Download Guide to the Alternating Current Circuits and more Assignments Electrical Engineering in PDF only on Docsity!
Alternating-Current Circuits
- Chapter
- 12.1 AC Sources 12-
- 12.2 Simple AC circuits............................................................................................ 12-
- 12.2.1 Purely Resistive load.................................................................................. 12-
- 12.2.2 Purely Inductive Load................................................................................ 12-
- 12.2.3 Purely Capacitive Load.............................................................................. 12-
- 12.3 The RLC Series Circuit 12-
- 12.3.1 Impedance 12-
- 12.3.2 Resonance 12-
- 12.4 Power in an AC circuit.................................................................................... 12-
- 12.4.1 Width of the Peak..................................................................................... 12-
- 12.5 Transformer 12-
- 12.6 Parallel RLC Circuit........................................................................................ 12-
- 12.7 Summary......................................................................................................... 12-
- 12.8 Problem-Solving Tips 12-
- 12.9 Solved Problems 12-
- 12.9.1 RLC Series Circuit 12-
- 12.9.2 RLC Series Circuit 12-
- 12.9.3 Resonance 12-
- 12.9.4 RL High-Pass Filter.................................................................................. 12-
- 12.9.5 RLC Circuit 12-
- 12.9.6 RL Filter 12-
- 12.10 Conceptual Questions 12-
- 12.11 Additional Problems 12-
- 12.11.1 Reactance of a Capacitor and an Inductor 12-
- 12.11.2 Driven RLC Circuit Near Resonance..................................................... 12-
- 12.11.3 RC Circuit 12-
- 12.11.4 Black Box............................................................................................... 12-
- 12.11.5 Parallel RL Circuit.................................................................................. 12-
- 12.11.6 LC Circuit............................................................................................... 12-
- 12.11.7 Parallel RC Circuit 12-
- 12.11.8 Power Dissipation 12-
- 12.11.9 FM Antenna 12-
- 12.11.10 Driven RLC Circuit 12-
Alternating-Current Circuits
12.1 AC Sources
In Chapter 10 we learned that changing magnetic flux can induce an emf according to
Faraday’s law of induction. In particular, if a coil rotates in the presence of a magnetic
field, the induced emf varies sinusoidally with time and leads to an alternating current
(AC), and provides a source of AC power. The symbol for an AC voltage source is
An example of an AC source is
V t ( ) = V 0 sin ω t (12.1.1)
where the maximum value V is called the amplitude. The voltage varies between and
since a sine function varies between +1 and −1. A graph of voltage as a function of
time is shown in Figure 12.1.1.
0 V 0
− V 0
Figure 12.1.1 Sinusoidal voltage source
The sine function is periodic in time. This means that the value of the voltage at time t
will be exactly the same at a later time t ′^ = t + T where T is the period. The frequency,
f , defined as f = 1/ T , has the unit of inverse seconds (s
− 1 ), or hertz (Hz). The angular
frequency is defined to be ω = 2 π f.
When a voltage source is connected to an RLC circuit, energy is provided to compensate
the energy dissipation in the resistor, and the oscillation will no longer damp out. The
oscillations of charge, current and potential difference are called driven or forced
oscillations.
After an initial “transient time,” an AC current will flow in the circuit as a response to the
driving voltage source. The current, written as
Figure 12.2.2 (a) Time dependence of ( ) R
I t and ( ) R
V t across the resistor. (b) Phasor
diagram for the resistive circuit.
The behavior of I (^) R ( ) t and can also be represented with a phasor diagram, as shown
in Figure 12.2.2(b). A phasor is a rotating vector having the following properties:
VR ( ) t
(i) length: the length corresponds to the amplitude.
(ii) angular speed: the vector rotates counterclockwise with an angular speed ω.
(iii) projection: the projection of the vector along the vertical axis corresponds to the
value of the alternating quantity at time t.
We shall denote a phasor with an arrow above it. The phasor has a constant
magnitude of. Its projection along the vertical direction is
VR 0
G
VR 0 VR 0 sin ω t , which is equal
to , the voltage drop across the resistor at time t. A similar interpretation applies
to
VR ( ) t
IR 0
G
for the current passing through the resistor. From the phasor diagram, we readily
see that both the current and the voltage are in phase with each other.
The average value of current over one period can be obtained as:
0 0 0 0 0
( ) ( ) sin sin 0
T T T R R R R
I t I t I t dt I t dt dt T T T T
This average vanishes because
0
sin sin 0
T t t T
ω = ω dt =
Similarly, one may find the following relations useful when averaging over one period:
0
0
2 2 2
0 0
2 2 2
0 0
cos cos 0
sin cos sin cos 0
sin sin sin 2
cos cos cos 2
T
T
T T
T T
t t dt T
t t t t dt T
t t t dt dt T T T
t t t dt dt T T T
From the above, we see that the average of the square of the current is non-vanishing:
2 2 2 2 2 2 0 0 0 0 0
( ) ( ) sin sin 2
T T T
R R R R
t (^) 2 0
I t I t dt I t dt I dt IR T T T T
It is convenient to define the root-mean-square (rms) current as
(^2 ) rms ( )^ 2
R R
I
I = I t = (12.2.7)
In a similar manner, the rms voltage can be defined as
(^2 ) rms ( )^ 2
R R
V
V = V t = (12.2.8)
The rms voltage supplied to the domestic wall outlets in the United States is
V rms (^) = 120 Vat a frequency f = 60 Hz.
The power dissipated in the resistor is
2 ( ) ( ) ( ) ( ) R R R R
P t = I t V t = I t R
from which the average over one period is obtained as:
2 (^2 2 2) rms 0 rms rms rms
R R R
V
P t I t R I R I R I V R
12.2.2 Purely Inductive Load
Consider now a purely inductive circuit with an inductor connected to an AC generator,
as shown in Figure 12.2.3.
frequencies the current changes more rapidly than it does at lower frequencies. On the
other hand, the inductive reactance vanishes as ω approaches zero.
By comparing Eq. (12.2.14) to Eq. (12.1.2), we also find the phase constant to be
The current and voltage plots and the corresponding phasor diagram are shown in the
Figure 12.2.4 below.
Figure 12.2.4 (a) Time dependence of I (^) L ( ) t and VL ( ) t across the inductor. (b) Phasor
diagram for the inductive circuit.
As can be seen from the figures, the current IL ( ) t is out of phase with VL ( ) t by φ = π/ 2;
it reaches its maximum value after VL ( ) t does by one quarter of a cycle. Thus, we say that
The current lags voltage by π / 2 in a purely inductive circuit
12.2.3 Purely Capacitive Load
In the purely capacitive case, both resistance R and inductance L are zero. The circuit
diagram is shown in Figure 12.2.5.
Figure 12.2.5 A purely capacitive circuit
Again, Kirchhoff’s voltage rule implies
( ) C ( ) ( ) 0
Q t V t V t V t C
which yields
Q t ( ) = CV t ( ) = CVC ( ) t = CVC 0 sin ω t (12.2.19)
where VC (^) 0 = V 0. On the other hand, the current is
0 0
( ) cos sin 2
C C C
dQ I t CV t CV t dt
where we have used the trigonometric identity
cos sin 2
t t
The above equation indicates that the maximum value of the current is
0 0 0
C C C C
V
I CV
X
where
X C
ω C
is called the capacitance reactance. It also has SI units of ohms and represents the
effective resistance for a purely capacitive circuit. Note that X (^) C is inversely proportional
to both C and ω , and diverges as ω approaches zero.
By comparing Eq. (12.2.21) to Eq. (12.1.2), the phase constant is given by
The current and voltage plots and the corresponding phasor diagram are shown in the
Figure 12.2.6 below.
0 sin
dI Q L IR V dt C
+ + = ω t (12.3.2)
Assuming that the capacitor is initially uncharged so that I = + dQ / dt is proportional to
the increase of charge in the capacitor, the above equation can be rewritten as
2
2 0 sin
d Q dQ Q L R V dt dt C
+ + = ω t (12.3.3)
One possible solution to Eq. (12.3.3) is
Q t ( ) = Q 0 cos( ω t − φ) (12.3.4)
where the amplitude and the phase are, respectively,
0 0 (^0 2 2 2 )
0
2 2
( L C )
V L V
Q
2 R L LC R L
V
R X X
C
and
tan
X L XC
L
R C R
The corresponding current is
( ) 0 sin( )
dQ I t I t dt
with an amplitude
0 (^0 0 ) ( (^) L C )
V
I Q
R X X
2
Notice that the current has the same amplitude and phase at all points in the series RLC
circuit. On the other hand, the instantaneous voltage across each of the three circuit
elements R , L and C has a different amplitude and phase relationship with the current, as
can be seen from the phasor diagrams shown in Figure 12.3.2.
Figure 12.3.2 Phasor diagrams for the relationships between current and voltage in (a)
the resistor, (b) the inductor, and (c) the capacitor, of a series RLC circuit.
From Figure 12.3.2, the instantaneous voltages can be obtained as:
0 0
0 0
0 0
( ) sin sin
( ) sin cos 2
( ) sin cos 2
R R
L L L
C C C
V t I R t V t
V t I X t V t
V t I X t V t
where
VR 0 = I R 0 , VL 0 = I X 0 L , VC 0 = I X 0 C (12.3.10)
are the amplitudes of the voltages across the circuit elements. The sum of all three
voltages is equal to the instantaneous voltage supplied by the AC source:
V t ( ) = V (^) R ( ) t + V (^) L ( ) t + VC ( ) t (12.3.11)
Using the phasor representation, the above expression can also be written as
V 0 = VR 0 + V L 0 + V C 0
G G G G
as shown in Figure 12.3.3 (a). Again we see that current phasor I 0
G
leads the capacitive
voltage phasor VC 0 by
G
π / 2 but lags the inductive voltage phasor VL 0
G
by π / 2. The three
voltage phasors rotate counterclockwise as time passes, with their relative positions fixed.
Figure 12.3.4 Diagrammatic representation of the relationship between Z , X (^) L and X (^) C.
The impedance also has SI units of ohms. In terms of Z , the current may be rewritten as
0 ( ) sin( )
V
I t t Z
Notice that the impedance Z also depends on the angular frequency ω, as do XL and XC.
Using Eq. (12.3.6) for the phase φ and Eq. (12.3.15) for Z , we may readily recover the
limits for simple circuit (with only one element). A summary is provided in Table 12.
below:
Simple
Circuit
R L C^ X^ L =^ ω L
1 X (^) C ω C
=
1 tan L^ C
X X
R
φ
− ⎛^ − ⎞ = (^) ⎜ ⎟ ⎝ ⎠
2 2 Z = R + ( X (^) L − XC )
purely
resistive
R^0 ∞ 0 0 0 R
purely
inductive
0 L ∞ X^ L 0 π / 2 XL
purely
capacitive
0 0 C 0 X^ C − π / 2 XC
Table 12.1 Simple-circuit limits of the series RLC circuit
12.3.2 Resonance
Eq. (12.3.15) indicates that the amplitude of the current I (^) 0 = V 0 (^) / Z reaches a maximum
when Z is at a minimum. This occurs when X L = XC , or ω L =1/ ω C , leading to
0
LC
The phenomenon at which I 0 reaches a maximum is called a resonance, and the
frequency ω 0 is called the resonant frequency. At resonance, the impedance
becomes Z = R , the amplitude of the current is
0 0
V
I
R
and the phase is
as can be seen from Eq. (12.3.5). The qualitative behavior is illustrated in Figure 12.3.5.
Figure 12.3.5 The amplitude of the current as a function of ω in the driven RLC circuit.
12.4 Power in an AC circuit
In the series RLC circuit, the instantaneous power delivered by the AC generator is given
by
2 0 0 0
2 0 2
( ) ( ) ( ) sin( ) sin sin( )sin
sin cos sin cos sin
V V
P t I t V t t V t t t Z Z
V
t t t Z
where we have used the trigonometric identity
sin( ω t − φ ) = sin ω t cos φ − cos ω t sinφ (12.4.2)
The time average of the power is
2 rms max rms^ rms
V
P I V
R
12.4.1 Width of the Peak
The peak has a line width. One way to characterize the width is to define ,
where
±
are the values of the driving angular frequency such that the power is equal to
half its maximum power at resonance. This is called full width at half maximum , as
illustrated in Figure 12.4.2. The width∆ ω increases with resistance R.
Figure 12.4.2 Width of the peak
To find∆ ω , it is instructive to first rewrite the average power P t ( ) as
2 2 0 0 2 2 2 2 2 2 0
V R V R
P t R L C R L
2
2 2 )
with
2 max^0
P t ( ) = V / 2 R. The condition for finding ω
±
is
2 2 2 0 0 max^2 2 2 2 2 0
V V R
P t P t R R L
which gives
2 2 2 2 ( 0 )
R
L
Taking square roots yields two solutions, which we analyze separately.
case 1: Taking the positive root leads to
2 2 0
R
L
Solving the quadratic equation, the solution with positive root is
2 2 0 2 4
R R
L L
Case 2: Taking the negative root of Eq. (12.4.10) gives
2 2 0
R
L
− − −^ = −^ (12.4.13)
The solution to this quadratic equation with positive root is
2 2 0 2 4
R R
L L
The width at half maximum is then
R
L
Once the width ∆ω is known, the quality factor Q (not to be confused with charge) can
be obtained as
0 0 L
Q
R
Comparing the above equation with Eq. (11.8.17), we see that both expressions agree
with each other in the limit where the resistance is small, and
2 2
ω ω 0 ( R / 2 L ) ω 0
12.5 Transformer
A transformer is a device used to increase or decrease the AC voltage in a circuit. A
typical device consists of two coils of wire, a primary and a secondary, wound around an
iron core, as illustrated in Figure 12.5.1. The primary coil, with turns, is connected to
alternating voltage source. The secondary coil has N
1
N
1
V t ( ) (^) 2 turns and is connected to a
“load resistance” 2
R. The way transformers operate is based on the principle that an
2 2 1 2 1 1
V N
I I 2
V N
I
1 1 1
1 1
Thus, we see that the ratio of the output voltage to the input voltage is determined by the
turn ratio. If , then , which means that the output voltage in the
second coil is greater than the input voltage in the primary coil. A transformer with
is called a step-up transformer. On the other hand, if
N 2 / N N 2 > N V 2 > V
N (^) 2 > N N 2 < N , then , and
the output voltage is smaller than the input. A transformer with
V 2 < V 1
N (^) 2 < N 1 is called a step-
down transformer.
12.6 Parallel RLC Circuit
Consider the parallel RLC circuit illustrated in Figure 12.6.1. The AC voltage source is
V t ( ) = V 0 sin ω t.
Figure 12.6.1 Parallel RLC circuit.
Unlike the series RLC circuit, the instantaneous voltages across all three circuit elements
R , L , and C are the same, and each voltage is in phase with the current through the
resistor. However, the currents through each element will be different.
In analyzing this circuit, we make use of the results discussed in Sections 12.2 – 12.4.
The current in the resistor is
0 0
R ( )^ sin^ sin
V t V I t t IR R R
= = ω = ω t (12.6.1)
where I (^) R 0 = V 0 (^) / R. The voltage across the inductor is
( ) ( ) 0 sin
L L
dI V t V t V t L dt
which gives
0 0 0 ( ) 0 sin ' ' cos sin 0 sin 2 2
t
L L L
V V V
I t t dt t t I t L L X
where I L 0 = V 0 / XL and X L = ω L is the inductive reactance.
Similarly, the voltage across the capacitor is VC ( ) t = V 0 sin ω t = Q t ( ) / C , which implies
0 ( ) 0 cos sin 0 sin 2 2
C C C
dQ V I t CV t t I t dt X
where I C 0 = V 0 / XC and X C = 1/ ω C is the capacitive reactance.
Using Kirchhoff’s junction rule, the total current in the circuit is simply the sum of all
three currents.
0 0 0
sin sin sin 2 2
R L C
R L C
I t I t I t I t
I t I t I t
The currents can be represented with the phasor diagram shown in Figure 12.6.2.
Figure 12.6.2 Phasor diagram for the parallel RLC circuit
From the phasor diagram, we see that
I 0 = I R 0 + IL 0 + IC 0
G G G G
and the maximum amplitude of the total current, I 0 , can be obtained as
2 2 0 0 0 0 0 0 0 0
2 2
(^0 2 )
R L C R C L
C L
I I I I I I I I
V C V
R L R X X
G G G G
