QUANTITATIVE TOOLS

CONCEPTS

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32

32.1 Alternating currents
32.2 AC circuits
32.3 Semiconductors

32.4 Diodes, transistors, and logic gates

32.5 Reactance
32.6 RC and RLC series circuits
32.7 Resonance
32.8 Power in AC circuits

Electronics


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32.1 AlternAting currents

I

n the preceding chapter, we discussed electric circuits
in which the current is steady. As noted in that chapter,
the steady flow of charge carriers in one direction only
is called direct current. Batteries and other devices that produce static electrical charge, such as van de Graaff generators, are sources of direct current. Although direct current

has many uses, it has several limitations as well. For example, in order to produce substantial currents, direct-current
sources must be quite large and are therefore cumbersome. More important, steady currents do not generate any
electromagnetic waves, which can be used to transmit information and energy through space, as we saw in Chapter 30.
Because of these and many other factors, most electric
and electronic circuits operate with alternating currents
(abbreviated AC)—currents that periodically change direction. The current provided by household outlets in the
United States, for instance, alternates in direction, completing 60 cycles per second (that is, with a frequency of 60 Hz),
and the currents in computer circuits change direction billions of times per second. It is no understatement to say that
contemporary society depends on alternating currents.
In this chapter we discuss the basics of both household currents and the electronics that lie at the heart of
computers.

32.1 Alternating currents

Figure 32.1 What happens when we connect an inductor to a charged
capacitor?

fully charged capacitor

E

inductor

32.1 (a) Just before the inductor is connected to the
charged capacitor, what type of energy is contained in the system comprising the two elements? (b) Once the two elements
are connected to each other, what happens to that energy? (c)
Once the capacitor is completely discharged, in what form is the
energy in the circuit?

As you saw in Checkpoint 32.1, when the capacitor is completely discharged, all of the energy in an LC circuit is contained in the magnetic field and this field reaches its maximum magnitude (Figure 32.2c). Because the magnetic energy

is proportional to the square of the current in the inductor
(Eq. 29.25), the current, too, reaches its maximum value at this
instant. Once the magnetic field and the current reach their
maximum values, the current begins to charge the capacitor
in the opposite direction (Figure 32.2d), and the charge on
the capacitor increases as the magnetic field in the inductor
decreases. When the magnetic field in the inductor is zero,
the current is also zero and the capacitor has again maximum
charge but with the opposite polarity (Figure 32.2e). The process then repeats itself with the current in the opposite direction (Figure 32.2f–h) until the capacitor is restored to its starting configuration. Then the cycle begins again.
Figure 32.3 on the next page shows the time dependence of
the electric potential energy U E stored in the capacitor and the
magnetic potential energy U B stored in the inductor. In the absence of dissipation, the energy in the circuit, U E + U B, must
stay constant. Therefore, when the capacitor is not charged and
U E drops to zero, U B must reach its maximum value, Umax.
There is always some dissipation in a circuit. Resistance
in the connecting wires gradually converts electrical energy
to thermal energy. Consequently, the oscillations decay in
the same manner as the damped mechanical oscillations we
considered in Section 15.8. Resistance therefore plays the
same role in oscillating circuits as damping does in mechanical oscillators.
Throughout this chapter we work with time-dependent
potential differences and currents. To make the notation as
concise as possible, we represent time-dependent quantities
with lowercase letters. In other words, vC is short for VC(t)
and i is short for I(t). We also need a symbol for the maximum value of an oscillating quantity—its amplitude (see
Section 15.1). For this we use a capital letter without the
time-dependent marker (t); thus VC is the maximum value
of the potential difference across a capacitor, and I is the
maximum value of the current in a circuit.
Unlike their counterparts in DC circuits, the potential

difference across the capacitor, vC, and the current in the LC
circuit, i, change sign periodically. So, when analyzing AC
circuits, we must carefully define what we mean by the sign
of these quantities. To analyze the LC circuit in Figure 32.2,
for example, we choose a reference direction for the current
i and let the potential difference vC be positive when the top
capacitor plate is at a higher potential than the bottom plate
(Figure 32.4a on page 863). Note that both of these choices
are arbitrary.

CONCEPTS

We have already encountered one example of an electrical
device that produces a changing current: a capacitor that
is either charging or discharging. Let’s consider what happens when we connect an inductor to a charged capacitor
(Figure 32.1). A circuit that consists of an inductor and a capacitor is called an LC circuit. As soon as the two circuit elements are connected, positive charge carriers begin to flow
clockwise through the circuit. The magnitude of the current
increases from its initial value of zero (Figure 32.2a on next
page) to a nonzero value (Figure 32.2b–d). The capacitor
discharges through the inductor, and the current causes a
magnetic field in the inductor. As the current in the inductor increases, the magnetic field also increases, causing an
induced emf (see Section 29.7) that opposes this increase
and prevents the current from increasing rapidly. Consequently, the capacitor discharges more slowly than it would
if we had connected it to a wire.

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Figure 32.2 A series of “snapshots” showing what happens when we connect an inductor to a charged capacitor.
I=0

E

I

E

I

(a)

B

E

B

(b)

(h)
I

I

B


B

(c)

(g)
I

I

E

B

(f)

I=0

E

B

(d)

CONCEPTS

E

(e)


Figure 32.3 Time dependence of the electric potential energy U E stored
in the capacitor and the magnetic potential energy U B stored in the inductor.
In the absence of dissipation, the energy in the circuit, U E + U B, is a
constant Umax.
electric potential energy
stored in capacitor
U

magnetic potential energy
stored in inductor

U E(t)

U B(t)

energy stored in circuit (U E + U B)

Umax

0
0

T

2T

t

Figure 32.4b shows graphs for vC and i with these choices. The potential difference across the capacitor vC is initially positive, representing the situation at Figure 32.2a.
During the first quarter cycle (Figure 32.2b), the capacitor

is discharging and positive charge carriers travel away from
the top plate of the capacitor in the chosen reference direction, and so i is positive. In the part of the cycle represented
by Figure 32.2f, where the capacitor is again discharging,
vC is negative (because the top plate is negatively charged)
and i is negative (because the direction of current is opposite the chosen reference direction), as shown in the time
interval 12T 6 t 6 34T in Figure 32.4b. (See if you can work
out the signs during the time intervals when the capacitor is


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863

32.2 Ac circuits
Figure 32.4 For the LC circuit shown in Figure 32.2, graphs of the timedependent potential difference across the capacitor (defined to be positive
when the top plate is at the higher potential) and the current in the circuit
(defined to be positive when positive charge carriers travel away from top
plate of the capacitor). One cycle is completed in a time interval T (the
period).
(a)

reference direction for current

vC is positive when
top plate is at higher
potential.

(b)

i
C


Exercise 32.1 AC source and resistor
Figure 32.6 shows a circuit consist-

ing of an AC source and a resistor.
The emf produced by the generator
varies sinusoidally in time. Sketch
the potential difference across the
resistor as a function of time and
the current in it as a function of
time.

Figure 32.6 Exercise 32.1.

ℰ

R

Solution Ohm’s law, the junction
rule, and the loop rule (see Chapter 31) apply to alternatingcurrent circuits just as they do to direct-current circuits. All I
need to remember here is that the potential differences and currents are time dependent. Applying the loop rule to this circuit
requires the time-dependent potential difference across the resistor vR to equal the emf ℰ of the AC source at every instant, so that
the sum of the potential differences around the circuit is always
zero. Consequently, vR oscillates just as ℰ oscillates, as shown in
Figure 32.7; VR is the maximum value of the potential difference
across the resistor.

L

vC


+VC

Figure 32.7
T

2T

t

−VC
i
+I
T

2T

t

−I

Ohm’s law requires the time-dependent current i in the resistor to be proportional to vR, which means that i also oscillates,
with the current at its maximum when vR is maximum. ✔

32.2 (a) Is energy dissipated in the resistor in the circuit
of Figure 32.6? (b) If so, why doesn’t the amplitude of the oscillations of vR and i (shown in Figure 32.7) decrease with time?

32.2 AC circuits
Figure 32.5 Symbol that represents an AC source in an electric circuit.
The AC source produces a sinusoidally varying emf ℰ across its terminals.

ℰ

The circuit discussed in Exercise 32.1 is an alternating
current, or AC circuit. Such circuits exhibit more complex
behavior when they contain elements that do not obey
Ohm’s law, so that the current is not proportional to the emf
of the source. For example, let’s consider the current in the
circuit shown in Figure 32.8 on the next page.

CONCEPTS

charging.) Both vC and i vary sinusoidally in time, with vC
at its maximum when i is zero, and vice versa.
Because of dissipation, the LC circuit in Figure 32.1 is not
a practical source of alternating current; instead, generators
are widely used to produce sinusoidally alternating emfs in
a circuit (see Example 29.6). The symbol for a source that
generates a sinusoidally alternating potential difference or
current is shown in Figure 32.5; such a source is called an
AC source. The time-dependent emf an AC source produces across its terminals is designated ℰ, and its amplitude is
designated ℰmax.


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Figure 32.8 AC circuit with a capacitor connected to an AC source.
reference direction for current

vC is positive when top
plate is at higher potential.

i
generator

ℰ

C

Figure 32.10 Time-dependent current in the circuit and potential difference across the capacitor for the circuit of Figure 32.9.
vC maximum,
current zero

vC minimum,
current zero

vC decreasing,
current negative

vC increasing,
current positive

capacitor
vC

i

To analyze the circuit we choose a reference direction
for the current i and let the potential difference vC again be

positive when the top capacitor plate is at a higher potential
than the bottom plate (Figure 32.8). Because the capacitor
is connected directly to the AC source, the time-dependent
potential difference across the capacitor vC equals the emf
of the AC source at any instant. What is the current in the
circuit? Let’s begin considering what happens when the capacitor is uncharged. As vC increases, the charge on the top
plate of the capacitor increases. This means that positive
charge carriers are moving toward the top plate, in the same
direction as the chosen reference direction for the current,
and so the current is positive (Figure 32.9a). When vC reaches its maximum, the capacitor reaches its maximum charge
and the current is instantaneously zero. As vC decreases, the
charge on the top plate of the capacitor decreases. Positive charge carriers now move away from the top plate and
the current is negative (Figure 32.9b). At some instant the top
plate becomes negatively charged (Figure 32.9c); vC continues to decrease until it reaches its minimum value and the
current is instantaneously zero. At that instant the capacitor
again reaches its maximum charge but with the opposite

CONCEPTS

Figure 32.9 The charging and discharging of the capacitor in the circuit
of Figure 32.8.
capacitor discharging, i 6 0

capacitor charging, i 7 0
(a)

(c)

i


(b)

i

vC positive,
increasing

vC positive,
decreasing

capacitor charging, i 6 0

capacitor discharging, i 7 0

i

vC negative,
decreasing

(d)

i

vC negative,
increasing

T

t


polarity. As vC begins to increase again, positive charge
carriers flow toward the top plate and the current is positive again (Figure 32.9d). When both plates are uncharged
again, the cycle is complete.
Figure 32.10 shows the time dependence of i and vC in
Figure 32.9. Note that i and vC are not simply proportional
to one another. Instead, the current maximum occurs onequarter cycle before the potential difference maximum.
For this reason, the current is said to lead the potential
difference:
In an AC circuit that contains a capacitor, the current in the capacitor leads the potential difference by
90° (a quarter of an oscillation cycle).

To describe the time dependence of a sinusoidally oscillating quantity, we must specify both the angular frequency of oscillation v and the instant at which the oscillating
quantity equals zero. As discussed in Chapter 15, a sinusoidally time-dependent quantity (such as the circuit potential
difference we are looking at here) can be written in the form
v = Vsin(vt + fi). The argument of the sine, vt + fi, is
the phase. At t = 0 the phase is equal to the initial phase
fi (Chapter 15). When the phase of an oscillating quantity
is zero, vt + fi = 0, the quantity is zero as well because
sin(0) = 0.
We can analyze phase differences in AC circuits with lots
of algebra, but the underlying physics is much clearer (and
the analysis much simpler!) if we use the phasor notation
developed in Chapter 15 to describe oscillatory motion.
Following the approach of Section 15.5, we can represent
an oscillating potential difference v by a phasor rotating in
a reference circle (Figure 32.11). Because the length of the
phasor equals the amplitude (maximum value) of v, the
phasor is labeled V. The phasor rotates counterclockwise at
angular frequency v. The magnitude of v at any instant is
given by the vertical component of the phasor; as the phasor

rotates, that component oscillates sinusoidally in time, as
shown in Figure 32.11. The angle measured counterclockwise from the positive horizontal axis to the phasor is the
phase vt + fi.


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32.2 Ac circuits

865

Figure 32.11 Phasor representation of a sinusoidally varying potential difference v. The phasor rotates
counterclockwise at the same angular frequency at which v oscillates. The instantaneous value of v equals
the length of the vertical component of the phasor.
f(t1) = vt1 + fi (phase at instant t1)

phasor representing v
v

v

V
V

V sin f(t1)

V sin fi
fi (phase at instant t = 0)

t1


T

t

v

Example 32.2 Phasors
Consider the oscillating emf represented in the graph of

Figure 32.12. Which of the phasors a–d, each shown at t = 0,

correspond(s) to this oscillating emf?

Figure 32.12 Example 32.2.
ℰ
ℰmax

c

a

b

d

ℰi

T

t


32.3 Construct a phasor diagram for the time-dependent
current and potential difference at t = 0 in the AC sourceresistor circuit of Figure 32.6.

Figure 32.13 Phasor diagram and graph showing time dependence of vR
and i from Figure 32.7.
vR

VR
I

i
vt1
t1

T

t

CONCEPTS

❶ GettinG Started I begin by observing from the graph that
the emf is negative at instant t = 0 and increases until it reaches
a maximum value ℰmax.
❷ deviSe plan To identify the correct phasor or phasors, I
can use the following information: (1) the length of the phasor
is equal to the amplitude of the oscillation, (2) the value of the
emf at any instant corresponds to the vertical component of
the phasor, and (3) the phasor rotates counterclockwise around
the reference circle.

❸ execute plan The amplitudes of phasors a and b are too
small and so I can rule these two out. The fact that the emf starts
out negative at t = 0 and then increases tells me that the phasor
representing it must be in the fourth quadrant (below the horizontal axis and to the right of the vertical axis), meaning the correct phasor must be d. ✔
❹ evaluate reSult I can verify my answer by tracing out the
projection of the phasor on the vertical axis as the phasors rotates counterclockwise. The initial value of the projection, initial
phase, and amplitude all agree with the values of these variables
represented in the graph.

We can generalize the result of this checkpoint to represent i and vR from Figure 32.7 at an arbitrary instant t1.
Because i and vR are in phase for a resistor, the two phasors for i and vR always have the same phase and so overlap
(Figure 32.13). Note that the initial phase fi is zero because
i and vR are zero at t = 0 (at that instant both phasors point
to the right along the horizontal axis).
The relative lengths of the I and VR phasors are meaningless because the units of i and vR are different. However,
for circuits with multiple elements (resistors, inductors, or
capacitors), the relative lengths of phasors showing the potential differences across different elements are meaningful
and will prove very useful in analyzing the circuit.
Phasors are most useful when we need to represent quantities that are not in phase. Figure 32.14 on the next page
shows the phasor diagram that corresponds to Figure 32.10
(at the instant represented by Figure 32.9a). As the phasor
diagram shows, the angle between VC and I is 90°, and so the
phase difference between the two phasors is p>2. Because
the phasors rotate counterclockwise, we see that current
phasor I is ahead of the potential difference phasor VC, in
agreement with our earlier conclusion that the current in a
capacitor leads the potential difference across the capacitor.


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Figure 32.14 Phasor diagram and graph showing time dependence of i
and vC corresponding to Figure 32.10.

reference direction for current

vC

VC
v t1

I

Figure 32.17 AC circuit consisting of an inductor connected across the
terminals of an AC source.

i
t1

vL is positive when upper
end is at higher potential.

i
T

ℰ


L

Example 32.3 Nonsinusoidal AC circuit
When a certain capacitor is connected to a nonsinusoidal source
of emf as in Figure 32.15a, the emf varies in time as illustrated in
Figure 32.15b. Sketch a graph showing the current in the circuit
as a function of time.
Figure 32.15 Example 32.3.
(a)

(b)
ℰ

CONCEPTS

ℰ

C

1

2

3

4

5

6


7

t (ms)

❶ GettinG Started From Figure 32.15b I see that the emf has
five distinct parts during the time interval shown. During each
part, the emf either is changing at a constant rate or is constant.
❷ deviSe plan I know that the current is proportional to the
rate at which the charge on the capacitor plates changes over
time. I also know that the emf is proportional to the charge on
the plates, and so the current is proportional to the derivative of
the emf with respect to time.
❸ execute plan Between t = 0 and t = 1 ms, the emf increases at a constant rate, so i = Cd ℰ>dt is constant and positive. Between t = 1 ms and t = 2 ms, the emf is constant,
so i = Cd ℰ>dt = 0. Between t = 2 ms and t = 4 ms, the
emf decreases at a constant rate, so i = Cd ℰ>dt is constant and
negative. Because the rate of decrease between t = 2 ms and
t = 4 ms is the same as the rate of increase between t = 0
and t = 1 ms, the magnitude of the current between t = 2 ms
and t = 4 ms should be the same as that between t = 0 and
t = 1 ms. The current is zero again during the next millisecond
(t = 4 ms to t = 5 ms) because here the emf is again constant.
After t = 5 ms, the emf increases again at the same constant
rate as between t = 0 and t = 1 ms, so the current has the same
positive value as between t = 0 and t = 1 ms. The graph representing these current changes is shown in Figure 32.16. ✔
Figure 32.16

❹ evaluate reSult When the current is positive, the emf is
increasing; when the current is negative, the emf is decreasing;
and when the current is zero, the emf is constant, as it should be.


Now let’s examine the behavior of an inductor connected to an AC generator (Figure 32.17). When the current
in the circuit is changing, an emf is induced in the coil,
in a direction to oppose this change (see Section 29.7).
The potential difference between the ends of the inductor,
which we’ll denote by vL, is proportional to the rate di>dt
at which the current changes (Eq. 29.19). If the current is
increasing in the reference direction for current indicated
in Figure 32.17, the upper end of the inductor must be at a
higher potential than the lower end to oppose the increase
in current. If we take vL to be positive when the upper end
of the coil is at a higher potential, vL must therefore be
positive when the current is increasing in the reference
direction for the current. This situation is represented in
Figure 32.18a.
When the current reaches its maximum value in the
cycle, vL is instantaneously zero. After this instant, the current begins to decrease and the lower end of the inductor

Figure 32.18 Current and magnetic field oscillations through the inductor of Figure 32.17.
i positive, increasing (di>dt 7 0)
(a)

i

i positive, decreasing (di>dt 6 0)

B

(b)


i

vL positive

vL negative

i negative, decreasing (di>dt 6 0)

i negative, increasing (di>dt 7 0)

(c)

(d)

i

vL negative

B

i

vL positive

B

B


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32.3 semiconductors

Figure 32.21 Schematic depiction of silicon, phosphorus, and boron
atoms, shown as an inner core surrounded by valence electrons.

Figure 32.19 Graph of time-dependent current in the circuit and potential difference across the inductor for the circuit in Figure 32.17.
current maximum,
vL zero

current minimum,
vL zero

current decreasing,
vL negative

(a) silicon

(b) phosphorus

(c) boron

no. inner electrons:

10

10

2


no. valence electrons:

4

5

3

core: nucleus plus
inner electrons

current increasing,
vL positive

valence electron

vL

i
t

T

32.3 Semiconductors

must be at a higher potential than the upper end to oppose
this decrease in current. The potential difference vL is now
negative (Figure 32.18b). In the second half of the cycle,
the current is in the opposite direction. As in the first part

of the cycle, vL has the same sign as di/dt (Figure 32.18c
and d).
Figure 32.19 illustrates the time dependence of i and vL in
Figure 32.18. Note that the current maximum occurs onequarter cycle after the potential difference maximum. For
this reason, the current is said to lag the potential difference:
In an AC circuit that contains an inductor, the
current in the inductor lags the potential difference
by 90°.

32.4 What are the initial phases for the phasors in Figures
32.13 and 32.20?

Figure 32.22 Schematic of a crystalline lattice of silicon atoms, showing
electrons participating in silicon-silicon bonds. (A real silicon crystal
exists in three dimensions, and not all of the silicon-silicon bonds lie in a
plane; this diagram illustrates only the essential idea that all of the valence
electrons participate in covalent bonds.)

Figure 32.20 Phasor diagram and graph showing time dependence of i
and vL corresponding to Figure 32.19.
vL

VL
I

i
vt1
t1

T


t

silicon core

bound electron

CONCEPTS

Figure 32.20 shows the phasor diagram that corresponds
to Figure 32.19 (at the instant represented by Figure 32.18a).
Just as with the capacitor, the angle between VL and I is 90°
and so the phase difference is p>2, but in this case the current phasor I is behind the potential difference phasor VL,
in agreement with our earlier conclusion that the current in
an inductor lags the potential difference across the inductor.

Most modern electronic devices are made from a class of
materials called semiconductors. Semiconductors have
a limited supply of charge carriers that can move freely;
consequently, their electrical conductivity is intermediate
between that of conductors and that of insulators. Semiconductors are widely used in the manufacture of electronic
devices such as transistors, diodes, and computer chips because their conductivity can be tailored chemically for particular applications layer by layer, even within a single piece
of semiconductor.
Semiconductors are of two main types: intrinsic and
extrinsic. Intrinsic semiconductors are chemically pure and
have poor conductivity. Extrinsic or doped semiconductors
are not chemically pure, have a conductivity that can be
finely tuned, and are widely used in the microelectronics
industry. The most widely used semiconductor is silicon, a
nonmetallic element that makes up more than one-quarter

of Earth’s crust. Figure 32.21a shows a schematic of a silicon
atom, which consists of a nucleus surrounded by fourteen
electrons. Ten of these electrons are tightly bound to the
nucleus—we’ll refer to these electrons plus the nucleus as
the core of the atom. The remaining outermost four electrons are called the atom’s valence electrons. Each valence
electron can form a covalent bond with a valence electron
of another silicon atom. These bonds hold many identical
silicon atoms together in a crystalline lattice (Figure 32.22).


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Figure 32.23 Schematic depiction of a crystalline lattice of silicon atoms
doped with phosphorus atoms. The only charge carriers that are free to
move in the crystal are the free electrons supplied by the phosphorus
dopant atoms.

silicon core

phosphorus core

silicon core

free electron

The electrons in a covalent bond are not free to move; consequently, pure silicon has a very low electrical conductivity
because all of its valence electrons form covalent bonds.

In extrinsic silicon, other types of atoms, such as boron
or phosphorus, replace some of the atoms in the silicon
lattice, introducing freely moving charge carriers into the
lattice. The substituted atoms are called either impurities
or dopants. For example, phosphorus has five valence electrons (Figure 32.21b). Because the silicon lattice structure
requires only four bonds from each atom, the fifth electron
from a phosphorus atom dopant is not involved in a bond
and is free to move through the solid (Figure 32.23).
If an electric field is applied to the doped semiconductor
of Figure 32.23, the free electrons move, creating a current
in the semiconductor (Figure 32.24). As free electrons leave
the semiconductor from one side, other free electrons enter
it on the opposite side. Because the semiconductor must remain electrically neutral, the number of free electrons in the
semiconductor at any given instant is always the same and it
is equal to the number of phosphorus atoms in the material.
If boron atoms, which have three valence electrons
(Figure 32.21c), are substituted for some silicon atoms in a
CONCEPTS

Figure 32.25 Schematic of crystalline lattice of silicon atoms with some
boron atoms substituted for silicon, showing both bonding electrons and
holes (missing electrons). The only free charge carriers in the crystal are
the holes caused by the boron impurities.

Figure 32.24 In an applied electric field, the free electrons in a phosphorous-doped semiconductor are free to move in the direction opposite the
field direction. Free electrons leave the semiconductor at the left, travel
through the circuit wire, and enter the semiconductor at the right.

boron core


silicon lattice, the “missing” fourth electron at each boron
leaves behind what is called a hole—an incomplete bond
(Figure 32.25). These holes behave like positive charge carriers
and are free to move through the lattice (Figure 32.25). The
holes therefore increase the ability of the silicon to conduct current, just as do the free electrons in phosphorus-doped silicon.
Keep in mind that the motion of holes involves electrons
moving to fill existing holes, leaving new holes in the previous positions of the electrons (Figure 32.26). The boron
Figure 32.26 Sequence of four snapshots showing how holes “move”
through a crystal by trading places with bonding electrons. In the presence
of an electric field, holes move in the direction of the field (opposite to
the directions in which the electrons move). To maintain continuity, free
electrons from attached metal wires enter at the right, recombining with
holes that accumulate there, and leave at the left.
S

E batt
hole

bound electron

battery terminal

Electron jumps to position of hole…

Second electron jumps to that hole…

S

E batt


hole

…leaving new hole.

…leaving new hole.

Effect is as though hole itself moves.
electrons out

electrons in
electrons out

motion of electrons in lattice

electrons in
motion of electrons

motion of holes


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32.4 diodes, trAnsistors, And logic gAtes

cores do not move! In the presence of an electric field, the
positively charged holes move in the direction of the field
as the negatively charged electrons move in the opposite direction. If the semiconductor is attached to metal wires on
either side, as in Figure 32.26, free electrons travel into the
semiconductor from the right (eliminating holes that reach
the right edge) and travel out of the semiconductor on the
left (producing holes on the left edge). Electrons thus flow

from right to left, making holes travel in the opposite direction. Unlike the electrons, however, the holes never leave the
semiconductor.
Doped semiconductors are classified according to the
nature of the dopant. In a p-type semiconductor, the dopant
has fewer valence electrons than the host atoms, contributing positively charged holes as the free charge carriers (thus
the p in the name). In an n-type semiconductor, the dopant
has more valence electrons than the host atoms, contributing negatively charged electrons as the free charge carriers
(thus the n in the name). Substituting as few as ten dopant
atoms per million silicon atoms produces conductivities appropriate for most electronic devices.
32.5 Is a piece of n-type silicon positively charged, negatively charged, or neutral?

869

(a) Pieces of p- and n-type doped silicon
p-type

n-type

electrically neutral

electrically neutral

(b) When the two are put in contact, a diode is formed
electron and hole
recombine

S

Edepl


electric field due
to recombination
in depletion zone
electrically neutral

electrically neutral

depletion zone: insulator
(c) Battery connected so as to produce electric field in same direction as
electric field in depletion zone; diode blocks current

S

Ebatt

32.4 Diodes, transistors, and logic gates

S

Edepl

electrically neutral

electrically neutral

Electric field due to battery broadens
depletion zone, so diode blocks current.
(d) Battery connected with the opposite polarity; diode conducts current

CONCEPTS


Although tailoring the conductivity of a single piece of
semiconductor can be a useful procedure, the most versatile
semiconductor devices combine doped layers that have different types of charge carriers. The simplest such device is a
diode, made by bringing a piece of p-type silicon into contact with a piece of n-type silicon (Figure 32.27a). Near the
junction where the two pieces meet, free electrons from the
n-type silicon wander into the p-type material, where they
end up filling holes. This recombination process turns free
electrons into bound electrons (that is, electrons not free to
roam around in the material) and eliminates the holes. Likewise, some of the holes in the p-type silicon wander into the
n-type silicon, where they recombine with free electrons.
As recombination events take place, a thin region containing no free charge carriers (neither free electrons nor
holes), called the depletion zone, develops at the junction.
Although there are no free charge carriers in this zone, the
trapping of electrons on the p-side of the junction causes
negative charge carriers that are nonmobile to accumulate
there. Similarly, positive nonmobile charge carriers accumulate on the n-side of the junction. As a result, the depletion
zone consists of a negatively charged region and a positively

S

Ebatt

electrons out

electrons in

Electric field due to battery eliminates
depletion zone, so diode conducts current.


Figure 32.27 How a diode transmits current in one direction but blocks it in the other. If the battery is
connected as shown in part d and produces a sufficiently strong electric field to compensate for the field
of the depletion zone, there is a steady flow of both electrons and holes. (Remember, though: The holes
never leave the semiconductor. Only the electrons enter and leave the semiconductor.)


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chApter 32  electronics

charged region, and an electric field points across the depletion zone from the n-side to the p-side (Figure 32.27b).
As this electric field in the depletion zone of the diode
increases, it becomes more difficult for free electrons and
holes to cross the junction and recombine because the electric field pushes free electrons back into the n-type silicon
and pushes holes back into the p-type silicon. Consequently, the depletion zone stops growing. Typically this region
is less than a micrometer wide. Because of the lack of free
charge carriers in it,
the depletion zone acts as an electrical insulator.

If we now connect the n-side of this diode to the positive
terminal of a battery and the p-side to the negative terminal,
the battery produces across the diode an electric field that
points in the same direction as the electric field in the depletion zone (Figure 32.27c). The electric field of the battery
pulls free electrons in the n-type silicon toward the positive terminal and pulls holes in the p-type silicon toward the
negative terminal, broadening the (nonconducting) depletion zone. Connecting the battery in this manner therefore
causes no flow of charge carriers in the diode.
When the battery is connected in the opposite direction,
however, the depletion zone narrows as the battery’s electric
field pushes free electrons and holes toward the junction

(Figure 32.27d). When the magnitude of the applied electric
field created by the battery equals that of the electric field
across the depletion zone, both types of free charge carriers
can reach the junction, resulting in a current in the device
carried both by free electrons and by holes.
As Figure 32.27 shows, a diode conducts current in one
direction only: from the p-type side to the n-type side. The
symbol for a diode is shown in Figure 32.28a; the triangle
points in the direction in which the diode conducts current
(from the p-side to the n-side).

CONCEPTS

32.6 In the diode of Figure 32.28a, which way do holes
travel? Which way do electrons travel?

An ideal diode acts like a short circuit for current in the
permitted direction and like an open circuit for current in
the opposite direction. (That is not exactly how a diode behaves, but it’s pretty close.)
32.7 Suppose a sinusoidally varying potential difference is
applied across a diode connected in series with a resistor. Sketch
the potential difference across the diode as a function of time,
and then, on the same graph, sketch the current in the resistor
as a function of time.
Example 32.4 Rectifier
Consider the arrangement of ideal diodes shown in Figure 32.29.
This arrangement, called a rectifier, converts alternating current (AC) to direct current (DC). Sketch a graph showing, for a
sinusoidally alternating source, the current in the resistor in the
direction from b to c as a function of time.
Figure 32.29 Example 32.4.

a
1

ℰ

2
R

b

c

4

3
d

❶ GettinG Started Because the source is alternating, the current in the circuit periodically reverses direction. During part of
the cycle the charge carriers creating the current flow clockwise
through the source, and during another part of the cycle they
flow counterclockwise. The diodes, however, conduct current
in one direction only. I begin by making a sketch of the current
between a and d, taking the direction from a to d to be positive
(Figure 32.30a).
Figure 32.30

Figure 32.28 (a) Circuit symbol for a diode. (b) Schematic of a diode
made using integrated-circuit technology.
diode


(a)

p n

aluminum pads

(b)
insulating layer
(SiO2)

p-type
n-type
p-n junction

❷ deviSe plan In an ideal diode, the charge carriers can flow
only in the direction in which the triangle in the diode symbol
points. I shall determine which diodes allow charge carriers to


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32.4 diodes, trAnsistors, And logic gAtes

flow when the current direction is clockwise and when it is counterclockwise. I can then determine in each case which way the
charge carriers flow through the resistor.
❸ execute plan When the current in the circuit is clockwise, only diodes 1 and 3 are conducting, so the current direction is abcd. When the current in the circuit is counterclockwise
(iad 6 0), only diodes 2 and 4 are conducting, so the current direction is dbca. At all instants, the current in the resistor points
in the same direction: from b to c. This means that ibc is positive
regardless of whether iad is positive or negative. Whenever iad is
negative, the diodes reverse the direction of the current in the
resistor, so ibc is always positive and my graph is as shown in Figure 32.30b. ✔

❹ evaluate reSult The arrangement of diodes keeps the current from b to c always in the same direction, even though the
current from a to d alternates in direction. It makes sense, then,
that this arrangement of diodes is called a rectifier.

Figure 32.28b shows how a diode may be constructed as
part of an integrated circuit (a computer chip, for example).
An aluminum pad (part of the metal wire connecting the
diode to the rest of the circuit) is in contact with a small
p-type region of silicon, which is surrounded by a larger ntype region that is in contact with a second aluminum pad.
The p-n junction forms at the interface between the p- and
n-type regions. A thin layer of silicon oxide (SiO2) insulates
the aluminum from the underlying silicon except where
electrical contact is needed. On a modern computer chip,
the entire device is only a few micrometers wide.
Another important circuit element in modern electronics is the transistor, a device that allows current control that
is more precise than the on/off control of a diode. A transistor consists of a thin layer of one type of doped semiconductor sandwiched between two layers of the opposite type
of doped semiconductor. Figure 32.31, for example, shows
an npn-type bipolar transistor—a thin layer of p-type silicon
sandwiched between two thicker regions of n-type silicon.*
If the p-type layer is thin, the depletion zone formed at the

n-type

p-type

S

Edepl

electrically neutral


n-type

left p-n junction merges with the depletion zone formed at
the right p-n junction. The merged depletion zones form
one wide depletion zone.
When a potential difference is applied across such a transistor (Figure 32.32a), the depletion zone across junction 1
disappears, but that across junction 2 grows, shifting the
depleted region toward the positive terminal of the battery.
While charge carriers can now cross junction 1 where the
depletion zone has disappeared, the (shifted) depletion zone
that still exists prohibits their movement, which means no
current in the transistor. For historical reasons, the n-type
region connected to the negative terminal is called the emitter, the n-type region connected to the positive terminal is
called the collector, and the p-type layer is called the base. If
the direction of the applied potential difference is reversed,
the roles of the emitter and the collector are also reversed,
and there is still no current in the transistor.

Figure 32.32 How an npn-type bipolar transistor works.
(a) Potential difference applied from collector to emitter only

S

E

emitter
(n-type)

1


electrically neutral

junction 1: Depletion
zone eliminated.

junction 2: Electric field due
to battery broadens depletion
zone. Current blocked.

(b) Potential difference also applied from base to emitter

base current

S

electrically neutral

Ib

flow of electrons
Ie

Ic
1

* Transistors in which a thin layer of n-type silicon is sandwiched between
pieces of p-type silicon, called pnp-type bipolar transistors, are also used.

collector

(n-type)

2

electrically neutral

Edepl

two merged depletion zones,
one from each p-n junction

base
(p-type)

CONCEPTS

Figure 32.31 Schematic of an npn-type bipolar transistor, showing
charge distribution and depletion zones for both p-n junctions.

871

2

Depletion zone narrow; electrons have
enough kinetic energy to pass through it.

collector
current



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chApter 32  electronics

Figure 32.33 Circuit symbol for an npn-type bipolar transistor.
npn-type bipolar transistor
collector

emitter

Figure 32.35 Schematic of an npn-type bipolar transistor made using
integrated-circuit technology.
emitter

insulating layer
(SiO2)

n

base
p-type

n-type

CONCEPTS

base

collector


The situation changes drastically when, in addition to
the potential difference between the emitter and the collector, a small potential difference is applied between the
emitter and the base (Figure 32.32b). Adding this potential
difference, called a bias or bias potential difference, makes
the depletion zone much thinner than it is in Figure 32.32a
because the formerly negatively charged region of this zone
is brought to a positive potential, restoring mobile holes to
that region. Because the emitter-base junction is conducting (remember, the depletion zone at junction 1 has disappeared), electrons now start flowing from the emitter toward the base. Once in the base, three things happen: (1) a
small fraction of the electrons recombine with holes in the
base, (2) electrons are attracted by the positive charge on the
collector and have sufficient kinetic energy to pass straight
through the very thin depletion zone, producing a collector
current Ic, and (3) electrons diffuse through the base toward
the positively charged end of the base, causing a small base
current Ib. In a typical bipolar transistor, the collector current is 10 to 1000 times greater than the base current.
The circuit symbol for an npn-type bipolar transistor is
shown in Figure 32.33.
Transistors are ubiquitous in modern electronics. In
most applications, the transistor functions as either a switch
or a current amplifier. If we consider Ib to be the input current and Ic the output current, the transistor acts as a switch
in which Ib turns on and controls Ic. As a current amplifier, a
small current Ib produces a much larger current Ic.
For electrical devices that draw large currents, it is useful
to switch the device on and off with a mechanical switch
wired in parallel with the device, rather than in series, so
that the current in the device does not have to pass through
the switch. Figure 32.34 shows a circuit that utilizes such
switching. When switch S is open, the base current is zero,


Figure 32.34 Circuit in which a bipolar transistor is used to turn a motor
on and off.
S

ℰ

motor
Ic = 0

R
Ib = 0

transistor
Ie

and so the collector current (and therefore the current in
the device) is zero. When switch S is closed, the small current from base to emitter causes a large current from collector to emitter that turns on the motor.
32.8 In a bipolar transistor, what relationship, if any, exists
among Ib, Ic, and the emitter current Ie?
Figure 32.35 shows how an npn-type bipolar transistor can be fabricated. A drawback of this type of transistor, however, is that a continuous small current through the
base is required to make the transistor conducting. For this
reason, another type of transistor, called the field-effect transistor, is used much more frequently. Figure 32.36a shows the
configuration of one. Two n-type wells are made in a piece
of p-type material. The p-type material between the two
wells is covered with a nonconducting oxide layer (typically
SiO2) and then with a metal layer called the gate. The two
n-type wells are called the source and the drain (the n-type
well that is kept at a higher potential is the drain).
Because of the depletion zones between the p-type and
n-type materials, no charge carriers can flow from the

source to the drain (or vice versa). The nonconducting layer
between the gate and the p-type material prevents charge
carriers from traveling between the gate and the rest of the
device.
If the gate is given a positive charge, as in Figure 32.36b,
the (positively charged) holes just underneath the gate are
pushed away, forming underneath the gate an additional
depletion zone that connects the depletion zones around
the two n-p junctions. If the positive charge is made large
enough, electrons from the source and from the drain are
pulled underneath the gate, forming an n-type channel below the gate (Figure 32.36c). This channel allows charge
carriers to flow between the source and the drain. The gate
thus controls the current between the source and the drain,
just as the base in an npn-type bipolar transistor controls
the current between the emitter and the collector. (The difference is that there is no current in the gate in a field-effect
transistor.) Applying a positive charge to the gate is often
referred to as putting a positive bias on the gate.
Figure 32.37a shows the circuit symbol for a field-effect
transistor, and Figure 32.37b shows how this type of transistor can be realized in an integrated circuit. This type of
transistor has two advantages over the bipolar transistor


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873

32.4 diodes, trAnsistors, And logic gAtes
Figure 32.36 How a field-effect transistor works.

Figure 32.37 (a) Circuit symbol for a field-effect transistor. (b) Schematic of a field-effect transistor made using integrated-circuit technology.


(a) Field-effect transistor with uncharged gate
source
(n-type well)
depletion
zone

gate (uncharged)
insulating layer

(a)

field-effect transistor
source

drain
(n-type well)

drain
gate

p-type
(b)

gate

source
insulating layer
(SiO2)

Uncharged gate: Separate depletion zones at p-n junctions.


n-type

drain
n-type

p-type

(b) Small positive charge on gate attracts electrons to gate and extends
depletion zone below gate

Small gate charge causes depletion zone to extend beneath gate.

when both inputs are at positive potential with respect to
ground. In an OR gate, the output potential is nonzero
when either input potential is positive. The symbols used
for these gates in circuit diagrams are shown in Figure 32.38;
the inputs are on the left, and the output is on the right. In
analyzing these circuits, we’ll make the simplifying assumption that a transistor is just a switch that is open (off) when
the potential of the gate is either at ground or negative with
respect to ground and is closed (on) when the gate is at a
positive potential.

(c) Large positive charge on gate attracts more electrons to gate and causes
n-type channel, which connects source and drain

Figure 32.38 Circuit symbols for AND and OR logic gates.
A
B


n-type channel

A
B

A B

OR

A B

32.9 Circuit diagrams for two logic gates are shown in

Strong gate charge pushes depletion zone away; conducting
n-type channel now connnects source and drain.

Figure 32.39. Which is the AND gate, and which is the OR gate?

Explain briefly how each one works.

CONCEPTS

shown in Figure 32.35. First, all the terminals in the fieldeffect transistor are on the same side of the chip, making
fabrication in integrated circuits much easier. Second, the
current between the source and the drain is controlled by
the charge on the gate, allowing a potential difference rather
than a current to be used to control the source-drain current. Because no current is leaving the gate, no energy is required to keep current flowing from the source to the drain.
Field-effect transistors are widely used in devices called
logic gates, which are the building blocks of computer processors and memory. A logic gate takes two input signals
and provides an output after performing a logic operation

on the input signals. For example, in a so-called AND gate,
the output potential is nonzero with respect to ground only

AND

Figure 32.39 Checkpoint 32.9
(a)

+5 V

(b)

A

transistor 1

A

B

transistor 2

B

output

+5 V

transistor 1


transistor 2
output


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Self-quiz
1. At the instant shown in Figure 32.40, the potential difference across the capacitor is half its maximum value and the
charge on the plates is increasing. Draw the direction of
the current and sketch the magnetic field at this instant. Is
the magnitude of current increasing or decreasing?
2. Construct a phasor diagram representing the current and
potential difference in Figure 32.10 at t = T>4, T>2, and
3T>4.
3. Figure 32.41 shows the time-varying potential difference
and current for the circuit of Figure 32.8. At the instant
labeled ta, what are the charge on the capacitor plates and
the direction of the current?

Figure 32.40

E

Figure 32.41
vC

i

T

ta

t

4. Is there any current in a diode connected as shown in Figure 32.42?
Figure 32.42
aluminum
SiO2

p-type
n-type

Answers:
1. Your sketch should show the current directed counterclockwise. The magnetic field in the center of the coil
points up the page according to the right-hand dipole rule (assuming we are looking down on the top of the
coil in Figure 32.40). Because the current is zero when the capacitor has maximum charge, the magnitude of the
current is decreasing at the instant shown in Figure 32.40.
2. See Figure 32.43. At t = T>4 the potential difference phasor VC points along the positive y axis because the
potential difference reaches its maximum positive value at this instant, and the current phasor I points along the
negative x axis because it leads the current by 90°. Each quarter cycle both phasors rotate 90° counterclockwise.

self-quiz

Figure 32.43

3. Because the potential difference across the capacitor is zero at instant ta, the charge on the plates must be zero.
The current is a maximum at this instant and is directed clockwise.
4. Yes. The holes in the p-type material move away from the positive terminal, and the electrons move toward it.

According to Figure 32.27d, this flow shrinks the depletion zone, the charge carriers can flow, and so there is a
current.


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875

32.5 reActAnce

32.5 Reactance
Let us now develop a mathematical framework for analyzing alternatingcurrent circuits. The instantaneous emf supplied by an AC source is customarily
written as
ℰ = ℰmax sin vt,

Figure 32.44 AC circuit consisting of a resistor
connected across the terminals of an AC source.
a

R

ℰ

(32.1)

where ℰmax is the maximum value of the emf, typically called the peak value or
amplitude (see Section 15.1), v = 2pf is the angular frequency of oscillation in
inverse seconds (Section 15.5), and ƒ is the frequency in hertz. Most generators
have frequencies of 50 Hz or 60 Hz. Audio circuits typically operate at kilohertz
frequencies, radio transmitters at 108 Hz, for instance, and computer chips at
109 Hz. It’s very important to remember to convert frequencies in hertz (cycles

per second) to angular frequencies in s-1 when v appears in the equations.
Note that the initial phase for the emf as written in Eq. 32.1 is zero. When
we make this choice, the source emf serves as the reference for phase in the
circuit.
Let’s begin by revisiting the circuit from Exercise 32.1—a resistor connected
to an AC source (Figure 32.44). At any instant, Ohm’s law relates the potential
difference across the resistor to the current in it, just as it does for DC circuits:
vR = iR.

b

(32.2)

The only difference between Eq. 32.2 and Ohm’s law for DC circuits (Eq. 31.11) is
that the potential difference and the current in Eq. 32.2 oscillate in time.
Applying the loop rule to this circuit gives the AC version of Eq. 31.23:
ℰ − iR = 0.

(32.3)

Equations 32.2 and 32.3 show that the potential difference across the load equals
the emf supplied by the source (as we would expect):
vR = ℰ = ℰmax sin vt.

(32.4)

32.10 (a) In Figure 32.44, is the potential at point a higher or lower than the potential at b when the current direction is clockwise through the circuit? (b) If we define such
a current to be positive, is ℰ positive or negative? Express vR in terms of the potential at
a and the potential at b. (c) Half a cycle later, when the current is negative, is ℰ positive
or negative? Express vR again in terms of the potential at a and the potential at b.


Figure 32.45 (a) Phasor diagram and (b)
graph showing time dependence of i and vR for
the circuit shown in Figure 32.44.
(a)

VR, ℰ
I

Using Eqs. 32.2 and 32.4, we can write the current in the resistor as
i=

vR
ℰmax sin vt
=
= I sin vt,
R
R

(32.5)

VR = IR.
Figure 32.45

vR and i.

(32.6)

shows the corresponding phasor diagram and time dependence of


(b)

vR

i

T

t

QUANTITATIVE TOOLS

where I = ℰmax >R is the amplitude of the current. Note that the current and the
potential difference both oscillate at angular frequency v and are in phase, as we
concluded in Exercise 32.1. If we write vR = VR sin vt, we see that the amplitudes
of the current and the potential difference satisfy the relationship


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chApter 32  electronics

Exercise 32.5 AC circuit with two resistors
In Figure 32.46, the resistances are R1 = 100 Ω and R2 = 60 Ω,
the amplitude of the emf is ℰmax = 160 V, and its frequency is
60 Hz. (a) What is the amplitude of the potential difference across
each resistor? (b) What is the instantaneous potential difference
across each resistor at t = 50 ms?


(a) Because the current and the emf are in phase, they reach their
maximum values at the same instant. As a result, the amplitude
(maximum value) of the current is given by the amplitude of the
emf divided by the resistance:
I=

Figure 32.46 Exercise 32.5.
R1
ℰ

ℰmax
= 1.0 A.
R1 + R2

The potential differences across the resistors are in phase with
the current, and so I calculate the amplitude of the potential differences from the amplitude of the current using Eq. 32.6:

R2

VR1 = IR1 = (1.0 A)(100 Ω) = 100 V
VR2 = IR2 = (1.0 A)(60 Ω) = 60 V. ✔

Solution I analyze this circuit just as I would analyze a DC cir-

cuit containing two resistors, except now I must keep in mind
that the current and potential differences are oscillating. The
resistance of the load is

(b) I can use Eq. 32.5 to calculate the instantaneous value of the
current:


Rload = R1 + R2,

i = (1.0 A) sin(2p # 60 Hz # 0.050 s) = 0.

and the instantaneous current in the load is
i=

ℰ
R load

ℰ
=
.
R1 + R2

Figure 32.47 AC circuit consisting of a capacitor connected across the terminals of an AC
source.

(In 50 ms, three full cycles at 60 Hz take place.) Because the
current is zero at 50 ms, the potential differences vR1 and vR2 at
50 ms are also zero. ✔

Next consider a capacitor connected to an AC source (Figure 32.47). Because
the capacitor and the source are connected to each other, we have
ℰ = vC,

(32.7)

and so the potential difference across the capacitor is

ℰ

C

vC = ℰmax sin vt = VC sin vt.

(32.8)

At any instant the potential difference across the capacitor and the charge on the
upper plate are related by (see Eq. 26.1)
q
= C,
vC

(32.9)

where the potential difference vC and the charge q on the plate oscillate in time.
The charge on the upper capacitor plate is thus
q = CvC = CVC sin vt,

(32.10)

QUANTITATIVE TOOLS

and the current is the rate of change of the charge on the plate:
i=

dq
d
= (CVC sin vt) = vCVC cos vt.

dt
dt

(32.11)

Using the identity cos a = sin (a + p2 ), we can rewrite this as
i = vCVC sin avt +

p
p
b = I sin avt + b.
2
2

(32.12)

We now see that vC and i are not in phase: i reaches its maximum value onequarter period before vC reaches its maximum value (Figure 32.48), as we found
in Section 32.2.


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32.5 reActAnce

The current in the capacitor of Figure 32.47 is not simply proportional to the
potential difference across the capacitor because the two are out of phase. However, the amplitude of the current is proportional to the amplitude of the potential
difference: I = vCVC. Rewriting this to express VC in terms of I gives
I
VC =
.
vC


1
,
vC

(32.14)

VC = IXC.

(32.15)

XC K
so Eq. 32.13 becomes

Figure 32.48 (a) Phasor diagram and (b)
graph showing time dependence of i and vC for
the circuit of Figure 32.47. The phasor diagram
shows the relative phase of i and vC .
(a)

VC , ℰ

(32.13)

Note how this expression differs from the expression for a circuit that consists of
only an AC source and a resistor, VR = IR (Eq. 32.6), where R is the proportionality constant between V and I. In Eq. 32.13, the proportionality constant is no longer a resistance (though it still has units of ohms). In circuits that contain capacitors and/or inductors, we use the general name reactance for the proportionality
constant between the potential difference amplitude and the current amplitude.
From Eq. 32.13 we see that this proportionality constant for a circuit that contains a capacitor is 1>vC, and we call this constant the capacitive reactance XC:

p>2

I

(b)

vC

i
t

T

Reactance is a measure of the opposition of a circuit element to a change in
current. Unlike resistance, reactance is frequency dependent. At low frequency,
the capacitive reactance XC is large, which means that the amplitude of the current is small for a given value of VC. At zero frequency, the current I = vCVC
is zero, as it should be. (There is no direct current in a capacitor because the
capacitor is just like an open circuit!) The higher the frequency of the source, the
smaller the capacitive reactance and the greater the current (the less the capacitor
opposes the alternating current).
Often, when analyzing AC circuits, the only things we are interested in are the
amplitudes of the currents and potential differences. The capacitive reactance allows us to calculate the amplitude of the current in the capacitor directly from the
amplitude of the potential difference across it—in this case, the emf of the source.
It is conventional to write the current in an AC circuit in the form
i = I sin (vt − f),

877

(32.16)

where f is called the phase constant. The negative sign in front of the phase
constant is chosen so that a positive f corresponds to shifting the curve for the

current to the right, in the positive direction along the time axis, and a negative f
corresponds to shifting the curve to the left, in the negative direction along this
axis (Figure 32.49).

Figure 32.49 Positive and negative phase constant.
(a)

(b)

ℰmax
f70
I
Positive f means
current lags emf
and c

t
i = I sin (vt − f)

ccurrent curve is shifted right relative to emf.

Negative f means
current leads emf
and c
I
f60
ℰmax

ℰ = ℰmax sin vt


t
i = I sin (vt − f)
ccurrent curve is shifted left relative to emf.

QUANTITATIVE TOOLS

ℰ = ℰmax sin vt


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Figure 32.50 AC circuit consisting of an
inductor connected across the terminals of an
AC source.
a

ℰ

L

b

The phase constant represents the phase difference between the source emf
and the current. It is measured from the current phasor to the source emf phasor
with the counterclockwise direction being positive (Figure 32.49). As a result,
when the current leads the source emf, f is negative; when the current lags the
source emf, f is positive.

Comparing Eqs. 32.12 and Eq. 32.16, we see that for the capacitor-AC source
circuit of Figure 32.47, f = - p>2, as shown in Figure 32.48a. The negative
phase constant means that the current leads the source emf. The curve for i is
shifted to the left relative to the curve for vC, as shown in Figure 32.48b. As you
can see from the figure, when the capacitor has maximum charge (vC maximum),
the current is zero because at that instant the current reverses direction as the
capacitor begins discharging. The current reaches its maximum value when the
capacitor is completely discharged (vC = 0).
32.11 As in the LC circuit discussed in Section 32.1, the current in the circuit
of Figure 32.47 oscillates. If we think of vC as corresponding to the position of the
simple harmonic oscillator described in Section 15.5, what property of the circuit of
Figure 32.47 corresponds to the velocity of the oscillator?

Finally, consider an inductor connected to an AC source (Figure 32.50). Because the inductor and the source are connected to each other, we have
ℰ = vL,

(32.17)

so the potential difference across the inductor is
vL = ℰmax sin vt = VL sin vt.

(32.18)

In Chapter 29 we saw that a changing current in an inductor causes an induced emf (Eq. 29.19):
di
ℰind = - L .
dt

(32.19)


The negative sign in this expression means that the potential decreases across the
inductor in the direction of increasing current. Consequently, in Figure 32.50,
the potential at b is lower than the potential at a when the current is increasing clockwise around the circuit. However, for consistency with Eq. 32.3, we always measure the potential difference vL from b to a, just as we did with the AC
source-resistor circuit of Figure 32.44. Therefore the sign of the potential difference across the inductor is the opposite of the sign in Eq. 32.19:
vL = L

di
.
dt

(32.20)

QUANTITATIVE TOOLS

We obtain the current in the circuit by substituting Eq. 32.18 into Eq. 32.20:
L

di
= VL sin vt
dt

di =

VL
sin vt dt.
L

(32.21)
(32.22)


To obtain the current, we integrate this expression:
i=

VL
VL
sin vt dt = cos vt.
L3
vL

(32.23)


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32.5 reActAnce

The amplitude of the current is thus
I=

VL
,
vL

(32.24)

i = I sin avt −

p
b.
2


VL, ℰ

p>2

(32.25)

I

The phase constant is f = +p>2, which means the current lags the source by
90°, as shown in Figure 32.51.
Just as we defined a capacitive reactance for a circuit that contains a capacitor,
we define the inductive reactance XL for a circuit that contains an inductor as the
constant of proportionality between the amplitudes VL and I in the circuit. From
Eq. 32.24 we see that this proportionality constant is vL:

so that

Figure 32.51 (a) Phasor diagram and (b)
graph showing time dependence of i and vL for
the circuit of Figure 32.50. The phasor diagram
shows the phase difference f = p>2 between
i and vL.
(a)

and using the identity cos vt = - sin (vt − p2 ), we get

879

XL K vL,


(32.26)

VL = IXL.

(32.27)

(b)

vL

i
t

T

Inductive reactance, like capacitive reactance, has units of ohms and depends on
the frequency of the AC source. However, XL increases with increasing frequency,
so, at a given potential difference, the amplitude of the current is greatest at zero
frequency and decreases as the frequency increases. This makes sense because
for a constant current, an inductor is just a conducting wire and does not impede
the current; as the frequency of the AC source increases, the emf induced across
the inductor increases.
Example 32.6 Oscillating inductor

I=

VL
.
vL


(1)

Substituting the capacitive reactance from Eq. 32.14,
XC = 1>vC, into Eq. 32.15, VC = IXC, I get
VC =

I
.
vC

(2)

Solving Eq. 2 for C and substituting Eq. 1 for I give
C=
=

VL
I
1
= 2
= 2
vVC
v VCL
vL
1

= 2.3 × 10
(2p # 60 Hz)2(3.0 H)

-6


F. ✔

❹ evaluate reSult To check my answer, I can calculate the
inductive and capacitive reactances from Eqs. 32.26 and 32.14,
respectively: XL = vL = (2p # 60 Hz)(3.0 H) = 1.1 kΩ and
XC = 1>vC = 1>(2p # 60 Hz)(2.3 × 10-6 F) = 1.1 kΩ. The
two are identical, as I expect given that they yield the same current amplitude for the same AC source.

QUANTITATIVE TOOLS

When a 3.0-H inductor is the only element in a circuit connected
to a 60-Hz AC source that is delivering a maximum emf of
160 V, the current amplitude is I. When a capacitor is the only
element in a circuit connected to the same source, what must the
capacitance be in order to have the current amplitude again be I?
❶ GettinG Started I begin by identifying the information
given in the problem statement: ℰmax = 160 V, angular frequency
v = 2p(60 Hz), and inductance L = 3.0 H. The problem asks
me to compare two circuits, one with an inductor connected to
an AC source and the other with a capacitor connected to the
same source. What I must determine is the capacitance value that
makes the current amplitude the same in the two circuits.
❷ deviSe plan For both circuits the potential difference
across the load equals the source emf, so ℰmax = VC = VL. I can
use Eqs. 32.26 and 32.27 to get an expression for VL in terms of I,
from which I can express I in the inductor circuit in terms of VL,
v, and L. Next I can use Eqs. 32.14 and 32.15 to get an expression for VC in terms of I. I can then substitute this into my first
expression for I and obtain an expression for C that contains
only known quantities.

❸ execute plan Substituting the inductive reactance from
Eq. 32.26, XL = vL, into Eq. 32.27, VL = IXL, I get VL = IvL, so
the amplitude of the current is


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Figure 32.52 An RC series circuit, consisting
of a resistor and a capacitor in series across the
terminals of an AC source.
R

ℰ

C

32.12 For the three circuits discussed in this section (AC source with resistor, capacitor, or inductor), sketch for a given emf amplitude (a) the resistance or reactance
as a function of angular frequency v and (b) the current amplitude in the circuit as a
function of v. Explain the meaning of each curve on your graphs.

32.6 RC and RLC series circuits
When an AC source is connected to multiple circuit elements, either in series
or in parallel, applying the loop rule becomes more complicated than for DC
circuits because we need to add several oscillating potential differences that may
be out of phase with one another. For example, suppose we have a resistor and
a capacitor in series with an AC source (Figure 32.52), known as an RC series
circuit. The loop rule states that

ℰ = vR + vC.

(32.28)

To compute the sum on the right side of this equation, we must add potential
differences that vary sinusoidally at the same angular frequency v but are out of
phase. The combined potential difference v of two potential differences v1 and v2
that oscillate at the same angular frequency is
v = V1 sin (vt + f1) + V2 sin (vt + f2),

(32.29)

where f1 and f2 are the initial phases of the two potential differences. Calculating this sum algebraically gets very messy, but using phasors to calculate it
simplifies things greatly.
Figure 32.53a shows the phasors that correspond to the two terms on the right
in Eq. 32.29. Recall that the instantaneous value of the quantity represented by a
rotating phasor equals the vertical component of the phasor (see Figure 32.11).
Therefore, v at any instant equals the sum of the vertical components of the phasors that represent v1 and v2. This sum is equal to the vertical component of the
vector sum V1 + V2 of the phasors, as shown in Figure 32.53b.
Note that the combined potential difference v oscillates at the same angular
frequency as v1 and v2. Consequently, the three phasors V1, V2, and V1 + V2 rotate as a unit at angular frequency v, as shown in Figure 32.54. The phase relationship among the three phasors is constant, as is the phase relationship among
the potential differences.

Figure 32.53 (a) Phasor diagram for a system of two oscillating potential differences v1 and v2. (b) Vector diagram
indicating that the vertical component of the vector sum of the phasors equals the sum of the vertical components of
the individual phasors.

QUANTITATIVE TOOLS

(a)


(b)

vertical component of V1 + V2
V1 + V2

v = v1 + v2

V2

v2 = V2 sin(vt + f2)

V2
v2

vertical component
of V2

v = v1 + v2
v1 = V1 sin(vt + f1)

f2

V1
f1

V1
v1

vertical component

of V1


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32.6 rc And rlc series circuits

881

Figure 32.54 Phasor diagram and graph showing time dependence of v1, v2, and v = v1 + v2 from Figure 32.53.
All three phasors rotate as a unit at angular frequency v.

V2

V1 + V2

v = v1 + v2
v2

v
V1

v1

t

The next example shows how to apply these principles to a specific situation.
To convince yourself that the phasor method is worthwhile, try adding the two
original trigonometric functions algebraically after solving the problem using
phasors!
Example 32.7 Adding phasors

Use phasors to determine the sum of the two oscillating potential differences v1 = (2.0 V) sin vt and v2 = (3.0 V) cos vt.
❶ GettinG Started I begin by making a graph showing the
time dependence of v1 and v2, and I draw the corresponding
phasors V1 and V2 to the left of my graph (Figure 32.55). I add
to my phasor diagram the phasor V1 + V2, which is the phasor
that represents the potential difference sum v1 + v2 that I must
determine. Using phasor V1 + V2, I can sketch the time dependence of the sum v1 + v2 by tracing out the projection of phasor
V1 + V2 onto the vertical axis of my V(vt) graph as this phasor
rotates counterclockwise from the starting position I drew.
Figure 32.55

to the length of the phasor V1 + V2, and from my sketch I see
that the initial phase f i is given by the angle between V1 + V2
and V1.
❸ execute plan The length of the phasor V1 + V2 is given by
the Pythagorean theorem applied to the right triangle containing
f i in my phasor diagram:
A = 2V 21 + V 22 = 2(3.0 V)2 + (2.0 V)2
= 213 V2 = 3.6 V.

The tangent of the angle between V1 + V2 and V1 is then
tan f i =

V2
3.0 V
=
= 1.5,
V1
2.0 V


so f i = tan-1(1.5) = 56°.
Now that I have determined A and f i, I can write the sum of the
two potential differences as
v1 + v2 = (3.6 V) sin (vt + 56°). ✔
❹ evaluate reSult The amplitude of the sinusoidal function I
obtained is 3.6 V, which is greater than the larger of the two phasors I added, as I expect. My answer shows that the sum of the two
potential differences reaches its maximum when vt + f i = 90°,
or when vt = 90° − f i = 34°. This conclusion agrees with my
phasor diagram: The phasor V1 + V2 reaches the vertical position
after it rotates through an angle of 90° − f i = 90° − 56° = 34°.
(I could also verify my answer by adding the two original sine
functions algebraically, but the trigonometry needed in that approach is tedious.)

QUANTITATIVE TOOLS

❷ deviSe plan To obtain an algebraic expression for v1 + v2,
I first write the oscillating potential differences in the form
v1 = V1 sin (vt + f 1) and v2 = V2 sin (vt + f 2). Comparing
these expressions with the given potential differences, I see that
V1 = 2.0 V, f 1 = 0, and V2 = 3.0 V. In order to determine f 2,
I use the trigonometric identity cos (vt) = sin (vt + p>2),
and so my given information v2 = (3.0 V) cos vt =
(3.0 V) sin (vt + p>2) tells me that f 2 = p>2. The sum
v1 + v2 is a sinusoidally varying function that can be written as v1 + v2 = A sin (vt + f i). The amplitude A is equal


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Figure 32.56 Steps involved in constructing a phasor diagram for the circuit in Figure 32.52. The diagram in
part d indicates the phase of the current relative to the source emf.
(a) Draw current phasor

(b) Add phasors for vR and vC

Phase and length
are arbitrary.
I

VR

(c) Add phasor for emf

vR is in phase
with current.

ℰmax: vector sum
of VR and VC

VR

I

I

f

ℰmax


VC

VC
vC lags current by 90°.

32.13 Suppose you need to add two potential differences that are oscillating at
different angular frequencies—say, 2 sin (vt) and 3 cos (2vt). Can you use the phasor
method described above to determine the sum? Why or why not?

Let us now return to the RC series circuit of Figure 32.52 and construct a
phasor diagram in order to determine the amplitude and phase of the current in
terms of the amplitude of the source emf and the resistance and capacitance of
the circuit elements. From the current, we can calculate the potential differences
across the circuit elements.
Because the circuit contains only one loop, the time-dependent current i is
the same throughout. Therefore, we begin by drawing a phasor that represents i
(Figure 32.56a). We are free to choose the phase of this phasor because we have
not yet specified the phase of any of the potential differences in the circuit. Also,
the length we draw for phasor I is unimportant because it is the only current phasor for this circuit.
Next, we draw the phasors for vR and vC, the potential differences across the
resistor and capacitor, respectively. We must get the relative phases right, and
the lengths of the phasors must also be appropriately proportioned. Because the
current is in phase with vR (Figure 32.45a), we draw the corresponding phasor as
shown in Figure 32.56b; its length is VR = IR.
What about the phasor for vC? We found previously that the current in a
capacitor leads the potential difference across the capacitor by 90° (Figure 32.48a),
which means we must draw the phasor for vC 90° behind the phasor for i, as it is
in Figure 32.56b. The length of this phasor is VC = IXC.
Finally, we need to draw the phasor for the emf supplied by the source. Phasor

addition with the loop rule for this circuit (Eq. 32.28) tells us that the phasor
ℰmax for the emf is the vector sum of the phasors VR and VC (Figure 32.56c). The
amplitudes of the potential differences are related by

QUANTITATIVE TOOLS

ℰ2max = V 2R + V 2C.

(32.30)

If we substitute VR = IR (Eq. 32.6) and VC = IXC (Eq. 32.15), this becomes
ℰ2max = (IR)2 + (IXC)2 = I 2(R2 + X 2C) = I 2 aR2 +
Solving for I gives

I=

ℰmax
2

2R + 1>v2C 2

.

1
vC

2 2 b.

(32.31)


(32.32)


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32.6 rc And rlc series circuits

883

Remembering that ℰmax = Vload, we see that even though this load includes both
resistive and reactive elements, I is still proportional to Vload! The constant of
proportionality is called the impedance of the load and is denoted by Z:
I=

ℰmax
.
Z

(32.33)

The impedance of the load is a property of the entire load. It is measured in ohms
and depends on the frequency for any load that contains reactive elements.
Impedance plays the same role in AC circuits that resistance plays in DC circuits. In fact, Eq. 32.33 can be thought of as the equivalent of Ohm’s law for AC
circuits. Equation 32.32 shows that, for an RC series circuit, Z depends on both
R and C:
ZRC K 2R2 + 1>v2C 2

(RC series combination).

(32.34)


To express VR and VC in terms of ℰmax, R, C, and v, we use Eq. 32.32:
VR = IR =
VC = IXC =

ℰmaxR

(32.35)

2R2 + 1>v2C 2
ℰmax >vC

2R2 + 1>v2C 2

.

(32.36)

To calculate the phase constant f, the geometry shown in Figure 32.56c gives
us, with Eqs. 32.6 and 32.15,
tan f = -

or

f = tan-1 a -

VC
IXC
1
==VR
IR

vRC

1
b
vRC

(RC series circuit).

(32.37)

(32.38)

The negative value of f indicates that the current in an RC series circuit leads
the emf, just as it does in an AC circuit with only a capacitor. As you can see in
Figure 32.56c, however, the phase difference between the emf and the current in
the RC series circuit is less than 90°.
Example 32.8 High-pass filter
Figure 32.57 Example 32.8.
C

ℰ

R

vout

❶ GettinG Started This circuit is the same as the one in
Figure 32.52, which I used to determine expressions for VR
(Eq. 32.35) and VC (Eq. 32.36) in terms of R and C, so I can use
(Continued)


QUANTITATIVE TOOLS

A circuit that allows emfs in one angular-frequency range to
pass through essentially unchanged but prevents emfs in other
angular-frequency ranges from passing through is called a filter.
Such a circuit is useful in a variety of electronic devices, including audio electronics. An example of a filter, called a high-pass
filter, is shown in Figure 32.57. Emfs that have angular frequencies above a certain angular frequency, called the cutoff angular
frequency vc, pass through to the two output terminals marked
vout, but the filter attenuates the amplitudes of emfs that have
frequencies below the cutoff value. (a) Determine an expression
that gives, in terms of R and C, the cutoff angular frequency vc
at which VR = VC. (b) Determine the potential difference amplitude vout across the output terminals for v W vc and for
v V v c.


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those results. From Figure 32.57 I see that the potential difference vout is equal to the potential difference across the resistor,
so Vout = VR.
❷ deviSe plan In order to determine the value of vc at which
VR = VC, I equate the right sides of Eqs. 32.35 and 32.36. The
resulting v factor in my expression then is the cutoff value vc.
For part b, I know that Vout = VR. Therefore I can use Eq. 32.35
to determine Vout and then determine how Vout behaves in the
limiting cases where v W vc and v V vc.
❸ execute plan (a) Equating the right sides of Eqs. 32.35 and

32.36, I get R = 1>vC. Solving for v yields the desired cutoff
angular frequency vc:
vc =

1
.✔
RC

(b) To obtain the values of Vout for v W vc and for v V vc, I
first rewrite Eq. 32.35 in a form that contains vc:
ℰmaxR

Vout = VR =

2

2R + 1>v2C 2
ℰmax
ℰmax
=
.
=
2 2 2
21 + 1>R v C
21 + v2c >v2

Figure 32.58 An RLC series circuit, consisting of a resistor, an inductor, and a capacitor in
series across the terminals of an AC source.
R


ℰ

L
C

(1)

For v W vc, the second term in the square root vanishes and
Eq. 1 reduces to Vout = ℰmax. ✔
For v V vc, the second term in the square root dominates,
so I can ignore the first term. Equation 1 then becomes
Vout = VR =
=

ℰmax
21 +

v2c >v2

ℰmaxv
= ℰmaxvRC.
vc

≈

ℰmax
2v2c >v2

In the limit that the angular frequency v approaches zero, Vout
approaches zero as well. ✔

❹ evaluate reSult The name high-pass filter makes sense because this circuit allows emfs with an angular frequency higher
than the cutoff angular frequency to pass through to the output
but attenuates emfs of angular frequency lower than the cutoff
angular frequency, preventing them from passing through to the
output. It is the capacitor that does the actual passing or blocking. It blocks low-angular-frequency emfs because for these emfs
the capacitive reactance, XC = 1>vC, is very high. For highangular-frequency emfs, XC approaches zero, and so the capacitor passes the emf undiminished.

32.14 Interchange the resistor and the capacitor in Figure 32.57, and then show
that the high-pass filter becomes a low-pass filter.

Filters can also be constructed by wiring an inductor and a resistor in series
with an AC source. Such a circuit is called an RL series circuit and can be analyzed
in exactly the manner we used to analyze an RC series circuit (see Example 32.9).
Finally, let’s analyze an RLC series circuit: a resistor, a capacitor, and an inductor all in series with an AC source (Figure 32.58). As with the RC series circuit, the
instantaneous current i is the same in all three elements, and the sum of all the
potential differences equals the emf of the source:
ℰ = vR + vL + vC.

(32.39)

The phasor diagram for this circuit is constructed in Figure 32.59 for the case
where VL 7 VC. As before, we begin with the phasors for i and vR, and then note
Figure 32.59 Steps involved in constructing a phasor diagram for the RLC series circuit in Figure 32.58.
The diagram in part c indicates the phase of the current relative to the source emf.

QUANTITATIVE TOOLS

(a) Begin with phasors for i and vR (in phase)

(b) Add VC and VL


(c) Add VL − VC and VR to obtain ℰmax
ℰmax

vL leads i by 90°

VR

VL
I

VR

VL − VC
I

VC
vC lags i by 90°

VL − VC

f

VR

I
If VL 7 VC , f is positive
and current lags emf.