Download Alternating Current (AC) Circuits

Physics 169
Kitt Peak National Observatory
Tuesday, April 19, 16
Luis anchordoqui
1
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
10.1 Alternating Current (AC) Voltage
Driven RLC Circuits
We learned that changing
magnetic flux could induce an emf according to Faraday’s law
12.1 AC Sources
An example of an AC source is
In Chapter 10of
we learned
that changing magnetic
If coil rotates in presence
magnetic
fieldflux could 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 emfVvaries
withω
time
and
to an alternating current
,for leads
(t)ofsinusoidally
= V0 sin(
t)varies
induced
emf
sinusoidally
with (12.1.1)
time and leads to AC
(AC), and provides a source
AC power.
The symbol
an AC voltage
source is
Symbol for AC voltage source ☛
where the maximum value
amplitude. The voltage varies between V0 and
V0 isof ancalled
An example
AC sourcethe
is
(12.1.1) as a function of
= V sin(−1.
ω t) , A graph of voltage
−V0 since a sine function varies between +1V (t)and
0
V (t) = V sin(!t)
the maximum value V is called the amplitude. 0
The voltage varies between V and
time is shown in Figurewhere
12.1.1.
The phase of the voltage
source is φV = ω t , (the phase
−V 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. The phase of the voltage source is φ = ω t , (the phase
constant is zero in Eq. (12.1.1)).
0
0
0
V
constant is zero in Eq. (12.1.1)).
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 = 2π f .
Figure 12.1.1 Sinusoidal voltage source
When a voltage source is connected to a 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.
! = 2⇡f = 2⇡/T
The sine function is periodic
in time. This means that the value of the voltage at time t
After an initial “transient time,” an AC current will flow in the circuit as a response to the
source.
The current
is also sinusoidal,
T is the period. The frequency,
will be exactly the samedriving
at avoltage
later
time
t ′ =intthe+circuit
T where
1
f , defined
unit
☛ as f = 1/ T , has the unit of inverse seconds ( s-1 ), or hertz12-2( Hz ). The angular
frequency is defined to be ω = 2π f .
s
Tuesday, April 19, 16
⌘ Hz
2
purely resistive circuit corresponds to infinite capacitance C = ∞ and
10.2 AC circuits
L = 0 .)with a source and one circuit element
Purely Resistive Load
We’d like to find current through resistor
IR (t) = I0,R sin(!t +
R)
Applying Kirchhoff’s loop rule yields
V (t)
IR (t)R = 0
V (t)
V0 sin(!t)
Figure
12.2.1 A purely resistive circuit
IR (t) =
=
= I0,R
sin(!t)
R
R
would
to find the
current
through
the resistor,
VR (t) = IRWe
(t)R
☛ like
instantaneous
voltage
drop
across the
resistor
I0,R
V0,R
V0,R
=
=
R
XR
I R (t) = I R0 sin(ω t − φ R ) .
Applying Kirchhoff’s loop rule yields
XR = R ☛ resistive reactance
V (t) − I R (t)R = 0 ,
and VR (t) are in phase with each other
R = 0 ☛ IR (t)
Tuesday, April 19, 16
where VR (t) = I R (t)R is the instantaneous voltage drop across
3
me dependence of the current and the voltage across the resistor is depict
Time dependence of VR (t) and IR (t)
Phasor diagram for resistive circuit
12.2.2(a).
(a)
(b)
Phasor ☛ rotating vector having following properties:
(i) length:
the length
correspondsof
to the
I R (tamplitude
) and VR (t ) across the resistor.
12.2.2
(a) Time
dependence
(ii) angular speed: the vector rotates counterclockwise with angular speed !
m for(iii)
theprojection:
resistivethe
circuit.
projection of vector along vertical axis corresponds to value
(b) P
t
ehavior of I R (t ) and VR (t ) can also be represented with a phasor diagram
Tuesday, April 19, 16
4
in Figure 12.2.2(b). A phasor is a rotating vector having the following proper
of the alternating quantity at time
Average value of current over one period
1
hIR (t)i =
T
Z
T
0
Z
1
IR (t) dt =
T
T
0
I0,R
I0,R sin(!t)dt =
T
Z
T
sin(2⇡t/T ) dt = 0
0
Average of the square of the current is non-vanishing
2
hIR
(t)i
1
=
T
Z
T
0
2
IR
(t)
1
dt =
T
Z
T
0
2
I
0,R
2
I0,R
sin2 (!t)dt =
T
Z
T
sin2 (2⇡t/T ) dt =
0
1 2
I0,R
2
q
I0,R
2
hIR (t)i = p
root-mean-square (rms) current ☛ Irms =
2
q
V0,R
2
rms voltage ☛ Vrms =
hVR (t)i = p
2
rms voltage supplied to domestic wall outlets in US ☛ Vrms = 120 V @ 60 Hz
It is convenient to define:
Power dissipated in the resistor ☛
2
PR (t) = IR (t)VR (t) = IR
(t)R
Average power over one period
2
1
V
2
2
2
hPR (t)i = hIR
(t)Ri = I0,R
R = Irms
R = Irms Vrms = rms
2
R
Tuesday, April 19, 16
5
shall see below, a purely inductive circuit corresponds to infinite capacitan
zero resistance R = 0 .
Purely Inductive Load
We’d like to find current in the circuit
IL (t) = I0,L sin(!t
L)
Applying Kirchhoff’s loop rule yields
V (t)
VL (t) = V (t)
dIL Figure 12.2.3 A purely inductive circuit
L
=0
dt
dIL
V (t)We would
V0,Llike to find the current in the circuit,
=
=
sin(!t)
dt
L
L
I L (t) = I L0 sin(ω t − φ L ) .
IL (t) =
Z
Z
V0,L
V0,L
V0,L
dILApplying
=
sin(!t)dt
=
cos(!t)
=
⇡/2) yi
the modified
Kirchhoff’s
rule
for inductors,
thesin(!t
circuit equation
L
!L
!L
V0,L
V0,L
I0,L =
=
!L
XL
XL = !L
☛ inductive reactance
L = +⇡/2 ☛ phase constant
Tuesday, April 19, 16
dI L
V (t ) − VL (t ) = V (t ) − L
=0.
dt
6
(1
φ L = π / 2 ; it reaches its maximum value one quarter of a cycle later than VL (t ) .
2
The current lags voltage by π / 2 in a purely inductive circuit
oltage plots and the corresponding phasor diagram are shown
w. The word “lag” means that the plot of I (t ) is shifted to the right of the plot of V (t ) in
L
L
Figure 12.2.4 (a), whereas in the phasor diagram the phasor I L (t) is “behind” the phasor

for VL (t) as they rotate counterclockwise in Figure 12.2.4(b).
12.2.3 Purely Capacitive Load
Consider now a purely capacitive circuit with a capacitor connected to an AC generator
with AC source voltage given by V (t) = V0 sin(ω t) . In the purely capacitive case, both
resistance R and inductance L are zero. The circuit diagram is shown in Figure 12.2.5.
(a)
(b)
Figure 12.2.4 (a) Time dependence of I L (t ) and VL (t ) across the inductor. (b) Pha
Time for
dependence
of I L (t ) and VL (t ) across the inductor. (b)
diagram
the inductive circuit.
ductive
circuit.
As can be seen from
Tuesday, April 19, 16
Figure
12.2.5 Athe
purely
capacitive
I L (circuit
t ) is out of phase with VL7(t )
the
figures,
current
Purely Capacity Load
We’d like to find current in circuit
IC (t) = I0,C sin(!t
C)
Again ☛ Kirchhoff’s loop rule yields
V (t)
VC (t) = V (t)
Charge on capacitor
Q(t)
=0
C
Figure 12.2.5 A purely capacitive circuit
Q(t) = CV (t) = CVC (t) = CV0,C sin(!t)
We would like to find the current in the circuit,
Current
dQ
IC (t) =
= !CV0,C cos(!t) = !CV0,C sin(!t + ⇡/2)
dt
V0,C
I0,C = !CV0,C =
XC
1
XC =
☛ capacitance reactance
!C
☛ phase constant
C = ⇡/2
Tuesday, April 19, 16
8
VC (t ) .
The current leads the voltage by π /2 in a capacitive circuit
The word “lead” means that the plot of I C (t) is shifted to the left of the plot of VC (t) in

Figure 12.2.6 (a), whereas in the phasor diagram the phasor I C (t) is “ahead” the phasor

for VC (t) as they rotate counterclockwise in Figure 12.2.6(b).
12.3 The RLC Series Circuit
Consider now the driven series RLC circuit with V (t) = V0 sin(ω t + φ ) shown in Figure
12.3.1.
(b)
(a)
gure 12.2.6 (a) Time dependence of I C (t ) and VC (t ) across the capacitor. (b)
agram for the capacitive circuit.
Tuesday, April 19, 16
Figure 12.3.1 Driven series RLC Circuit
9
RLC circuit
series
V (t)what
= V0 sin(
I = + dQ now
/ dt , the
Differentiate Eq. (12.3.3), using Consider
anddriven
divide
through
by L ,with
yields
isω t + φ
12.3.1.
called aSingle
second order
damped
driven
differential equation,
10.3
Loop
RLClinear
AC
Circuit
We’d like to find current in 2circuit
I(t) = I0 sin(!t
ωV0
d I R dI
I
+
+
=
cos(ω t + φ ) .
2)
L dt LC
L
dt
(12.3.4)
Kirchhoff’s loop rule yields
We shall find the amplitude, I 0 , of the current, and phase constant φ which is the phase
the voltage
Vshift
(t)between
VR (t)
VL source
(t) and
VCthe
(t)current
= 0 by examining the phasors associates with
the three circuit elements R , L and C .
VThe
(t)instantaneous
= VR (t) voltages
+ VL (t)
+ Veach
C (t)
across
of the three circuit elements
R , L , and C has a
different amplitude and phase compared to the current, as can be seen from the phasor
Figure 12.3.1 Driven series RLC Circuit
Using phasor
representation
diagrams
shown in
Figure 12.3.2.
~0 = V
~0,R (a)
~0,L + V
~0,C
V
+V
(b)
(c)
Figure 12.3.2 Phasor diagrams for the relationships between current and voltage in (a) 10
Tuesday, April 19, 16
q
~ 0 = |V
~ 0 | = |V
~0,R + V
~0,L + V
~0,C | = V 2 + (V0,L V0,C )2
V
0,R
q
p
2 + (X
2 + (I X Eq.I(12.3.2)
2 =beIwritten
2
=Using(I
X
)
the0 X
phasor
R ) representation,
0 L
0 XC ) can
0 Xas
L
c
R
current amplitude
V0
I0 = p
 


V0 = VR 0 + VL 0 + VC 0
(12.3.5)
V0
q
=
2 12.3.3(a).
2 see that current phasor
as shown in Figure
Again
we
X
+
(X
X
)
L
C
 R
R2 + !L

I 0 1leads2 the capacitive

phase voltage phasor VC 0 by π / 2 but lags the inductive voltage phasor VL!C
0 by π / 2 . The three
✓ phasors rotate
◆ counterclockwise
✓
◆as time increases, with✓ their relative◆positions
voltage
XL XC
1
1
1
1 1
fixed.
tan =
=
!L
! = tan
!L
XR
Note that ☛
R
!C
V0 6= V0,R + V0,L + V0,C
R
!C
Figure 12.3.3 (a) Phasor diagram for the series RLC circuit. (b) voltage relationship
Tuesday, April 19, 16
11
The relationship between Z , X R , X L , and X C can be represented by th
in Figure
q12.3.4: X L − X C
Impedance ☛
Z=
2 + (X
XR
l
XC ) 2
V0
I(t) = sin(ω t)
Z
(12.3.13)
0
Z also depends on the angular frequency ω , as do X L and
Notice that the V
impedance
I(t) =
sin(!t)
Z
XC .
Vector
representation
of the
between
Z , we may
φ and
Using Eq. (12.3.9) for theFigure
phase 12.3.4
constant
Eq. (12.3.12)
forrelationship
readilyZ , X R
recover theSimple-circuit
limits for simple circuit
(with
only series
one element).
A summary is provided in
limits
of
the
RLC
circuit
The impedance also has SI units of ohms. In terms of Z , the current (Eq
Table 12.1 below:
(12.3.7)) may be rewritten as
Simple
Circuit
purely
resistive
purely
inductive
purely
capacitive
⎛ X − XC ⎞
φ = tan ⎜ L
X R ⎟⎠
⎝
Z=
X R 2 + ( X L − X C )2
R
L
C
X L = ωL
1
XC =
ωC
R
0
∞
0
0
0
XR
0
L
∞
XL
0
π /2
XL
0
0
C
0
XC
−π / 2
XC
−1
Table 12.1 Simple-circuit limits of the series RLC circuit
Tuesday, April 19, 16
12
10.4 Resonance
In driven RLC series circuit
☛ amplitude of current has a maximum value: a resonance
which occurs at the resonant angular frequency
Because current amplitude is inversely proportional to impedance
φ =0.
!0
maximum
occurs
when
A qualitative plot I
of0 the
amplitude
of the
current
as a functi
Z is minimum
frequency for two driven RLC circuits, with different values o
This occurs at angular frequency !0 such that XL = XC
illustrated in Figure 12.3.5. The amplitude is larger for small
resistance.
1
1
!0 L =
) !0 = p
!0 C
LC
V0
At resonance ☛ Z = R ) I0 =
R
Tuesday, April 19, 16
13
Figure 12.3.5 The amplitude of the current as a function of ω in t
10.5 Power in AC Circuits
In series RLC circuit
☛
instantaneous power delivered by AC generator is given by
V0
V02
P (t) = I(t)V (t) =
sin(!t
)V0 sin(!t) =
sin(!t) sin(!t
Z
Z
V02
=
[sin2 (!t) cos( ) sin(!t) cos(!t) sin( )]
Z
)
we have used trigonometric identity
sin(!t
) = sin(!t) cos( )
cos(!t) sin( )
Time average of the power is
Z
1 T V02
hP (!)i =
sin2 (!t) cos( )dt
T 0 Z
2
Vrms
=
cos = Irms Vrms cos
Z
power factor
Tuesday, April 19, 16
☛
cos
1
T
Z
T
0
V02
1 V02
sin(!t) cos(!t) sin( )dt =
cos
Z
2 Z
R
=
Z
14
In Figure 12.4.1, we plot the time-averaged power as a function of th
frequency ω for two driven RLC circuits, with different values of
2
hP (!)i = Irms
(!)R
V0
1
Irms (!) = p q
2 R2 + !L
1 2
!C
= 1power
R
Maximum @ resonance condition
☛ cos
or Z
Figure 12.4.1
Average
as =
a function
of frequency in a driven
We see that P(ω ) attains the maximum value when cos φ = 1 , o
resonance condition. At resonance,
we have
2
hP (!0 )i = Irms Vrms = Vrms R
P(ω 0 ) = I rmsVrms =
Tuesday, April 19, 16
2
Vrms
R
.
15
Width of the peak
12.4.1 Width of the Peak
The peak has a line width
hasisa to
linedefine
width. One
the width is to defi
One way to characterizeThe
thepeak
width
! way
= !to+characterize
!
where ω ± are the values of the driving angular frequency such that the
!± ☛ values of driving
angular frequency
halfthat
its power
maximum
powerto athalf
resonance.
This power
is called
width at h
such
is equal
its maximum
at full
resonance
illustrated in Figure 12.4.2. The width Δω increases with resistance R .
This is called full width at half maximum
Width
! increases with resistance R
Tuesday, April 19, 16
Figure 12.4.2 Width of the peak
To find Δω , it is instructive to first rewrite the average power P(ω16) a
To find
!
it is instructive to first rewrite the average power as
1
V02 R
hP (!)i =
2 R2 + !L
1 2
!C
1
V02 R! 2
=
2 ! 2 R2 + L2 (! 2
!02 )2
V02
hP (!0 )i =
2R
condition for finding
!±
is
2
V02 R!±
1
V0
1
hP (!0 )i = hP (!± )i )
=
2 R2 + L2 (! 2
2
4R
2 !±
±
after some algebra
2
☛ (!±
!02 )2
!02 )2 = (R!± /L)2
Taking square roots yields two solutions which we analyze separately
Tuesday, April 19, 16
17
Case 1: Taking the positive root leads to
2
!+
!02
R!+
R
=+
) !+ =
+
L
2L
s✓
Case 2: Taking the negative yields
!
2
!02
=
R!
L
width at half maximum
)! =
☛
R
+
2L
! = !+
R
4L
s✓
◆2
R
4L
+ !02
◆2
+ !02
R
! =
L
Quality factor
Qqual
Tuesday, April 19, 16
!0
!0 L
=
=
!
R
18
10.6 Transformer
Transformer is device used to increase or decrease the AC voltage in a circuit
Typical device consists of two coils of wire
primary and secondary wound around an iron core
Primary coil with
N1 turns is connected to alternating voltage source V (t)
Secondary coil has N2 turns and is connected to a load with resistance R2
Figure 12.5.1 A transformer
The way transformers operate is based on the principle that
Inalternating
the primary
circuit,
neglecting
in the
Faraday’scoil
law of
current
in primary
coil the
will small
induce resistance
an alternating
emfcoil,
on secondary
induction implies
due to their mutual inductance
dΦB
Tuesday, April 19, 16
19
V1 = − N1
,
(12.5.1)
Neglecting small resistance in coil
V1 =
B
☛
N1
Faraday’s law of induction implies
d
B
dt
☛ magnetic flux through one turn of primary coil
Iron core extends from primary to secondary coils
Iron core serves to increase magnetic field produced by current in primary coil
and ensures that nearly all magnetic flux through primary coil
also passes through each turn of the secondary coil
Voltage (or induced emf) across secondary coil is
☛ V2 =
N2
d
B
dt
Ideal transformer ☛ power loss due to Joule heating can be ignored
so that power supplied by primary coil is completely transferred to secondary coil
I 1 V1 = I 2 V2
In addition
☛ if
no magnetic flux leaks out from iron core
through each turn is same in both primary and secondary coils
V2
N2
=
V1
N1
Tuesday, April 19, 16
20
Transformation of currents in the two coils reads
V2
N2
I1 =
I2 =
I2
V1
N1
Ratio of output voltage to input voltage is determined by turn ratio
N 2 > N 1 ) V 2 > V1
N2
N1
output voltage in secondary coil is greater than input voltage in primary coil
Transformer with N2 > N1
☛
step-up transformer
N 1 > N 2 ) V 1 > V2
output voltage is smaller than input
Transformer with N1 > N2
☛
step-down transformer
For home safety ☛ we would like LOW emf supply
Tuesday, April 19, 16
21
Why do we use high voltages?
As electricity flows down a metal wire
electrons carrying its energy jiggle through the metal structure
That's why wires get hot when electricity flows through them
(useful for electric toasters and other appliances that use heating elements)
Electricity that comes from power plants is sent dow wires at extremely high voltages
to save energy
2
hP (!)i = Irms
R
2
For power transmission ☛ we’d like to keep Irms at minimum
The higher the voltage and the lower the current ☛ the less energy is wasted
Tuesday, April 19, 16
22