voltmeter across R, L, and C, we would find that
V ≠ VR + VL + VC
(28.8)
The effective voltages do not add up algebraically because the sinusoidal voltages
are not in phase with one another.
Phase Relations
The different phase relations of the current and voltages in an AC circuit can be
seen with the help of figure 28.3. To understand these phase relations let us look at
28-6
Chapter 28 Alternating Current Circuits
the instantaneous voltage drop across each circuit component.
Figure 28.3 Phase relations in a series RLC circuit.
Phase Relationship for a Resistor The instantaneous voltage drop across the
resistor is given by
VR = iR
(28.9)
But the current in an AC circuit was given by equation 27.35 as
Hence
i = imax sin ωt
VR = imax R sin ωt
(28.10)
Equation 28.10 gives the instantaneous voltage drop across the resistor as a function
of time. Notice that VR varies with the sin ωt, just as the current does, and hence
the voltage drop across the resistor VR is everywhere in phase with the current i in
the circuit. Figure 28.3(a) shows the AC current i in the circuit, while part b of the
diagram shows the voltage drop VR across the resistor. Notice that the voltage VR is
28-7
Chapter 28 Alternating Current Circuits
everywhere in phase with the current i. That is, the location of the maximum and
minimum values of the voltage VR coincide with the maximum and minimum values
of the current i. This was to be expected since the current i was multiplied by the
constant value R.
Phase Relationship for an Inductor Recall that the voltage drop across an
inductor was given in chapter 27 as
VL = Ldi
(28.11)
dt
Using equation 27.35 for the current i, this becomes
VL = Ld (imax sin ωt)
dt
VL = L imax d (sin ωt) = L imax cos ωt (ω)
dt
VL = (ωL)imax cos ωt
(28.12)
Equation 28.12 gives the voltage drop across the inductor as a function of time, and
is plotted in figure 28.3(c). Since the cosine function is 900 out of phase with the
sine function, the voltage across the inductor will be 900 out of phase with the
current in the circuit. Notice in the figure that the voltage across the inductance VL
leads the voltage across the resistance VR and the current i by 900. In other words,
VL leads VR or i, because the peak of the VL wave occurs before the peak of the VR
wave. To emphasize this difference in phase between the cosine and sine function
the cosine term 1 is sometimes written as
cos ωt = sin(ωt + π/2)
Hence equation 28.12 can also be written in the form
VL = (ωL)imax sin(ωt + π/2)
(28.13)
In this form it is very obvious that VL is 900 or π/2 rad out of phase with the current
i in the circuit
Phase Relationship for a Capacitor The potential drop across the capacitor is
11From the trigonometric formula
sin(ωt + θ) = sin ωt cos θ + cos ωt sin θ
we get
sin(ωt + π/2) = sin ωt cos π/2 + cos ωt sin π/2
sin(ωt + π/2) = sin ωt (0) + cos ωt (1)
sin(ωt + π/2) = cos ωt
28-8
Chapter 28 Alternating Current Circuits
The current i in the circuit is given by
hence
VC = q
C
(28.14)
i = dq/dt
dq = i dt
and the charge q is found by integrating, that is,
=
q
q
dq ∫=
idt ∫ i
∫=
max
0
sin ωtdt
i
i
q =max ∫ sin ωt (ω ) dt =max ( − cos ωt ) + C
ω
ω
Where C is a constant of integration that would depend upon some initial charge on
the capacitor if there was one. Since we assumed that there is no initial charge on
the capacitor, we set this constant equal to zero. We then obtain
i
q = − max
 cos t
(28.15)
Combining equations 28.14 and 28.15 we get
VC = −
1 i
cos t
C max
(28.16)
Equation 28.16 gives the voltage drop VC across the capacitor as a function of time
and is plotted in figure 28.3(d). Again, since the cosine function is 900 out of phase
with the sine function, the voltage across the capacitor is also 900 out of phase with
the current in the circuit. Hence, we see that the VC curve is one-quarter of a cycle
behind the current curve. That is, the voltage drop across the capacitor VC lags
behind the voltage across the resistance VR and the current i in the AC circuit by 900.
In other words, VC lags VR or i, because the peak of the VC wave occurs after the
peak of the VR wave. Also notice that because of the minus sign in equation 28.16
the voltage VC across the capacitor is 1800 out of phase with the voltage across the
inductor VL.
We can again emphasize this difference in phase between the cosine and sine
function by writing the cosine term 2 as
From the trigonometric formula
sin(ωt − θ) = sin ωt cos θ − cos ωt sin θ
we get
sin(ωt − π/2) = sin ωt cos π/2 − cos ωt sin π/2
sin(ωt − π/2) = sin ωt (0) − cos ωt (1)
sin(ωt − π/2) = − cos ωt
2
28-9
Chapter 28 Alternating Current Circuits
−
cos ωt = sin(ωt − π/2)
Hence equation 28.16 can also be written in the form
VC =
1 i
sin(t − /2 )
C max
(28.17)
From equation 28.13 we see that VL is + π/2 rad or +900 before i, while equation
28.17 shows that VC is − π/2 rad or − 900 after i. In these forms it is also very obvious
that VC and VL are 1800 or π rad out of phase with each other.
An old mnemonic device to remember whether the emf across an inductor or
capacitor leads or lags the current in an AC circuit, is the phrase “ELI the ICE
man.” Meaning, in an inductor L, the emf E leads the current I in the circuit,
whereas in a capacitor C, the emf E lags behind the current I in the circuit.
Ohm’s Law for an RLC Series Circuit
Equation 28.8 stated that the applied effective voltage in the circuit was not equal
to the algebraic sum of the effective voltages, that is,
V ≠ VR + VL + VC
(28.8)
If we were to place a voltmeter across the circuit, we would not read the algebraic
sum of these values, because, as just seen, the voltage across the inductor VL and
the voltage across the capacitor VC are out of phase with the voltage across the
resistor VR. The instantaneous values of VR, VL, and VC do add up by Kirchhoff’s
rule. It is the phase relations that causes the effective values not to add up
algebraically. Since VL and VC are 1800 out of phase with each other, as seen in
figure 28.3, and VL is 900 out of phase with VR, we can treat each of these voltages
as though they were vectors, as shown in figure 28.4(a). Taking the x-axis as the
reference direction, we place VR along the x-axis. VL then lies along the positive yaxis, which, if we think of a counterclockwise rotation of the vector, is +900 before
the x-axis, and VC lies along the negative y-axis, which is −900 after the x-axis. The
sum of these voltages becomes the vector sum shown in figure 28.4(b) and its
magnitude is given by the Pythagorean theorem as
Figure 28.4 Vector diagram of the voltages in an AC circuit.
28-10
Chapter 28 Alternating Current Circuits
V = V 2R + (V L − V C ) 2
(28.18)
The voltage across RLC is, of course, also equal to the impressed voltage from the
AC source. The angle φ between the applied voltage V and the voltage across the
resistor VR is called the phase angle, and is found from the figure to be
 = tan −1
VL − VC
VR
(28.19)
Recall that the voltage across the resistor is in phase with the current in the circuit.
Hence, the phase angle is a measure of how much the applied voltage V leads or
lags the AC current in the circuit. If φ is above the x-axis, φ is positive, VL is greater
than VC, and the voltage leads the current by φ degrees. A circuit that has a positive
value of φ is called an inductive circuit. If φ lies below the x-axis, φ is negative, VC is
greater than VL, and the voltage lags the current by φ degrees. A circuit that has a
negative value of φ is called a capacitive circuit.
The potential drop across the resistor is given by Ohm’s law as
VR = iR
(28.20)
and we saw the voltage drop across the resistor was also given by
VR = imax R sin ωt
(28.10)
In the same way, the voltage drop across the inductor was shown to be
which can also be written as
which looks like the form
VL = (ωL)imax cos ωt
(28.12)
VL = (ωL)i
VR = R i
Hence the quantity ωL can be viewed as some type of AC resistance. Let us define it
as the inductive reactance and designate it as XL = ωL. But since ω = 2πf this
becomes
XL = ωL = 2πfL
(28.21)
Equation 28.21 gives the inductive reactance of the inductor in the AC circuit. The
inductive reactance XL is the inductive analogue to resistance, that is, it also has
the effect of impeding current in a circuit. The inductive reactance, which is
measured in ohms, is a type of AC resistance but one that varies with the
inductance L of the circuit and the frequency f of the alternating current in the
circuit. The inductance L of the circuit is measured in henries (H).
28-11
Chapter 28 Alternating Current Circuits
The effective voltage drop across the inductor can now be given by
VL = iXL
(28.22)
Note that in practice, most inductors also have some resistance. We will assume
here that the resistance is small enough to be ignored. In the laboratory, it is a good
idea to measure the resistance of the coil you are using to see if the assumption is
valid.
Using the same analogy, we found the voltage drop across the capacitor as
VC = −
which can also be written as
which again looks like the form
1 i
cos t
C max
VC =
(28.16)
1 i
C
VR = R i
Hence the quantity (1/ωC) can also be viewed as another type of AC resistance. Let
us define it as the capacitive reactance and designate it as XC = 1/ωC. But since ω =
2πf this becomes
XC = 1 = 1
(28.23)
ωC
2πfC
Equation 28.23 gives the capacitive reactance XC of the capacitor in the AC
circuit. The capacitive reactance XC is the capacitive analogue to resistance, that is,
it also has the effect of impeding current in a circuit. The capacitive reactance,
which is also measured in ohms, is a type of AC resistance but one that varies with
the capacitance C of the circuit and the frequency f of the alternating current in the
circuit. The capacitance C of the circuit is measured in Farads (F).
The potential drop across the capacitor, using the same analogy, is now given
by
VC = iXC
(28.24)
yields
or
Substituting equations 28.20, 28.22, and 28.24 back into equation 28.18,
V = (iR) 2 + (iX L − iX C ) 2
V = R 2 + (X − X ) 2
L
C
i
Just as V/i was defined as the resistance R of a DC circuit, we will now define the
ratio as the impedance Z of an AC circuit, that is,
Z = R 2 + (X L − X C ) 2
28-12
(28.25)
Chapter 28 Alternating Current Circuits
With this definition, Ohm’s law for an AC circuit takes the simple form
i= V
Z
(28.26)
Therefore, in order to solve for the current in an AC circuit, we reduce it to
the same simple form as a DC circuit. Instead of simply knowing the resistance, as
in a DC circuit, we must now know the impedance of an AC circuit for its solution.
The impedance is a function of the frequency of the AC circuit, through its inductive
and capacitive reactances. It is often helpful to draw an impedance diagram, which
can be deduced from the voltage diagram, and is shown in figure 28.5.
Figure 28.5 Impedance diagram for an AC circuit.
A complete analysis of the series RLC circuit is given in example 28.3.
Interactive tutorial 56 at the end of the chapter will not only give a complete
analysis of the series RLC circuit but it will also plot all the voltages and current to
show the phase relationships in the circuit.
Example 28.3
An RLC series circuit. The RLC series circuit shown in figure 28.6 has a resistance
R = 400 Ω, an inductor L = 5.00 H, a capacitor C = 3.00 µF, and they are connected
to a 110-V, 60.0-Hz line. Find (a) the inductive reactance XL, (b) the capacitive
reactance XC, (c) the impedance Z of the circuit, (d) the current i in the circuit,
(e) the voltage drop VR across R, (f) the voltage drop VL across L, (g) the voltage drop
VC across C, (h) the total voltage V across RLC, and (i) the phase angle φ.
Solution
a. The inductive reactance, found from equation 28.21, is
28-13
Chapter 28 Alternating Current Circuits
Figure 28.6 An example of an AC circuit.
XL = 2πfL
V/(A/s)
= 2 60.0 1
s (5.00 H)
H
V 
= 1890 A V/A
= 1890 Ω
b. The capacitive reactance, found from equation 28.23, is
XC =
1 =
1
( F )
−6
2πfC
2π(60.0 1/s)(3.00  10 F) (C/V)
V C/s

V/A
= 884 C/s A
= 884 Ω
c. The impedance of the circuit, found from equation 28.25, is
Z = R 2 + (X L − X C ) 2
= (400 ) 2 + (1890  − 884 ) 2
= 1080 Ω
d. The effective current i in the circuit, found from equation 28.26, is
i= V
Z
28-14
Chapter 28 Alternating Current Circuits
= 110 V
1080 Ω
= 0.102 A
e. The voltage drop across R, found from equation 28.20, is
VR = iR
= (0.102 A)(400 Ω)
= 40.8 V
f. The voltage drop across L, found from equation 28.22, is
VL = iXL
= (0.102 A)(1890 Ω)
= 193 V
g. The voltage drop across C, found from equation 28.24, is
VC = iXC
= (0.102 A)(884 Ω)
= 90.2 V
h. The total voltage across R, L, and C in series, found from equation 28.18, is
V = V 2R + (V L − V C ) 2
2
= (40.8 V ) + (193 V − 90.2 V) 2
= 110 V
which is, of course, equal to the applied voltage. Notice that the voltages are added
vectorially and not algebraically.
i. The phase angle, found from equation 28.19, is
V − VC
 = tan −1 L
VR
−1 193 V − 90.2 V
= tan
40.8 V
= 68.40
This means that the applied voltage leads the current in the circuit by 68.40, and
the phase relation is shown in figure 28.7. Since φ is a positive angle the circuit is
called an inductive circuit.
28-15
Chapter 28 Alternating Current Circuits
Figure 28.7 The phase angle between the applied voltage and the current in the
AC circuit.
To go to this Interactive Example click on this sentence.
Example 28.4
An RC series circuit. A 110-V, 60.0-Hz, AC line is connected across a resistance of
1000 Ω and a capacitor of 1.00 µF, as shown in figure 28.8. Find (a) the capacitive
reactance, (b) the impedance, (c) the current in the circuit, (d) the voltage drop VR
across the resistor, (e) the voltage drop VC across the capacitor, (f) the total voltage
across R and C, and (g) the phase angle φ between the voltage and current.
Figure 28.8 An RC series circuit.
Solution
a. The capacitive reactance XC, found from equation 28.23, is
XC =
1 =
1
2πfC
2π(60.0 1/s)(1.00  10−6 F)
= 2650 Ω
b. The impedance Z is found from equation 28.25. Because there is no induction in
this RC circuit, XL = 0. Therefore, the impedance becomes
Z = R 2 + X 2C
28-16
Chapter 28 Alternating Current Circuits
2
Z = (1000  ) + (2650 ) 2
= 2830 Ω
c. The effective current i in the circuit comes from Ohm’s law, equation 28.26, and
is
i = V = 110 V
Z 2830 Ω
= 3.89  10−2 A
d. The voltage drop VR across the resistor, found from equation 28.20, is
VR = iR = (3.89  10−2 A)(1000 Ω)
= 38.9 V
e. The voltage drop VC across the capacitor, found from equation 28.24, is
VC = iXC
= (3.89  10−2 A)(2650 Ω)
= 103 V
f. The total voltage drop across R and C in series is found from equation 28.18.
Since there is no inductance in this circuit, VL = 0. Therefore,
V = V 2R + V 2C
2
= (38.9 V ) + (103 V) 2
= 110 V
Note that the voltage across R and C in series is the same as the applied voltage,
which it should be. Because of the phase difference of the voltages, they add as
vectors rather than as algebraic quantities.
g. The phase angle φ between the voltage and the current in the circuit is found
from equation 28.19 with VL = 0. Therefore,
−V C
 = tan −1
VR
−1 −103 V
= tan
38.9 V
= −69.30
This phase angle is represented in figures 28.10(a) and 28.10(b). The voltage in the
circuit lags the current in the circuit by 69.30. Since φ is a negative quantity, the
circuit is called a capacitive circuit.
28-17
Chapter 28 Alternating Current Circuits
Figure 28.9 Phase relation for an RC circuit.
To go to this Interactive Example click on this sentence.
Example 28.5
An RL series circuit. A 110-V, 60-Hz, AC line is connected across a resistance of
1000 Ω and an inductor of 5.00 H, as shown in figure 28.10. Find (a) the inductive
reactance, (b) the impedance, (c) the current in the circuit, (d) the voltage drop VR
across the resistor, (e) the voltage drop VL across the inductor, (f) the total voltage
drop V across R and L, and (g) the phase angle between the voltage V and the
current I.
Solution
a. The inductive reactance XL, found from equation 28.21, is
XL = 2πfL = 2π(60 1/s)(5.00 H)
= 1890 Ω
Figure 28.10 An RL series circuit.
b. The impedance Z is found from equation 28.25, with XC = 0. Therefore,
28-18
Chapter 28 Alternating Current Circuits
Z = R2 + XL 2
= (1000 ) 2 + (1890 ) 2
= 2140 Ω
c. The current i in the circuit, found from Ohm’s law, equation 28.26, is
i = V = 110 V
Z
2140 Ω
= 5.14  10−2 A
d. The voltage drop VR across the resistor, found from equation 28.20, is
VR = iR = (5.14  10−2 A)(1000 Ω)
= 51.4 V
e. The voltage drop VL across the inductor, found from equation 28.22, is
VL = iXL = (5.14  10−2 A)(1890 Ω)
= 97.1 V
f. The total voltage drop V across R and L is found from equation 28.18 with VC = 0.
Therefore,
V = V 2R + V L 2
2
= (51.4 V ) + (97.1 V) 2
= 110 V
Note that the voltage drop across R and L is the same as the applied voltage V, as is
expected.
g. The phase angle φ between the current and voltage in the circuit is found from
equation 28.19 with VC = 0. Therefore,
V
 = tan −1 L
VR
−1 97.1 V
= tan
51.4 V
= 62.10
Because this is a positive angle it means that the voltage leads the current in the
circuit by 62.10 and the circuit is an inductive circuit. This is shown in figures
28.12(a) and 28.12(b).
28-19
Chapter 28 Alternating Current Circuits
Figure 28.11 Phase relations in an RL circuit.
To go to this Interactive Example click on this sentence.
28.4 Resonance in an RLC Series Circuit
Not only does the instantaneous value of the AC current change with time, but the
effective value of the AC current even changes with the frequency of the AC source.
When is this current a maximum in the RLC series circuit? The current is given by
Ohm’s law as
i= V =
V
(28.27)
2
2
R + (X L − X C )
Z
Notice from the form of the equation that the current i is a maximum whenever the
impedance Z is a minimum. This condition is called resonance. As we can see from
equation 28.27, Z is a minimum whenever XL = XC. In that case the impedance Z is
equal to the resistance R. To determine the frequency at which resonance occurs,
we set the inductive reactance equal to the capacitive reactance and get
XL = XC
2πfL = 1
2πfC
Solving for the resonant frequency we get
f0 =
1
2 LC
(28.28)
where f0 is the frequency at which resonance occurs. Notice that it depends on the
value of the inductance L and capacitance C. At this frequency, the maximum
current that can flow in the circuit is obtained. If we plot the current in the circuit
as a function of the frequency of the impressed voltage source, we obtain figure
28-20
Chapter 28 Alternating Current Circuits
28.12. At the frequency f0, the maximum current is obtained. At resonance, VL and
VC are equal, but are 1800 out of phase. Resonance is used in the tuning circuit of a
radio and in variable oscillators such as are found in physics laboratories.
Figure 28.12 Resonant frequency of an RLC series circuit.
Example 28.6
Resonant frequency. Find the resonant frequency of the RLC series circuit in
example 28.3.
Solution
The resonant frequency, found from equation 28.28, is
f0 =
1
2π
1
2 LC
(5.00 H ) (3.00 × 10−6 F )
H
F
V/ ( A/s ) C/V
A C/s
s
C A
= 41.1
1
= 41.1 s 2
= 41.1 cycles/s = 41.1 Hz
To go to this Interactive Example click on this sentence.
28.5 Power in an AC Circuit
A current in an inductor causes a magnetic field to exist within that coil as
discussed in chapter 26. In chapter 27 we saw that the work done in moving a
charge through a coil shows up as stored energy in the magnetic field of the coil. If
the coil is a perfect inductor, that is, its resistance is zero or at least negligible, all
the energy applied to the coil is stored in the magnetic field of the coil and no
28-21
Chapter 28 Alternating Current Circuits
energy is lost in the coil. When the applied emf goes to zero, the energy in the
magnetic field of the coil is returned to the circuit by a current. Hence, there is no
energy lost in a perfect inductor. Whatever energy the inductor absorbs in one half
of a cycle, it returns to the circuit in the next half-cycle.
The capacitor acts similarly. As charge is stored on the plates of the
capacitor, the energy used to place that charge on the capacitor is stored in the
electric field between the plates of the capacitor. The energy absorbed by the
electric field in one half-cycle is returned to the circuit in the next half-cycle.
Therefore, there is no power consumed by the inductor or the capacitor in an AC
circuit. Only across the resistance R of the circuit, is power dissipated by Joule
heating.
The power lost across the resistor is given by
But as we can see in figure 28.4,
P = iVR
VR = V cos φ
where φ is the phase angle between the voltage and the current i in the circuit.
Therefore, the power dissipated in an AC circuit is
P = iV cos φ
(28.29)
The quantity cos φ is called the power factor of the circuit, and is given by
PF = cos φ
(28.30)
We can obtain further insight into the meaning of the power factor from the
geometry in figure 28.5 as
PF = cos φ = R
(28.31)
Z
If we multiply both numerator and denominator of equation 28.31 by i2, we get
PF = cos φ = i2R = i2R
i2Z
iVapplied
The quantity i2R is the power dissipated by the circuit, whereas iVapplied is the
power supplied to the circuit by the AC generator. Therefore,
PF = cos φ = Power consumed
Power supplied
(28.32)
Hence, the power factor is the ratio of how much power is actually used in
the circuit to the total power that must be supplied to the circuit. The power factor is
28-22
Chapter 28 Alternating Current Circuits
quite often expressed as a percentage. For example, if the phase angle φ = 30.00, the
power factor is
PF = cos 30.00 = 0.866 = 86.6%
This means that 86.6% of the power supplied to the circuit is actually consumed in
the circuit. The rest is returned to the generator. A power factor close to 1 (or 100%)
is usually desired for AC circuits. If a circuit has a power factor of 50%, and 500 W
of power are actually used in the circuit, the AC generator would have to be able to
deliver 1000 W. Thus, a larger and more expensive generator would be necessary to
supply energy to a circuit with a low power factor. If the power factor was 1 (or
100%) a generator rated at 500 W would be adequate for the circuit.
Example 28.7
Power factor in an AC circuit. In the RLC series circuit of example 28.3, find (a) the
power factor, (b) the power consumed, and (c) the total power that must be supplied
to the circuit.
Solution
a. In example 28.3 the phase angle φ was found to be 68.40. The power factor of the
circuit, found from equation 28.30, is
PF = cos φ = cos 68.40
= 0.369 = 36.9%
b. The power dissipated in the circuit is
P = i2R = (0.102 A)2(400 Ω)
= 4.16 W
c. The power applied to the circuit is
As a check, note that
Papp = iVapp = (0.102 A)(110 V)
= 11.2 W
Power consumed = 4.16 W = 0.371 = 37.1%
Power supplied
11.2 W
which agrees with the original value, within round-off errors of the calculations.
To go to this Interactive Example click on this sentence.
28-23
Chapter 28 Alternating Current Circuits
28.6 An RLC Parallel Circuit
An AC circuit with a resistor, inductor, and capacitor in parallel, called an
RLC parallel circuit, is shown in figure 28.13. Since R, L, and C are in parallel
Figure 28.13 A parallel RLC circuit.
the voltage across each of them is the same. This means that the voltages are not
only equal, but they are also in phase with one another. Thus,
V = VR = VL = VC
Because all these voltages are in phase, it is reasonable to expect that some of the
currents will not be in phase. The impressed voltage V, or the voltage across the
resistor VR, will be used as the reference to measure all phase differences. The
current IR through R is in phase with V as shown in figures 28.14(a) and 28.14(b).
The current through the inductor is out of phase with the current through the
resistor. We can see this phase difference by observing the phase relations in the
series circuit of figure 28.3, where the voltage VL leads VR and the current i by 900.
Therefore, if VL leads i by 900, then we can equally well say that i lags VL by
900. Hence, the current in the inductor part of the series circuit lags the voltage
across the inductor by 900. For the parallel circuit, the current IL through the
inductor must also lag the voltage across the inductor by 900. But since the voltage
across the inductor is in phase with the voltage across the resistance, which, in
turn, is in phase with the current IR in the resistor, it follows that IL, the current
through the inductor, must lag the current IR in the resistor by 900. This is shown
in figure 28.14(c).
The current through the capacitor is also out of phase with the current
through the resistor. For the series circuit, the voltage VC lags VR and i by 900, as
observed in figure 28.3. Therefore, the current in the series circuit leads the voltage
across the capacitor by 900. Thus, in the parallel circuit of figure 28.13, the current
through the capacitor must also lead the voltage across the capacitor. Since the
28-24
Chapter 28 Alternating Current Circuits
Figure 28.14 Phase relations in a parallel RLC circuit.
voltage across the capacitor is in phase with the voltage across the resistor, which is
in phase with the current IR through the resistor, then the current through the
capacitor IC must lead the current through the resistor IR by 900. This is shown in
figure 28.14(d). Figure 28.14 can be summarized by saying that
IL lags IR by 900
IC leads IR by 900
IC is 1800 out of phase with IL
Because the currents in the parallel RLC circuit are out of phase with each other,
they can be drawn as vectors in figure 28.15(a). The total resultant current IT,
found from the geometry of figure 28.15(b), is
I T = I 2R + (I C − I L )
28-25
2
(28.33)
Chapter 28 Alternating Current Circuits
Figure 28.15 Currents in a parallel circuit are added vectorially.
The individual currents, found by Ohm’s law, are
IR = V
R
IL = V
XL
IC = V
XC
(28.34)
(28.35)
(28.36)
The phase angle φ that the total current leads or lags the current in the
resistor, and hence the applied voltage, found from figure 28.15(b), is
tan φ = IC − IL
IR
−1 I C − I L
 = tan
IR
(28.37)
If we substitute equations 28.34, 28.35, and 28.36 into equation 28.37, we also
obtain
V − V
X
XL
 = tan −1 C
V/R
(28.38)
R
−1 R
−
 = tan
XC XL
The total impedance of the circuit, found from Ohm’s law, is
Z= V
IT
28-26
(28.39)
Chapter 28 Alternating Current Circuits
Example 28.8
An RLC parallel circuit. A resistor of 400 Ω, an inductor of 5.00 H, and a capacitor
of 3.00 µF are connected in parallel to a 110-V, 60.0-Hz line. Find (a) the inductive
reactance, (b) the capacitive reactance, (c) the current through the resistor IR,
(d) the current through the inductor IL, (e) the current through the capacitor IC,
(f) the total current in the circuit IT, (g) the phase angle φ, and (h) the total
impedance of the circuit.
Solution
a. The inductive reactance is
b. The capacitive reactance is
XL = 2πfL
= 2π(60.0 1/s)(5.00 H)
= 1890 Ω
=
XC = 1
2πfC
1
2π(60.0 1/s)(3.00  10−6 F)
= 884 Ω
c. The current through the resistor IR, found from equation 28.34, is
IR = V
R
= 110 V
400 Ω
= 0.275 A
d. The current IL through the inductor, found from equation 28.35, is
IL = V
XL
= 110 V
1890 Ω
= 5.82  10−2 A
e. The current IC through the capacitor, found from equation 28.36, is
IC = V
XC
= 110 V
884 Ω
28-27
Chapter 28 Alternating Current Circuits
= 0.124 A
f. The total current IT in the circuit, found from equation 28.33, is
2
I T = I 2R + (I C − I L )
= (0.275 A) 2 + (0.124 A − 0.0582 A) 2
= 0.283 A
g. The phase angle φ, found from equation 28.37, is
I −I
 = tan −1 C L
IR
−1 0.124 A − 0.0582 A
 = tan
0.275 A
= 13.50
The total current in the circuit leads the applied voltage by 13.50.
h. The total impedance of the circuit, found from equation 28.39, is
Z= V
IT
= 110 V
0.283 A
= 389 Ω
To go to this Interactive Example click on this sentence.
Example 28.9
An RC parallel circuit. A 110-V, 60.0-Hz, AC line is connected in parallel to a
resistor of 1000 Ω and a capacitor of 1.00 µF, as shown in figure 28.16. Find (a) the
capacitive reactance, (b) the current IR through the resistor, (c) the current IC
through the capacitor, (d) the total current in the circuit, (e) the phase angle φ, and
(f) the total impedance of the circuit.
Solution
a. The capacitive reactance is found to be
XC =
1
2πfC
28-28
Chapter 28 Alternating Current Circuits
Figure 28.16 A parallel RC circuit.
=
1
2π(60.0 1/s)(1.00  10−6 F)
= 2650 Ω
b. The current in the resistor, found from equation 28.34, is
IR = V
R
= 110 V
1000 Ω
= 0.110 A
c. The current IC through the capacitor, found from equation 28.36, is
IC = V
XC
= 110 V
2650 Ω
= 4.15  10−2 A
d. The total current in the circuit is found from equation 28.33 with IL = 0, that is,
I T = I 2R + I 2C
= (0.110 A) 2 + (0.0415 A) 2
= 0.118 A
e. The phase angle φ is found from equation 28.37 with IL = 0, that is,
I
 = tan −1 C
IR
−1 0.0415
= tan
0.110
= 20.70
28-29
Chapter 28 Alternating Current Circuits
f. The total impedance of the circuit, found from equation 28.39, is
Z= V
IT
= 110 V
0.118 A
= 932 Ω
To go to this Interactive Example click on this sentence.
28.7 The Transformer
The transformer is a device, based on the principle of mutual induction studied in
chapter 25, that changes electrical energy at a particular voltage and current in one
coil into a different voltage and current in a second coil. The simplest transformer is
the coaxial solenoids shown in figure 27.6 and reproduced here as figure 28.17(a).
The schematic diagram for the transformer is shown in figure 28.17(b). The first
solenoid coil is called the primary coil, whereas coil 2 is called the secondary.
If a varying emf is applied across coil 1 causing a changing current in coil 1
and hence a changing magnetic flux, an emf is induced in coil 2 by this changing
magnetic flux ∆ΦC, and is given by
Figure 28.17 A transformer.
E2 = −N2 dΦC
dt
If we were to change magnetic flux in coil 2 by the same amount, this would induce
an emf in coil 1 given by
28-30
Chapter 28 Alternating Current Circuits
E1 = −N1 dΦC
dt
Since the changing magnetic flux is the same in each coil
Hence,
dΦC = −E2 = −E1
dt
N2
N1
E2 = E1
N2 N1
(28.40)
Therefore, the emf induced in the secondary of the transformer is given by
E 2 = N2 E1
N1
(28.41)
If N2, the number of turns in the secondary, is greater than N1, the number of turns
in the primary, then the ratio of N2/N1 is greater than 1. Therefore, the induced emf
in the secondary E2 is greater than the impressed emf in the primary E1. When N2
is greater than N1, the transformer is called a step-up transformer because it “stepsup” or increases the voltage from the primary to the secondary. When N2 is less
than N1, the ratio N2/N1 is less than 1, and the induced emf in the secondary is less
than the applied emf across the primary. In this case, the transformer is called a
step-down transformer because it has reduced the voltage from the primary to the
secondary.
The law of conservation of energy states that the power put into coil 1 is
equal to the power taken out of coil 2. Because the product of an emf E with the
current i represents power, we must have
E 2i2 = E 1i1
(28.42)
If an alternating emf and current are impressed across coil 1 by an AC generator,
then an alternating emf and current are found in coil 2, where the relation between
the emfs and currents is given by equation 28.42. (Of course, the emfs and currents
are given by their effective values.) Solving for the ratio of the emf of coil 2 to the
emf of coil 1 gives
E2 = i1
(28.43)
E1
i2
Equation 28.43 shows that the ratio of the emfs in a transformer is reciprocal to the
ratio of the currents.
Note, that if equation 28.40 is written in the form
28-31
Chapter 28 Alternating Current Circuits
E2 = N2
E1 N1
then we can combine it with equation 28.43 as
E2 = i1 = N2
E 1 i2 N 1
(28.44)
Equation 28.44 shows that the current in the secondary winding of a transformer is
given by
i2 = N 1 i 1
(28.45)
N2
Notice, that if N2 is greater than N1 (a step-up transformer), the current in the
secondary is less than the current in the primary. Comparing equations 28.41 and
28.45 shows that the currents are reciprocal to the emfs. This means that in a stepup transformer, E2 is greater than E1, but the current i2 is less than i1. Hence, we do
not get something for nothing from the transformer. The increase in voltage comes
with a decrease in current, such that the power into the circuit is equal to the
power out. A picture of a typical transformer is shown in figure 28.17(c).
Example 28.10
The voltage in a step-up transformer. A 120-V line from an AC generator is
connected to the primary of a transformer of 50 turns. If the secondary of the
transformer has 100 turns, what voltage will be found across the secondary of the
transformer?
Solution
The voltage across the secondary, found from equation 28.41, is
E2 = N2 E1 = 100 (120 V) = 240 V
N1
50
To go to this Interactive Example click on this sentence.
Example 28.11
The current in a step-up transformer. If the current in the primary of example 28.10
is 3.00 A, what is the current in the secondary?
28-32
Chapter 28 Alternating Current Circuits
Solution
The current in the secondary, found from equation 28.45, is
i2 = N1 i1 = 50 (3.00 A) = 1.50 A
N2
100
To go to this Interactive Example click on this sentence.
Example 28.12
A step-down transformer. You wish to make a transformer that can take the 120 V
from your wall outlet and drop it down to 24.0 V to operate a toy electric train. If
there are 100 turns of wire in the primary how many turns do you need for the
secondary?
Solution
The number of turns in the secondary can be found from equation 28.44 as
N2 = E2
N1
E1
N2 = E2 N1
E1
 24.0 V 
= 
 (100 turns )
 120 V 
= 20.0 turns
To go to this Interactive Example click on this sentence.
Having analyzed an AC and a DC circuit, it is obvious that the AC circuit is
more difficult to deal with than a DC circuit. What then is the great advantage of
AC power over DC power that it is used so extensively? Two of the main reasons for
the desirability of AC power are the ease in generating AC power with the AC
generator described in chapter 25 and, second, the ease of transmitting this AC
power from the power station to the home by means of transformers. Power is lost
in transmission lines by Joule heating in the form P = I2R. The smaller the value of
the current in the line the smaller is the energy loss. As an example, suppose a
particular power station generated 100,000 W by having a current of 20.0 A at 5000
28-33
Chapter 28 Alternating Current Circuits
V. If the transmission line has a resistance of 20.0 Ω, the power lost to heat during
transmission is
P = I2R = (20.0 A)2(20.0 Ω) = 8000 W
If the power output from the power station is connected to a step-up transformer
that has a turns ratio of 1:10 from the primary to the secondary, then the voltage at
the secondary becomes
E2 = N2 E1
N1
10
 
=   ( 5000 V )
1 
= 50,000 V
The current in the secondary is
i2 = N 1 i 1
N2
1 
=   ( 20.0 A )
 10 
= 2.00 A
The power lost in the transmission line due to Joule heating is now
P = i2R
= (2.00 A)2(20.0 Ω)
= 80.0 W
Therefore, instead of transmitting the 100,000 W of power at 20.0 A and 5000
V, with a loss of 8000 W, the same 100,000 W can be transmitted at 2.00 A and
50,000 V with a power loss of only 80.0 W, a saving of 7920 W. When the
transmission line reaches the local substation another transformer steps-down this
voltage to lower values, which are then transmitted to your local electric pole.
Another transformer steps-down this voltage further to give you the desired current
and voltage for your home.
Before we leave this chapter it is perhaps appropriate to compare the effects
of electromagnetic induction when used in a DC circuit and in an AC circuit. In
chapter 27 we saw that a changing current in one coil induced an emf in the second
coil. In most cases in chapter 27, we considered a DC current rising from zero to
some maximum value, remaining constant for a time, and then decreasing to zero.
The induced emf existed only for the short time that the current was rising or
falling. The transformer reaches its maximum utilization in an AC circuit because
the instantaneous value of an AC current is continually changing with time. Hence,
there is a continuously alternating emf and current induced in the second coil. The
transformer is thus an excellent practical example of electromagnetic induction
when used in an AC circuit.
28-34
Chapter 28 Alternating Current Circuits
Have You Ever Wondered?
An Essay on the Application of Physics
Metal Detectors at Airports
Have you ever wondered how airport security can determine if a terrorist
passenger is trying to sneak a gun or a bomb aboard an aircraft? To determine if a
gun, or any piece of iron, is being carried by a passenger, each passenger is required
to walk through a metal detector, figure 1(a).
The metal detector is essentially a coil of wire that the passenger walks
through. The coil is connected to an AC source, as shown in figure 1(b). An AC
voltmeter is connected across the coil to record the voltage VLA across the coil. The
subscript A has been placed on VL to indicate that this is the voltage across the coil
when air is in the space within the coil. The voltage drop across the coil, found by
equation 28.22, is
VLA = iXLA
(28H.1)
where XLA is the inductive reactance given by equation 28.21 as
XLA = 2πfLA
(28H.2)
Here LA is the inductance of the coil with air in the space within the coil. For
simplicity in our discussion, let us assume that the coil of the metal detector can be
Figure 1 Every passenger must walk through the metal detector.
considered to be a solenoid with air in the space inside the coil. The inductance for
such a coil is found from equation 23.39 as
LA = µ0Aln2
28-35
(28H.3)
Chapter 28 Alternating Current Circuits
where µ0 is the permeability of free space or air,
A is the area of the coil,
l is the length of the coil, and
n is the number of turns of wire per unit length.
If an iron rod is now placed within the coil, the permeability µ0 in equation
28H.3 must be replaced with the permeability µ of the iron rod. Recall that when
dealing with electric fields ε0 was the permittivity of free space, and when dealing
with different materials the permittivity of the medium was given by ε = κε0. In the
same way, the permeability of a medium is given by
µ = Km µ0
(28H.4)
where Km is called the relative permeability of the medium. For soft iron Km ≈ 5000,
and the inductance of the coil with iron in it is now given by
LI = µAln2 = (5000µ0)Aln2 = 5000LA
(28H.5)
The inductive reactance of the coil with iron in the space within the coil now
becomes XLI,
XLI = 2πfLI = 2πf(5000LA) = (5000)(2πfLA) = (5000)XLA
The voltage drop across the iron core solenoid is now
VLI = iXLI = i(5000XLA)
= 5000(iXLA) = 5000VLA
(28H.6)
Thus by placing an iron rod into the solenoid the voltage drop across the solenoid
has increased by a factor of 5000 over the voltage across the solenoid when air was
in the space. Hence moving iron into the coil is relatively easy to detect. This is the
principle of the metal detector. In practice, the iron to be detected does not fill up
the entire space within the coil and hence the voltage drop across the coil is not as
great as in this simplified model. However, the principle is still the same and the
introduction of iron into the coil produces an increase in voltage across the coil. In
our simplified description we have used a voltmeter to detect the presence of metal.
This increased voltage could also be used to trigger an alarm or a bell to indicate
the presence of the metal. Hence, if a terrorist passenger tries to carry a concealed
gun through the detector coil, the increase in the permeability of the coil due to the
presence of the gun causes an alarm to sound, indicating the presence of concealed
metal on the passenger.
28-36
Chapter 28 Alternating Current Circuits
The Language of Physics
Effective current
That constant current that generates heat in a resistor at the same rate as an
alternating current. The effective value of an alternating current is equal to 70.7%
of the maximum or peak value of the AC current. An AC ammeter measures the
effective current in the circuit. The effective current is sometimes called the rms
current (p ).
Effective voltage
The constant value of the voltage that produces the same effect as an alternating
voltage. The effective value of an alternating voltage is equal to 70.7% of the
maximum or peak value of the AC voltage. An AC voltmeter measures the effective
voltage. The effective voltage is sometimes called the rms voltage (p ).
RLC series circuit
A circuit containing a resistor, an inductor, and a capacitor in series. The voltage
across the resistor is in phase with the current in the circuit. The voltage across the
inductor in the circuit leads the current in the circuit by 900, whereas the voltage
across the capacitor lags the current in the circuit by 900 (p ).
Phase angle
A measure of how much the applied voltage leads or lags the current in an AC
circuit (p ).
Inductive reactance
The inductive analogue to resistance in an AC circuit; that is, an inductive
reactance tends to impede the flow of charge in an AC circuit. The inductive
reactance is proportional to the frequency of the AC source; a high-frequency AC
source causes a high reactance, whereas a low-frequency source causes a low
reactance (p ).
Capacitive reactance
The capacitive analogue to resistance in an AC circuit; that is, a capacitive
reactance tends to impede the flow of charge in an AC circuit. The capacitive
reactance is inversely proportional to the frequency of the AC source; a highfrequency AC source causes a low reactance, whereas a low-frequency source causes
a high reactance (p ).
Impedance
A measure of the opposition to the flow of charges in an AC circuit. It is composed of
the resistance, inductive reactance, and the capacitive reactance of the circuit. It is
the AC analogue to resistance in a DC circuit (p ).
28-37
Chapter 28 Alternating Current Circuits
Resonance
That condition in an AC circuit where the maximum current is obtained. It occurs
when the inductive reactance is equal to the capacitive reactance (p ).
Resonant frequency
The frequency of an AC circuit that causes resonance in the circuit. It depends on
the inductance and capacitance of the circuit (p ).
Power factor
The ratio of the power consumed in an AC circuit to the power applied to the circuit.
It is also equal to the cosine of the phase angle (p ).
RLC parallel circuit
An AC circuit in which the resistor, inductor, and capacitor are placed in parallel
with each other. In such a circuit, the voltages are all in phase but the currents are
out of phase with each other. The current in the capacitor leads the current in the
resistor by 900, whereas the current in the inductor lags the current in the resistor
by 900 (p ).
Transformer
A device, based on the principle of mutual induction, that changes electrical energy
at a particular voltage and current in the primary coil into a different voltage and
current in the secondary coil (p ).
Summary of Important Equations
AC voltage
E = E max sin ωt
V = Vmax sin(2πft)
(27.32)
(28.6)
AC current
i = imax sin ωt
i = imax sin(2πft)
(27.35)
(28.2)
Relation of angular speed and frequency of sine wave
ω = 2πf
Effective current in an AC circuit
ieff = 0.707imax
Effective voltage in an AC circuit
Veff = 0.707Vmax
(28.1)
(28.5)
(28.7)
Inductive reactance
XL = 2πfL
(28.21)
Capacitive reactance
XC =
(28.23)
28-38
1
2πfC
Chapter 28 Alternating Current Circuits
Z = R 2 + (X L − X C ) 2
Impedance of an AC series circuit
Ohm’s law for an AC circuit
(28.25)
(28.26)
Voltage drop across a resistor
i=V
Z
VR = iR
Voltage drop across an inductor
VL = iXL
(28.22)
Voltage drop across a capacitor
VC = iXC
(28.24)
Total voltage in an AC series circuit
V = V 2R + (V L − V C ) 2
(28.18)
Phase angle between voltage and
current in an AC series circuit
φ = tan−1VL − VC
VR
Resonant frequency of an AC series circuit
Power dissipated in an AC circuit
Power factor
Power factor
(28.20)
f0 =
(28.19)
1
2 LC
(28.28)
P = iV cos φ
(28.29)
PF = cos φ = R
Z
PF = Power consumed
Power supplied
(28.31)
I T = I 2R + (I C − I L )
Total current in a parallel RLC circuit
Current in resistor for parallel RLC circuit
Current in inductor for parallel RLC circuit
Current in capacitor for parallel RLC circuit
Phase angle for parallel RLC circuit
IR = V
R
IL = V
XL
IC = V
XC
φ = tan−1 IC − IL
IR
Total impedance for parallel RLC circuit
Z= V
IT
Transformer
E2 = i1 = N2
E1
i2
N1
28-39
(28.32)
2
(28.33)
(28.34)
(28.35)
(28.36)
(28.37)
(28.39)
(28.44)
Chapter 28 Alternating Current Circuits
Questions for Chapter 28
*1. When Americans visit Europe, they are told to get an adapter for their
personal appliances such as electric razors, hair driers, and the like. Why is this
necessary? What is the difference between European electricity and American
electricity?
2. What does doubling the frequency do to the inductive reactance? The
capacitive reactance?
3. Can you use the d’Arsonval galvanometer to measure AC currents or
voltages?
*4. How can you measure an AC current or voltage?
*5. What does increasing the area of the plates of a capacitor do to the
resonant frequency in an RLC series circuit? How is this used in a variable
capacitor of a tuned circuit?
6. What is the phase angle of an RLC series circuit when the inductive
reactance equals the capacitive reactance? What is the power factor in this case?
*7. How is an LC circuit like a mass attached to a vibrating spring? Compare
the equation for resonance for a vibrating spring and the LC circuit. Can you make
an analogue between them?
*8. How can an RLC circuit be used to filter signals of a special frequency?
*9. In many transformers, the wires are wrapped around an iron core rather
than a hollow air core. Why do you think this is done?
Problems for Chapter 28
28.2 The Effective Current and Voltage in an AC Circuit
1. The effective voltage in an AC circuit is measured as 50.0 V. What is the
maximum value of the voltage?
2. The peak-to-peak voltage of an AC voltage is measured on an oscilloscope
to be 100 V. What is the effective voltage?
3. The effective current in an AC circuit is measured to be 5.00 A. What is the
maximum value of the current?
4. An AC current varies from −5.65 A to +5.65 A. Find the effective value of
the current. What DC current would give the same effect?
28.3 An RLC Series Circuit
5. Find the inductive reactance of a 5.00-H coil when a 400-Hz AC voltage is
impressed across it.
6. At what frequency will a 50.0-mH inductor have a reactance of 800 Ω?
7. A 110-V, 60.0-Hz, AC line is connected to a 6.55-mH coil. Find the current
through the coil.
8. A coil has an impedance of 800 Ω and an inductive reactance of 600 Ω.
Find the resistance of the coil.
28-40
Chapter 28 Alternating Current Circuits
9. Find the capacitive reactance of a 10.0-µF capacitor at 60.0 Hz.
10. A 55.5-V, 400-Hz, AC line is connected to a 6-pF capacitor. Find the
current in the circuit.
11. At what frequency will the inductive reactance of a 2.00-H inductor be
(a) 20 Ω, (b) 200 Ω, and (c) 2000 Ω? Is it easier for an inductor to pass low-frequency
or high-frequency signals?
12. At what frequency will the capacitive reactance of a 5.00-µF capacitor be
(a) 10 Ω, (b) 100 Ω, and (c) 1000 Ω? Is it easier for a capacitor to pass low-frequency
or high-frequency signals?
13. A 2.00-mH inductor is connected to a 110-V, 60.0-Hz line. Find (a) the
inductive reactance and (b) the current through the inductor.
14. A 2.00-µF capacitor is connected to a 50.0-V, 40.0-Hz, AC line. Find the
current flowing in the capacitor circuit.
15. Find the impedance of a series circuit of R = 1000 Ω, L = 5.00 mH, and C
= 10.0 µF, if the AC source is at a frequency of 60.0 Hz.
16. A resistor R = 500 Ω, an inductor L = 20.0 mH, and a capacitor C = 6.00
µF are connected in series. Find the impedance if the source is 110 V at 400 Hz.
*17. An RLC series circuit has R = 1800 Ω, L = 4.89 mH, C = 4.78 µF, and f =
60.0 Hz. Find (a) the phase angle between the applied voltage and the current in
the circuit, (b) the phase angle between the voltage across the inductor and the
applied voltage, and (c) the phase angle between the voltage across the capacitance
and the applied voltage.
18. If the impedance of an RLC series circuit of R = 800 Ω, L = 50.0 mH, and
C = 3.00 µF is 1178 Ω, find the phase angle between the current in the circuit and
the applied voltage.
19. In an RLC series circuit, R = 200 Ω, C = 10.0 µF, and the frequency f =
70.0 Hz. What inductance would result in the potential across the RLC combination
leading the current by 30.00?
20. A 240-V, 50.0-Hz, AC line is connected in series with a resistor of 958 Ω
and a capacitor of 9.50 µF. Find (a) the impedance of the circuit and (b) the phase
angle of the circuit.
21. A 120-V, 50.0-Hz, AC line is connected to an RL series circuit of R = 280
Ω and L = 5.75 mH. Find the impedance of the circuit and the phase angle between
the applied voltage and the current in the circuit.
22. In an RLC series circuit, VL = 12.0 V, VC = 24.0 V, and R = 160 Ω. If the
potential across the RLC combination lags the current by 42.00, find the current in
the circuit.
23. A 400-Ω resistor is connected in series with an unknown inductor to a
110-V, 60.0-Hz, AC line. If the current in the circuit is 0.194 A, what is the value of
the inductance of the inductor?
28.4 Resonance in an RLC Series Circuit
24. A 2.00-mH inductor is connected in series with a 10.0-µF capacitor to an
AC line of variable frequency. At what frequency will resonance occur?
28-41
Chapter 28 Alternating Current Circuits
25. What value of inductance is necessary to tune an RLC circuit to a
frequency of 106.2 MHz if the capacitor has a value of 5.00 pF?
26. At what frequency is the inductive reactance equal to the capacitive
reactance if XL = 850 Ω and C = 6.00 µF?
27. In an RLC series circuit, E = 120 V, C = 15.0 pF, and the resonant
frequency is 50.0 Hz. (a) Find the inductance in the circuit. (b) If the current is 10.0
mA at resonance, find the resistance in the circuit. (c) If the resonant frequency is
changed to 60.0 Hz by varying the capacitance, find the new value of the
capacitance.
28. If R = 800 Ω, L = 50.0 mH, and C = 3.00 µF in an RLC series circuit of E =
120 V, plot the current I as a function of the frequency, for f = 100 to 800 Hz in
intervals of 100 Hz. Find the resonant frequency from the graph and compare with
the resonant frequency obtained from the equation.
29. A local FM radio station transmits at a frequency of 94.3 MHz. What
value of capacitance is needed in a LC series circuit of L = 1.00 H in order to have
resonance?
28.5 Power in an AC Circuit
30. A 5.00-A current flows in a series RLC circuit and leads the 120 V applied
voltage by 40.00. Find the power factor of the circuit and the power.
31. The power factor in an RLC series circuit is 0.80. If the resistance of the
circuit is 60.0 Ω, find the difference between the reactances, XL − XC.
28.6 An RLC Parallel Circuit
*32. A 220-V, 40.0-Hz, AC line is connected across a parallel RLC circuit of R
= 900 Ω, L = 7.00 H, and C = 8.00 µF. Find (a) the inductive reactance, (b) the
capacitive reactance, (c) the current through the resistor, (d) the current through
the inductor, (e) the current through the capacitor, (f) the total current in the
circuit, (g) the phase angle φ, and (h) the total impedance.
*33. A 110-V, 60.0-Hz, AC line is connected across a resistance of 1000 Ω,
which is in parallel with an inductor of 5.00 H. Find (a) the inductive reactance,
(b) the current IR through the resistor, (c) the current IL through the inductor,
(d) the total current in the circuit, (e) the phase angle φ, and (f) the total impedance
of the circuit.
*34. A 110-V, 60.0-Hz, AC line is connected across a 1.00-µF capacitor in
parallel with a 5.00-H inductor. Find (a) the capacitive reactance, (b) the inductive
reactance, (c) the current IC through the capacitor, (d) the current IL through the
inductor, (e) the total current IT in the circuit, (f) the phase angle φ, and (g) the total
impedance of the circuit.
*35. A 110-V, 60.0-Hz, AC line is connected across a resistance of 500 Ω in
parallel with a capacitor of 4.00 µF. Find (a) the capacitive reactance, (b) the
current through R, (c) the current through C, (d) the total current in the circuit,
(e) the phase angle φ, and (f) the total impedance of the circuit.
28-42
Chapter 28 Alternating Current Circuits
28.7 The Transformer
36. A transformer has 100 turns in the primary coil and 250 turns in
secondary. If the primary of the transformer is connected to an alternating voltage
generator of 400 V, what voltage will appear across the secondary?
37. A transformer has 100 turns in the primary coil and 250 turns in the
secondary. If there is an alternating current in the primary of the transformer of
3.00 A, what current will appear in the secondary?
38. If there are 200 turns of wire in the primary of a transformer connected to
a 120-V outlet, how many turns do you need in the secondary in order to produce
6.00 V there?
39. A transformer has a current in its primary winding of 2.50 A and a
current in its secondary winding of 0.750 A. (a) If there are 20 turns in the primary
coil, find the number of turns in the secondary coil. (b) If the resistance of the
secondary coil is 100 Ω, find the applied emf in the primary coil and the induced
emf in the secondary coil. (Assume 100% efficiency.)
40. Find the turns ratio for the transformer in the diagram such that a spark
is given off at the points AB. The distance from point A to point B is 5.00 mm.
Diagram for problem 40.
41. A power station generates 200 kW at 100 A and 2000 V. If the
transmission line has a resistance of 10.0 Ω, how much power will be lost in
transmission? If a step-up transformer of 1:10 is used, what will the power loss be
then?
Additional Problems
42. RC and RL circuits are sometimes used as timing circuits; show that the
units of RC and L/R have the unit of time.
*43. A 220-V, 40.0-Hz, AC line is connected across a series RLC circuit of R =
900 Ω, L = 7.00 H, and C = 8.00 µF. Find (a) the inductive reactance; (b) the
capacitive reactance; (c) the impedance of the circuit; (d) the current in the circuit;
(e) the voltage drop across R, L, and C; (f) the total voltage across the circuit; (g) the
phase angle φ; and (h) the resonant frequency.
*44. A 60.0-V, 100-Hz, AC line is connected across a series RLC circuit of R =
1000 Ω, L = 10.0 H, and C = 2.00 µF. Find (a) the inductive reactance, (b) the
capacitive reactance, (c) the impedance of the circuit, (d) the current in the circuit,
(e) the voltage drop across R, (f) the voltage drop across L, (g) the voltage drop
across C, (h) the total voltage drop across the circuit, (i) the phase angle, (j) the
power factor, and (k) the resonant frequency.
28-43
Chapter 28 Alternating Current Circuits
*45. A 110-V, 60.0-Hz, AC line is connected across a resistor of 1800 Ω, in
series with a capacitor of 3.00 µF. Find (a) the capacitive reactance, (b) the
impedance, (c) the current in the circuit, (d) the voltage drop across R, (e) the
voltage drop across C, (f) the total voltage measured across R and C in series, (g) the
phase angle φ, and (h) the power factor.
*46. A 120-V, 60.0-Hz, AC line is connected across a series circuit of a
resistance of 500 Ω and an inductor of 10.0 H. Find (a) the inductive reactance,
(b) the impedance, (c) the current in the circuit, (d) the voltage drop across R, (e) the
voltage drop across L, (f) the total voltage drop across the circuit, (g) the phase
angle φ, and (h) the power factor.
*47. Three inductors are connected in series to an AC source as shown in the
diagram. The inductors are shielded so that a changing flux in one inductor does
not cause a mutual inductance in another inductor. Show that the equivalent
inductance of the inductors in series is given by
L = L 1 + L2 + L3
Diagram for problem 47.
Diagram for problem 48.
*48. Three shielded inductors are connected in parallel as shown in the
diagram. The inductors are shielded so that a changing flux in one inductor does
not cause a mutual inductance in another inductor. Show that the equivalent
inductance of the inductors in parallel is given by
1 = 1 + 1 + 1
L
L1 L2 L3
*49. In the circuit shown R = 1300 Ω, C1 = 5.00 µF, C2 = 8.00 µF, L = 6.78
mH, and the applied voltage is 110 V at 60.0 Hz. Find the current through each
circuit element.
28-44
Chapter 28 Alternating Current Circuits
Diagram for problem 49.
Diagram for problem 50.
*50. In the circuit shown R1 = 978 Ω, R2 = 560 Ω, L1 = 6.78 mH, L2 = 3.25 mH,
C = 5.00 µF, and the applied voltage is 110 V at 60.0 Hz. Find the current through
each circuit element.
*51. In the circuit shown R1 = 535 Ω, R2 = 350 Ω, L1 = 3.25 mH, L2 = 2.56 mH,
C1 = 5.00 µF, C2 = 7.50 µF, and the applied voltage is 110 V at 60.0 Hz. Find the
current through each circuit element.
Diagram for problem 51.
Diagram for problem 52.
52. The diagram is an example of a low-pass filter. Show that very low-input
frequencies pass through the circuit while high frequencies do not. That is, show
that high-output voltages are obtained across AB for low-input frequencies, while
very low-output voltages are obtained for high-input frequencies.
53. The diagram is an example of a high-pass filter. Show that very highinput frequencies pass through the circuit while low frequencies do not. That is,
show that high-output voltages are obtained across AB for high-input frequencies,
while very low-output voltages are obtained for low-input frequencies.
28-45
Chapter 28 Alternating Current Circuits
Diagram for problem 53.
Diagram for problem 54.
*54. Find the turns ratio between the primary and secondary coils in the
diagram, to produce a spark across AB, a distance of 5.00 mm, and find the value of
C to give a spark rate of 5.00 Hz, 10.0 Hz, and 30.0 Hz. The value of the inductance
is 5.00 H, R = 500 Ω, and the peak voltage is 1200 V.
Interactive Tutorials
55. Effective current in an AC circuit. An alternating current of frequency f =
60 Hz varies from −2.54 A to 2.54 A. Find the effective value of this current. Plot the
current and show the effective current on the plot.
56. RLC series circuit. An RLC series circuit has a resistance R = 800 Ω,
inductance L = 2.00 H, capacitance C = 5.55  10−6 F, and they are connected to an
applied voltage V = 110 V, operating at a frequency f = 60 Hz. Find (a) the inductive
reactance XL, (b) the capacitive reactance XC, (c) the impedance Z of the circuit,
(d) the current i in the circuit, (e) the voltage drop VR across the resistor, (f) the
voltage drop VL across the inductor, (g) the voltage drop VC across the capacitor,
(h) the total voltage drop across RLC, (i) the phase angle φ, and (j) the resonant
frequency f0. (k) Plot the curves of the voltage across R, L, and C individually and
the voltage across the combination. Show the phase relations for all these voltages.
57. Resonance. An RLC series circuit has a resistance R = 800 Ω, inductance
L = 2.00 H, capacitance C = 5.55  10−6 F, and they are connected to an applied
voltage V = 110 V, operating at a frequency f = 60 Hz. Plot the current flow i as a
function of the frequency f of the AC source. From this graph pick out the resonant
frequency of the circuit and compare it to the calculated value for the resonant
frequency.
58. RLC parallel circuit. An RLC parallel circuit has a resistance R = 800 Ω,
inductance L = 2.00 H, capacitance C = 5.55  10−6 F, and they are connected to an
applied voltage V = 110 V, operating at a frequency f = 60 Hz. Find (a) the inductive
reactance XL, (b) the capacitive reactance XC, (c) the current IR through the resistor,
(d) the current IL through the inductor, (e) the current IC through the capacitor,
28-46
Chapter 28 Alternating Current Circuits
(f) the total current IT in the circuit, (g) the phase angle φ, and (h) the impedance Z.
(i) Plot the curves of the current IR, IL, and IC individually and the total current IT.
Show the phase relations for all these currents.
59. The transformer. A 120-V line from an AC generator is connected to the
primary of a transformer of N1 = 150 turns. (a) If the secondary of the transformer
has N2 = 25 turns, find the voltage V2 that will be found across the secondary of the
transformer. (b) If the current in the primary i1 = 0.05 A, find the current i2 in the
secondary.
To go to these Interactive Tutorials click on this sentence.
To go to another chapter, return to the table of contents by
clicking on this sentence.
28-47