Download 34.2. AC Resistor Circuits

Chapter 34
Physics 122
34. AC CIRCUITS
34.1.
Alternating Current
The current from a 110-V outlet is an oscillating function of time. This type is called
Alternating Current or AC. A source of AC is symbolized by a wavy line enclosed in a circle
(see Figure 34.1). The time dependence of the AC or the emf of the AC source is of the form
(t) =  max sin( t)
(34.1)
where max is the maximum amplitude of the oscillating emf and  is the angular frequency.

Figure 34.1. Symbol of AC source.
34.2.
AC Resistor Circuits
Figure 34.2 shows a single-loop circuit with a source of alternating emf and a resistor.
The current through the resistor will be a function of time. The magnitude of this current can be
obtained via Kirchhoff’s second rule which implies that
(t) - I(t) R = 0
(34.2)
The current I is thus equal to
R

Figure 32.2. Single-loop AC resistor circuit.
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Chapter 34
Physics 122

 max sin( t)
(t)
I(t) =
=
R
R
(34.3)
Equation (34.3) shows that the current oscillates in phase with the emf.
The power dissipated in the resistor depends on the current through and the voltage
across the resistor and is therefore also a function of time:
2
2

 max sin ( t)
P(t) = I(t) (t) =
R
(34.4)
The average power dissipated in the resistor during one cycle is equal to

T
1
P =
T
T
0
1
P(t) dt =
T
 max
R
2
2
sin ( t) dt =
0
1  max
=
T R
2
1  max
=
T R
2
T
0
1
(1 - cos(2  t)) dt =
2
 max
1
sin( T)
T =
2
4
2R
(34.5)
2
In the last step of the derivation of eq.(34.5) we used the relation between the period T and the
angular frequency  (T
=
2/).
Often, eq.(34.5) is written in terms of the
root-mean-square voltage rms which is defined as
 max

 rms =
2
(34.6)
In terms of rms we can rewrite eq.(34.5) as
  2
rms
P =
R
(34.7)
The root-mean-square voltage rms of the AC source is the value of the DC voltage that
dissipates the same power in the resistor as the AC voltage with a maximum voltage equal to
max. The household voltage of 115 Volt is the root-mean-square voltage; the actual peak
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Chapter 34
Physics 122
voltage coming out of a household outlet is 163 V.
34.3.
AC Capacitor Circuits
Figure 34.3 shows a capacitor connected to a source of alternating emf. The charge on
the capacitor at any time can be obtained by applying Kirchhoff’s second rule to the circuit
shown in Figure 34.3 and is equal to
Q(t)
 = C (t) = C  max sin( t)
(34.8)
The current in the circuit can be obtained by differentiating eq.(34.8) with respect to time
dQ

I(t) =
=  C  max cos( t)
dt
(34.9)
The current in the circuit is 90 out of phase with the emf. Since the maxima in the current
occur a quarter cycle before the maxima in the emf, we say that the current leads the emf.
It is customary to rewrite eq.(34.9) as
 max

I(t) =
cos( t)
XC
(34.10)
where

XC =
1
C

(34.11)
C
Figure 34.3. AC capacitor circuit.
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Chapter 34
Physics 122
is called the capacitive reactance. Note that eq.(34.10) is very similar to eq.(34.3) if the
resistance R is replaced by the capacitive reactance XC. The power delivered to the capacitor is
equal to
2
2

 max
 max
P(t) = I(t) (t) =
cos( t) sin( t) =
sin(2  t)
XC
2 XC
(34.12)
The power fluctuates between positive and negative extremes, and is on average equal to zero.
These fluctuations corresponds to periods during which the emf source provides power to the
battery (charging) and periods during which the battery provides power to the emf source
(discharging).
34.4.
AC Inductive Circuit
Figure 34.4 shows a circuit consisting of an inductor and a source of alternating emf.
The self-induced emf across the inductor is equal to LdI/dt. Applying Kirchhoff’s second rule
to the circuit shown in Figure 34.4 we obtain the following equation for dI/dt:
 max sin( t)
dI

=
=
dt
L
L
(34.13)
The current I can be obtained from eq.(34.13) by integrating with respect to time and requiring
that the magnitude of the DC current component is equal to zero:

I(t) =
 max cos( t)
dI
dt = dt
L
(34.14)
L

Figure 34.4. AC Inductor Circuit.
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Chapter 34
Physics 122
The current is again 90 out of phase with the emf, but this time the emf leads the current.
Equation (34.14) can be rewritten as

 max cos( t)
I(t) = XL
(34.15)
X L =  L
(34.16)
where
is called the inductive reactance. The power delivered to the inductor is equal to
2
2

 max
 max
P(t) = I(t) (t) = cos( t) sin( t) = sin(2  t)
XL
2 XL
(34.17)
and the average power delivered to the inductor is equal to zero.
Example: Problem 34.10
Consider the circuit shown in Figure 34.5. The emf is of the form 0 sin(t). In terms
of this emf and the capacitance C and the inductance L, find the instantaneous currents through
the capacitor and the inductor. Find the instantaneous current and the instantaneous power
delivered by the source of emf.
The circuit shown in Figure 34.5 is a simple multi-loop circuit. The currents in this
circuit can be determined using the loop technique. Consider the two current loops I1 and I2
L
C

I2
I1
Figure 34.5. Problem 34.10.
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Chapter 34
Physics 122
indicated in Figure 34.5. Applying Kirchhoff’s second rule to loop number 1 we obtain
 dI 1
-L
= 0
dt
(34.18)
Applying Kirchhoff’s second rule to loop number 2 we obtain
-
Q 2
C
(34.19)
= 0
Equation (34.18) can be used to determine I1:

I1 =
0

dt =
L
L
sin( t) dt = -
0
L
cos( t)
(34.20)
Equation (34.19) can be differentiated with respect to time to obtain I2:
 dQ 2
d
I2 =
=
C  0 sin( t) =  C  0 cos( t)
dt
dt
(34.21)
The current delivered by the source of emf is the sum of I1 and I2

1
I(t) = I 1(t) + I 2 (t) =  0  C cos( t)
L
(34.22)
The power delivered by the source of emf is equal to

1
2
P(t) = I(t) (t) =  0  C cos( t) sin( t)
L
34.5.
(34.23)
LC Circuits
Figure 34.6 shows a single-loop circuit consisting of an inductor and a capacitor.
Suppose at time t = 0 s the capacitor has a charge Q0 and the current in the circuit is equal to
zero. The current in the circuit can be found via Kirchhoff’s second rule which requires that
-L
dI Q
= 0
dt C
- 6 -
(34.24)
Chapter 34
Physics 122
L
C
Figure 34.6. LC circuit.
The current I(t) can be obtained from Q(t) by differentiating Q with respect to time:
dQ
dt
I =
(34.25)
Substituting eq.(34.25) into eq.(34.24) we obtain
2
-L
d Q
dt
2
Q
= 0
C
(34.26)
Q
= 0
LC
(34.27)
-
or
2
d Q
dt
2
+
A solution of eq.(34.27) is

Q(t) = Q 0 cos
1
t+
LC
(34.28)
where  is a phase constant that must be adjusted to fit the initial conditions. The current in the
circuit can be obtained by substituting eq.(34.28) into eq.(34.25):
 dQ(t)
Q0
I =
= sin
dt
LC
- 7 -
1
LC
t+
(34.29)
Chapter 34
Physics 122
The initial conditions for the circuit shown in Figure 34.6 are
I(t = 0) = 0 A
(34.30)
Q(t = 0) = Q 0
(34.31)
These boundary conditions are satisfied if  = 0. In this case, the charge and the current in the
LC circuit are given by
1
t
LC
Q(t) = Q 0 cos
(34.32)
and
I(t) = -
Q0
LC
1
sin
LC
(34.33)
t
The energy stored on the capacitor is a function of time since the charge on it is a function of
time. The energy stored is equal to
2
UC
2
1 Q (t)
1 Q0
2
=
=
cos
2 C
2 C
1
t
LC
(34.34)
The energy stored in the inductor is also time dependent since the current through it is a function
of time. The energy stored is equal to
2
UL
1
1 Q0
2
2
= L I (t) =
sin
2
2 C
1
t
LC
(34.35)
Equation (34.34) and eq.(34.35) show that the maximum energy is stored in the inductor when
the energy stored in the capacitor is zero and vice-versa. The total energy of the circuit can be
obtained by summing the energy stored in the capacitor and the energy stored in the inductor:
U = UC+ UL
1 Q0
=
2 C
2
(34.36)
Equation (34.36) shows that the energy stored in the circuit is conserved. This is expected since
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Chapter 34
Physics 122
no Joule heat will be generated in a circuit in which none of the elements has any resistance.
In practice, the circuit shown in Figure 34.6 will have some resistance (even good
conductors will have a finite resistance). A realistic LRC circuit is shown in Figure 34.7.
Applying Kirchhoff’s second rule to the circuit shown in Figure 34.7 we obtain
-L
dI
Q
- IR = 0
dt
C
(34.37)
Since the current I is equal to dQ/dt we can rewrite eq.(34.37) as
2
d Q
dt
2
+
R dQ
Q
+
= 0
L dt
LC
(34.38)
A solution of the differential equation shown in eq.(34.38) is
 = Q  -  t / 2 cos  t
Q(t)
0
(34.39)
The constant  can be determined by substituting eq.(34.39) into eq.(34.38):

1
R
2 R
-  t/ 2
-  t/ 2
- +
Q0
cos  t +   Q0
sin  t = 0
4
2L LC
L
2
(34.40)
This equation has to be satisfied at all times. This will only be the case if the terms within the
parenthesis are equal to zero:
L
C
R
Figure 34.7. LRC Circuit.
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Chapter 34
Physics 122
 
R
1
2
-  +
= 0
4
2L
LC
(34.41)
 R
 = 0
L
(34.42)
 R
 =
L
(34.43)
2
The constant  is determined by eq.(34.42)
The angular frequency  can be obtained from eq.(34.41) by substituting eq.(34.43) for 

 =
1
CR
1 LC
4L
(34.44)
Equation (34.39) shows that the presence of the resistor in the circuit will produce damped
harmonic motion. The damping constant  is proportional to the resistance R (see eq.(34.43)).
The change in the energy of the system can be studied by looking at the maximum charge on the
capacitor. At time t = 0 s the capacitor is fully charged with a charge equal to Q0 and the energy
stored in the capacitor is equal to
2
2
1 Q (t = 0)
1 Q0
U(t = 0) =
=
2
C
2 C
(34.45)
After one cycle (t = 2/) the maximum charge on the capacitor has decreased. This implies
that also the energy stored on the capacitor has decreased

2
1
U(t =
) =

2
2
Q (t =
2
)

C
2
1 Q 0 -2   / 
=
e
2 C
The relative change in the electrical energy of the system is therefore equal to
- 10 -
(34.46)
Chapter 34
Physics 122

2
²U
=
U
2
1 Q 0 - 2  / 1 Q 0
e
2 C
2 C
2
=
1 Q0
2 C
= e
 -
- 2  / 
- 1 
(34.47)
2 R
 L
The loss of electrical energy in a LRC circuit is usually expressed in terms of the quality
Q-value”

Q = 2
U
²U
= 0
L
R
(34.48)
A high quality factor indicates a low resistance and consequently a small relative energy loss per
cycle.
As a result of the damping in a LRC circuit the amplitude of the oscillations will
gradually decrease. In order to sustain an oscillation in a LRC circuit, energy needs to be
supplied, for example by connecting an oscillating source of emf to the circuit. Consider the
circuit shown in Figure 34.8 consisting of an alternating source of emf, a resistor R, a capacitor
C, and an inductor L. Suppose the emf has an angular frequency  and a maximum amplitude
max:
L
C
R
Figure 34.8. Driven LCR circuit.
- 11 -
Chapter 34
Physics 122
(t) =  max sin( t)
(34.49)
Applying Kirchhoff’s second rule to the circuit shown in Figure 34.8 produces the following
relation
dI 1

 - IR - L
dt C
(34.50)
I dt = 0
Under steady-state conditions, the current in the circuit will oscillate with the same angular
frequency  as the source of emf, but not necessarily in phase. The most general solution for
the current is therefore
I(t) = I max sin( t - )
(34.51)
where  is called the phase angle between the current and the emf. The maximum current Imax
and the phase angle  can be determined by substituting eq.(34.51) in eq.(34.50):

I max
 max sin( t) - R I max sin( t - ) -  L I max cos( t - ) +
cos( t - ) = 0 (34.52)
C
Equation (34.52) can be rewritten using trigonometric identities as

 max - I max R cos() +  L + I max
1
sin()
C
sin( t) +
1
R sin() -  L cos() cos( t) = 0
C
(34.53)
This equation can only be satisfied if the expressions in the brackets are equal to zero. This
requires that

1
 max - I max R cos() +  L sin() = 0
C
(34.54)

1
I max R sin() -  L cos() = 0
C
(34.55)
and
- 12 -
Chapter 34
Physics 122
Eq.(34.55) can be used to determine the phase angle:

tan() =
1
XL- XC
C
=
R
R
L -
sin()
=
cos()
(34.56)
Equation (34.54) can be used to determine the maximum current:

I max =
 max
1
R cos() +  L sin()
C
 max
=
(34.57)
R cos() + X L - X C sin()
Substituting eq.(34.56) into eq.(34.57) we obtain for the maximum current

I max =
 max
1
2
R + LC
2
 max
=
2
R + X L - XC
2
(34.58)
The quantity

Z =
2
R + XL- XC
2
1
R + LC
2
=
2
(34.59)
is called the impedance of the LCR circuit.
Equation (34.58) shows that the maximum amplitude is achieved when

L =
1
C
(34.60)
 max

I max =
R
(34.61)
The maximum amplitude of the current is
The system will reach its maximum amplitude when the driving frequency  of the applied emf
is equal to

 =
1
LC
- 13 -
(34.62)
Chapter 34
Physics 122
This frequency is the natural frequency of the LC circuit discussed previously. When the
system is driven at the natural frequency, it is said to be in resonance.
34.6.
The Phasor Diagram
A short cut that can be used to determine the amplitude and phase of current in an AC
circuit is the phasor diagram. In a phasor diagram the amplitude of a sinusoidal function is
represented by a line segment of length equal to its amplitude. The phase is represented by the
angle between the line segment and the horizontal axis. The sum of voltage drops across the
components of the circuit is then equivalent to the vector sum of the phasors. To illustrate the
use of phasor diagrams we determine the amplitude and phase of the LCR circuit just discussed.
The applied emf and induced current are given by the following equations:
 = 
(t)
sin( t)
max
(34.63)
I(t) = I max sin( t - )
The voltages across the resistor, the capacitor and the inductor are equal to

V R(t) = R I max sin( t - )
V C(t) = -
I max
C
cos( t - )
(34.64)
V L (t) =  L I max cos( t - )
The three phasors corresponding to these three voltages are shown in the phasor
Figure 34.9. The voltage drop across the resistor has the same phase as the current.
sum of these three vectors is also indicated and should be equal to the applied
amplitude of the vector sum of the three phasors must be equal to the amplitude of
emf. Thus
- 14 -
diagram in
The vector
emf. The
the applied
Chapter 34
Physics 122
Vr
Vl


Vc
Figure 34.9. Phasor circuit for LCR circuit.

 max = I max
2
R + L-
1
C
2
(34.65)
The phase of the vector sum of the phasors in Figure 34.9 is equal to t, and the angle between
the current (and the phasor representing the voltage drop across the resistor) and the vector sum
of the phasors is equal to the phase angle . From Figure 34.9 it is obvious that

1
C
R
L-
tan() =
(34.66)
Example: Problem 34.26
Consider the circuit shown in Figure 34.10. The oscillating source of emf delivers a
sinusoidal emf of amplitude 0.80 V and frequency 400 Hz. The inductance is 5.0 x 10-2 H, and
the capacitances are 8.0 x 10-7 F and 16.0 x 10-7 F. Find the maximum instantaneous current in
each capacitor.
Consider first the two capacitors. The emf across each of the capacitor must always be
the same. This implies that
- 15 -
Chapter 34
Physics 122
Q1
C1
=
Q2
(34.67)
C2
Rewriting eq.(34.67) in terms of the current I1 through capacitor C1 and the current I2 through
capacitor C2 we obtain
1
C1
I 1 dt =
1
C2
(34.68)
I 2 dt
or
I1
C1
-
I2
C2
(34.69)
dt = 0
Equation (34.69) can only be true at all times if the integrand is equal to zero. This requires that
I1
C1
=
I2
(34.70)
C2
In order to determine the maximum current in the circuit we use the phasor technique just
discussed. Consider the phasor diagram shown in Figure 34.11. The phasor labeled I indicates
the current in the circuit. The voltages across the inductor and the capacitor are 90 degrees out
of phase with the current and are indicated in Figure 34.11 by the phasors labeled V L and VC.
The total voltage drop across the circuit elements (vector sum of VL and VC) is also 90 degrees
out of phase with the current. Since the total voltage drop across the circuit elements must be
L
C2
C1
Figure 34.10. Problem 34.26.
- 16 -
Chapter 34
Physics 122

Vl


Vc
Figure 34.11. Phasor diagram for Problem 34.26.
equal to the applied emf, we conclude that the phase angle between the current and the emf is ±
90 degrees. The sign depends on the values of inductance, the capacitance and the angular
frequency of the emf.
The magnitude of the vector sum of the voltages across the inductor and the capacitor
must be equal to the magnitude of the emf. Thus

I max
 L I max =  max
C
(34.71)
Equation (34.71) can be used to determine the maximum current in the circuit:

I max =
 max
1
L C
(34.72)
The capacitance C used in eq.(34.72) is the net capacitance of the parallel network consisting of
capacitor C1 and capacitor C2 (C = C1 + C2). The sum of the currents flowing through
capacitors is equal to the maximum current in eq.(34.73). To determine the current through
capacitor C1 and capacitor C2 we can combine eq.(34.72) and eq.(34.70). In this manner we
obtain
- 17 -
Chapter 34
Physics 122

I1 =
I max
1+
C1
 max
=
1+
C2
C1
C2
(34.73)
1
L -
 C1 + C2
and

I2 =
I max
1+
C2
 max
=
1+
C1
C2
C1
L -
(34.74)
1
 C1 + C2
Example: Problem 34.32
An RC circuit consists of a resistor with R = 0.80  and a capacitor with C = 1.5 x 10-4 F
connected in series with an oscillating source of emf. The source generates a sinusoidal emf
with emax = 0.40 V and angular frequency equal to 9 x 103 rad/s. Find the maximum current in
the circuit. Find the phase angle of the current and draw a phasor diagram, with the correct
lengths and angles for the phasors. Find the average dissipation of power in the resistor.
The applied emf and the potential drops across the circuit elements in the RC circuit are
listed in eq.(34.75).
(t) =  max sin( t)
(34.75a)
V R(t) = R I max sin( t - )
(34.75b)

I max
V C(t) = cos( t - )
C
(34.75c)
The phasors representing the voltage drops across the resistor and across the capacitor are shown
in Figure 34.12. The vector sum of these phasors is also indicated. The magnitude of the
vector sum of the phasors must be equal to the magnitude of the applied emf. Thus

2
R I max +
I max
C
2
= I max
- 18 -
1
R +
C
2
2
=  max
(34.76)
Chapter 34
Physics 122
Vr

Vr + Vc

Vc
Figure 34.12. Phasor diagram for problem 34.32.
The maximum current is thus equal to

I max =
 max
1
R +
C
2
2
(34.77)
The phase angle  can be calculated easily (see Figure 34.12). It is determined by

1
C
1
tan() =
=
R
RC
34.7.
(34.78)
The Transformer
A transformer consists of two coils wound around an iron core (see Figure 34.13). The
iron core increases the strength of the magnetic field in its interior by a large fraction (up to
5000) and as a consequence, the field lines must concentrate in the iron. One of the coils, the
primary coil, is connected to a source of alternating emf.
The emf induced in the primary coil is related to the rate of change of magnetic flux
(Faraday’s law of induction):
- 19 -
Chapter 34
Physics 122

d 1
 1 ind = dt
(34.79)
Applying Kirchhoff’s second rule to the primary circuit, we conclude that the induced emf in the
coil must be equal to the applied emf. Thus

d 1
1 = dt
(34.80)
All field lines that pass through a winding of coil 1 will also pass through a winding of coil 2.
The flux through each winding of the primary coil is therefore equal to the flux through each
winding of the secondary coil. If the primary coil has N1 windings and the secondary coil has
N2 windings, then the total flux through the two coils are related
 1
N1
=
2
N2
(34.81)
or

N2
2 =

N1 1
(34.82)
The change in the enclosed flux of the primary coil will be related in the same way to the change
of flux in the secondary coil:

Figure 34.13. The transformer.
- 20 -
Chapter 34
Physics 122
d
 2
dt
=
N 2 d 1
N 1 dt
(34.83)
The emf induced across the secondary coil can be obtained using Faraday’s law and can be
expressed in terms of the emf in the primary circuit:

d 2
N 2 d 1
N2
2 = ==

dt
N 1 dt
N1 1
(34.84)
This emf is available to the various loads in the secondary circuit.
If the secondary circuit is open, no current will flow in it, and the primary circuit is
nothing else than a single-loop circuit with an alternating source of emf and an inductor. The
average power dissipated by the emf in such a circuit is zero, and consequently the transformer
does not consume any electric power.
If the secondary circuit is connected to a load, a current will flow. This induced current
will change the magnetic flux in the transformer and induce a current in the primary coil. If this
occurs, the primary circuit will consume power. In an ideal capacitor, the power delivered by
the source of emf in the primary circuit equal the power that the secondary circuit delivers to its
load. Thus
 1 I 1 =  2 I 2
- 21 -
(34.85)