# 6/15/2011 Chapter 31 ```6/15/2011
Chapter 31
A horizontal power line carries a current of 5000 A
from south to north. Earth's magnetic eld (60 &micro;T) is
directed towards north and inclined down-ward 50
degrees to the horizontal. Find the magnitude and
direction of the magnetic force on 100 m of the line
due to the Earth's field.
(1)23 N, west
(2) 23 N, east
(3) 30N, west
(4) 30N, east (5) none of these
Electromagnetic Oscillations and Alternating
Current
In this chapter we will cover the following topics:
-Electromagnetic oscillations in an LC circuit
-Alternating current (AC) circuits with capacitors
-Resonance in RCL circuits
-Power in AC-circuits
-Transformers, AC power transmission
I
50
N
(31 - 1)
B
and the magnetic flux through it is 5Wb. If the page is turned
by 30◦ around an edge the flux through it will be:
A car travels northward at 75 km/h along a straight road in a
region where Earth’s magnetic field has a vertical component of
0.50 &times; 10−4 T. The emf induced between the left and right
side, separated by 1.7m, is:
A. 2.5Wb
B. 4.3Wb
C. 5Wb
D. 5.8Wb
E. 10Wb
A. 0
B. 1.8mV
C. 3.6mV
D. 6.4mV
E. 13mV
LC Oscillations
L
C
d 2q 1
d 2q  1 
+ q = 0→ 2 +
 q = 0 (eqs.1)
2
dt
C
dt
 LC 
C
This is a homogeneous, second order, linear differential equation
L
L
The circuit shown in the figure consists of a capacitor C
and an inductor L. We give the capacitor an initial
chanrge Q and then abserve what happens. The capacitor
which we have encountered previously. We used it to describe
will discharge through the inductor resulting in a time
the simple harmonic oscillator (SHO)
d 2x
+ ω 2 x = 0 (eqs.2)
dt 2
with solution: x(t ) = X cos(ωt + φ )
q (t ) = Q cos (ωt + φ )
dependent current i.
We will show that the charge q on the capacitor plates as well as the current i
1
LC
The total energy U in the circuit is the sum of the energy stored in the electric field
in the inductor oscillate with constant amplitude at an angular frequency ω =
of the capacitor and the magnetic field of the inductor. U = U E + U B =
q 2 Li 2
+
.
2C
2
The total energy of the circuit does not change with time. Thus
dU
=0
dt
dU q dq
di
dq
di d 2 q
d 2q 1
=
+ Li = 0. i =
→ = 2 →L 2 + q=0
dt C dt
dt
dt
dt dt
dt
C
(31 - 2)
ω=
1
LC
If we compare eqs.1 with eqs.2 we find that the solution to the differential
equation that describes the LC-circuit (eqs.1) is:
q(t ) = Q cos (ωt + φ )
The current i =
where
ω=
1
, and φ is the phase angle.
LC
dq
= −ωQ sin (ωt + φ )
dt
(31 - 3)
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2
L
2
4
t = 3T / 8
2
q
Q
=
cos 2 (ωt + φ )
2C 2C
The energy stored in the magnetic field of the inductor
UE =
3
t =T /4
t =T /8
The energy stored in the electric field of the capacitor
C
8
6
t = 7T / 8
1
Q2
T
3T
at t = 0, , T , ,...
2C
2
2
Q2
T 3T 5T
The energy of the magnetic field has a maximum value of
at t = , , ,...
2C
4 4 4
Note : When U E is maximum U B is zero, and vice versa
(31 - 4)
t = 5T / 8
3
7
5
The energy of the electric field has a maximum value of
7
2
4
6
t = 3T / 4
8
(31 - 5)
q (t )
Damped oscillations in an RCL circuit
dU
= −i 2 R
dt
2
2
q
Li
dU q dq
di
+
U = UE +UB =
→
=
+ Li = −i 2 R
2C
2
dt C dt
dt
t =T /2
t =0
 Q2 
Q2
2
2
U =
 cos ( ωt + φ ) + sin ( ωt + φ )  =
2C
 2C 
The total energy is constant; energy is conserved
If we add a resistor in an RL cicuit (see figure) we must
modify the energy equation because now energy is
5
1
Li 2 Lω 2Q 2
Q2
=
sin 2 ( ωt + φ ) =
sin 2 ( ωt + φ )
UB =
2
2
2C
The total energy U = U E + U B
q (t ) = Qe− Rt / 2 L cos (ω ′t + φ )
Qe− Rt / 2 L
Q
q (t )
ω′ =
being dissipated on the resistor.
2
Qe− Rt / 2 L
2
dq
di d q
d q
dq 1
→ = 2 → L 2 + R + q = 0 This is the same equation as that
dt
dt dt
dt
dt C
d 2x
dx
of the damped harmonics oscillator: m 2 + b + kx = 0 which has the solution:
dt
dt
i=
−Q
x(t ) = xme −bt / 2 m cos (ω ′t + φ )
The angular frequency ω ′ =
k
b2
−
m 4m 2
For the damped RCL circuit the solution is:
q(t ) = Qe − Rt / 2 L cos ( ω ′t + φ )
The angular frequency ω ′ =
The equations above describe a harmonic oscillator with an exponetially decaying
amplitude Qe − Rt / 2 L . The angular frequency of the damped oscillator
1
R2
1
−
of the
is always smaller than the angular frequency ω =
LC 4 L2
LC
2
1
R
undamped oscillator. If the term
we can use the approximation ω ′ ≈ ω
4 L2
LC
ω′ =
1
R2
−
LC 4 L2
(31 - 7)
(31 - 6)
Three Simple Circuits
Our objective is to analyze the circuit shown in the
figure (RCL circuit). The discussion will be greatly
simplified if we examine what happens if we connect
each of the three elements (R, C , and L) separately
Alternating Current
A battery for which the emf is constant generates
a current that has a constant direction. This type
of current is known as &quot;direct current &quot; or &quot;dc&quot;
E = Em sin ωt
1
R2
−
LC 4 L2
In chapter 30 we encountered a different type
to an ac generator.
of sourse (see figure) whose emf is:
A convention
E = ω NAB sin ωt = Em sin ωt where Em = ω NAB, A is the area of the generator
From now on we will use the standard notation for ac circuit
analysis. Lower case letters will be used to indicate the
windings, N is the number of the windings, ω is the angular frequency of the
rotation of the windings, and B is the magnetic field. This type of generator
L
C
is known as &quot;alternating current&quot; or &quot;ac&quot; because the emf as well as the current
change direction with a frequency f = 2πω. In the US f = 60 Hz.
will be used to indicate the constant amplitudes of ac quantities.
Example: The capacitor charge in an LC circuit was written as:
q = Q cos ( ωt + φ )
Almost all commercial electrical power used today is ac even though the
analysis of ac circuits is more complicated than that of dc circuits.
The symbol q is used for the instantaneous value of the charge
The reasons why ac power was adapted will be discussed at the end of this
The symbol Q is used for the constant amplitude of q
chapter.
(31 - 8)
instantaneous values of ac quantities. Upper case letters
(31 - 9)
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VR = I R R
In fig.a we show an ac generator connected to a resistor R
From KLR we have: E − iR R = 0 → iR =
E Em
= sin ωt
R R
Em
R
across R is equal to Em sin ωt
The current amplitude I R =
The voltage vR
The voltage amplitude is equal to Em
The relation between the voltage and
current amplitudes is: VR = I R R
The resistor voltage vR and the resistor current iR are represented
In fig.b we plot the resistor current iR and the
resistor voltage vR as function of time t.
Both quantities reach their maximum values
at the same time. We say that voltage and
current are in phase.
(31 - 10)
2
2
⟨ P⟩ =
E
Em
Vrms ≡ m
2R
2
⟨ P⟩ =
2
⟨ P⟩ =
2
1
P (t ) dt
T ∫0
P=
Em
sin 2 ωt →
R
2. The length of each phasor is proportional to tha ac quantity amplitude
3. The projection of the phasor on the vertical axis gives the instantaneous
value of the ac quantity.
4. The rotation angle for each phasor is equal to the phase of the
ac quantity (ωt in this example)
(31 - 11)
Em 1
sin 2 ωtdt
R T ∫0
T
2
→ ⟨ P⟩ =
In fig.a we show an ac generator connected to a
capacitor C
XC
q
From KLR we have: E − C = 0 →
C
qC = EC = EmC sin ωt
T
1
1
sin 2 ωtdt =
2
T ∫0
by rotating vectors known as phasors using the following conventions:
1. Phasors rotate in the counterclockwise direction with angular speed ω
Average Power for R
T
A convenient method for the representation of ac
quantities is that of phasors
Em
2R
dqC
= EmCω cos ωtdt = sin ( ωt + 90&deg; )
dt
The voltage amplitude VC equal to Em
ω
O
iC =
1
XC =
ωC
VC
1/ ωC
The quantity X C = 1 / ωC is known as the
The current amplitude IC = ωCVC =
capacitive reactance
We define the &quot;root mean square&quot; (rms) value of V as follows:
In fig.b we plot the capacitor current iC and the capacitor
2
Em
V2
→ ⟨ P⟩ = rms
The equation looks the same
R
2
as in the DC case. This power appears as heat on R
voltage vC as function of time t. The current leads the
Vrms ≡
(31 - 12)
Average Power for C
PC = 0
2
E
P = VC I C = m sin ωt cos ωt
XC
2
2sin θ cos θ = sin 2θ
T
⟨ P⟩ =
→ P=
2
voltage by a quarter of a period. The voltage and
(31 - 13)
current are out of phase by 90 &deg;.
Em
sin 2ωt
2XC
T
E 1
1
P (t ) dt = m
sin 2ωtdt = 0
T ∫0
2 X C T ∫0
Note : A capacitor does not dissipate any power
on the average. In some parts of the cycle it absorbes
energy from the ac generator but at the rest of the cycle
it gives the energy back so that on the average no
power is used!
In fig.a we show an ac generator connected to an inductor L
di
di
E E
From KLR we have: E − L L = 0 → L = = m sin ωt
dt
dt L L
E
E
E
iL = ∫ diL = ∫ m sin ωtdt = − m cos ωtdt = m sin (ωt − 90&deg; )
L
ωL
ωL
The voltage amplitude VL equal to Em
XL
V
The current amplitude I L = L
ωL
ω
The quantity X L = ω L is known as the
O
inductive reactance
X L = Lω
In fig.b we plot the inductor current iL and the
inductor voltage vL as function of time t.
The current lags behind the voltage by a
quarter of a period. The voltage and
current are out of phase by 90 &deg;.
(31 - 14)
(31 - 15)
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PL = 0
Average Power for L
SUMMARY
2
Power P = VL I L = −
Em
sin ωt cos ωt
XL
2
2sin θ cos θ = sin 2θ
→ P=−
T
⟨ P⟩ ≡
2
Em
sin 2ωt
2X L
T
E 1
1
P (t ) dt = − m
sin 2ωtdt = 0
T ∫0
2 X L T ∫0
Circuit
element
Average
Power
Resistor
R
⟨ PR ⟩ =
Reactance
R
Note : A inductor does not dissipate any power
Capacitor
on the average. In some parts of the cycle it absorbes
energy from the ac generator but at the rest of the cycle
C
⟨ PC ⟩ = 0
XC =
Inductor
L
⟨ PL ⟩ = 0
X L = ωL
it gives the energy back so that on the average no
power is used!
Voltage amplitude
Current is in phase
with the voltage
2
Em
2R
Phase of current
1
ωC
VR = I R R
by a quarter of a
period
VC = I C X C =
Current lags behind
voltage by a quarter
of a period
VL = I L X L = I Lω L
(31 - 16)
IC
ωC
(31 - 17)
(31 - 18)
A
i = I sin (ωt − φ )
B
Z = R2 + ( X L − X C )
O
The series RCL circuit
Em
Z
Kirchhoff's loop rule (KLR) for the RCL circuit: E = vR + vC + vL . This equation
An ac generator with emf E = Em sin ωt is connected to
is represented in phasor form in fig.d. Because VL and VC have opposite directions
an in series combination of a resistor R, a capacitor C
we combine the two in a single phasor VL − VC . From triangle OAB we have:
and an inductor L, as shown in the figure. The phasor
Em2 = VR2 + (VL − VC ) = ( IR ) + ( IX L − IX C ) = I 2  R 2 + ( X L − X C )  →


Em
I=
The denominator is known as the &quot;impedance&quot; Z
2
R2 + ( X L − XC )
2
for the ac generator is given in fig.c. The current in
i = I sin (ωt − φ )
I=
2
this circuit is described by the equation: i = I sin ( ωt − φ )
The current i is common for the resistor, the capacitor and the inductor
The phasor for the current is shown in fig.a. In fig.c we show the phasors for the
voltage vR across R, the voltage vC across C , and the voltage vL across L.
of the circuit.
I=
The voltage vR is in phase with the current i. The voltage vC lags behind
2
2
Z = R 2 + ( X L − X C ) → The current amplitude I =
2
Em
Z
Em
1 

R + ωL −
ωC 

2
2
the current i by 90&deg;. The voltage vL leads ahead of the current i by 90&deg;.
2
(31 - 19)
i = I sin (ωt − φ )
(31 - 20)
1
XC =
ωC
A
B
Z = R2 + ( X L − X C )
tan φ =
O
2
X L − XC
R
X L = ωL
1. Fig.a and b: X L &gt; X C → φ &gt; 0
V −V
IX − IX C X L − X C
From triangle OAB we have: tan φ = L C = L
=
VR
IR
R
The current phasor lags behind
the generator phasor. The circuit is more
We distinguish the following three cases depending on the relative values
inductive than capacitive
of X L and X L .
1. X L &gt; X C → φ &gt; 0 The current phasor lags behind the generator phasor.
The circuit is more inductive than capacitive
2. X C &gt; X L → φ &lt; 0 The current phasor leads ahead of the generator phasor
The circuit is more capacitive than inductive
3. X C = X L → φ = 0 The current phasor and the generator phasor are in phase
2. Fig.c and d: X C &gt; X L → φ &lt; 0 The current phasor leads ahead of the generator
phasor. The circuit is more capacitive than inductive
3. Fig.e and f: X C = X L → φ = 0 The current phasor and the generator phasor are
(31 - 21)
in phase
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ω=
1
LC
I res =
2
Pavg = I rms
R
Resonance
Em
R
(31 - 23)
Power in an RCL ciruit
the angular frequency ω of the ac generator can
be varied continuously. The current amplitude
We already have seen that the average power used by
a capacitor and an inductor is equal to zero. The
power on the average is consumed by the resistor.
in the circuit is given by the equation:
Em
I=
1 

R2 + ω L −
ωC 

2
The current amplitude
This occurs when ω =
The instantaneous power P = i 2 R =  I sin (ωt − φ )  R
2
T
has a maximum when the term ω L −
The average power Pavg =
1
=0
ωC
Em
R
A plot of the current amplitude I as function of ω is shown in the lower figure.
This plot is known as a &quot;resonance curve&quot;
(31 - 22)
Transmission lines
Εrms =735 kV , I rms = 500 A
(31 - 24)
The transformer is a device that can change
the voltage amplitude of any ac signal. It
consists of two coils with different number
R = 220Ω
of turns wound around a common iron core.
l = 1000 km
Power Station
Energy Transmission Requirements
The resistance of the power line R =
2
Heating of power lines Pheat = I rms
R
ρl
. R is fixed (220 Ω in our example)
A
This parameter is also fixed
(31 - 25)
The transformer
home
T2
T1
The coil on which we apply the voltage to be changed is called the &quot;primary&quot; and
it has N P turns. The transformer output appears on the second coils which is known
as the &quot;secondary&quot; and has N S turns. The role of the iron core is to insure that the
magnetic field lines from one coil also pass through the second. We assume that
if voltage equal to VP is applied across the primary then a voltage VS appears
( 55 MW in our example)
Power transmitted Ptrans = Erms I rms
power consumed by the circuit is maximum
when φ = 0
Step-down
transformer 110 V
Step-up
transformer
1
Pdt
T ∫0
1 T
 I 2R
2
2
Pavg = I 2 R  ∫ sin (ωt − φ ) dt  =
= I rms
R
2
T
 0

E
R
Pavg = I rms RI rms = I rms R rms = I rmsErms = I rms Erms cos φ
Z
Z
The term cosφ in the equation above is known as
the &quot;power factor&quot; of the circuit. The average
1
LC
The equation above is the condition for resonance. When its is satisfied I res =
(368 MW in our example)
In our example Pheat is almost 15 % of Ptrans and is acceptable
To keep Pheat we must keep I rms as low as possible. The only way to accomplish this
is by increasing Erms . In our example Erms = 735 kV. To do that we need a device
that can change the amplitude of any ac voltage (either increase or decrease)
VS
V
= P
NS NP
Pavg = I rms Erms cos φ
In the RCL circuit shown in the figure assume that
dΦ P
dB
= − NP A
(eqs.1)
dt
dt
d ΦS
dB
Φ S = N S BA → VS = −
= −NS A
(eqs.2)
dt
dt
If we divide equation 2 by equation 1 we get:
Φ P = N P BA → VP = −
dB
− NS A
VS
dt = N S → VS = VP
=
VP − N A dB
NP
NS NP
P
dt
on the secondary coil. We also assume that the magnetic field through both coils
is equal to B and that the iron core has cross sectional area A. The magnetic flux
dΦP
dB
= −NP A
(eqs.1)
dt
dt
dΦS
dB
= −NS A
The flux through the secondary Φ S = N S BA → VS = −
dt
dt
through the primary Φ P = N P BA → VP = −
IS
IP
VS
V
= P
NS NP
(eqs.2)
IS NS = I P NP
We have that:
VS
V
= P
NS NP
→ VS N P = VP N S
(eqs.1)
If we close switch S in the figure we have in addition to the primary current I P
a current I S in the secondary coil. We assume that the transformer is &quot;ideal&quot;
The voltage on the secondary VS = VP
NS
NP
If N S &gt; N P →
NS
&gt; 1 → VS &gt; VP We have what is known a &quot;step up&quot; transformer
NP
If N S &lt; N P →
NS
&lt; 1 → VS &lt; VP We have what is known a &quot;step down&quot; transformer
NP
Both types of transformers are used in the transport of electric power over large
distances.
(31 - 26)
i.e. it suffers no losses due to heating then we have: VP I P = VS I S
(eqs.2)
VP I P
V I
If we divide eqs.2 with eqs.1 we get:
= S S → I P N P = I S NS
VP N S
VS N P
IS =
NP
IP
NS
In a step-up transformer (N S &gt; N P ) we have that I S &lt; I P
In a step-down transformer (N S &lt; N P ) we have that I S &gt; I P
(31 - 27)
5
6/15/2011
Hitt
A generator supplies 100 V to the primary coil of a
transformer. The primary has 50 turns and the
secondary has 500 turns. The secondary voltage is:
A. 1000 V
B. 500 V
C. 250 V
D. 100 V
E. 10V
hitt
The main reason that alternating current replaced
direct current for general use is:
A. ac generators do not need slip rings
B. ac voltages may be conveniently transformed
C. electric clocks do not work on dc
D. a given ac current does not heat a power line as
much as the same dc current
E. ac minimizes magnetic effects
6
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