Chapter 36 Viewgraphs
AC Circuits
Introduction
Most currents and voltages vary in time.
The presence of circuit elements like capacitors and inductors
complicates the relation between currents and voltage when
these depend on time.
Resistive element
-I&V proportional
V(t) = I(t)R
Reactive elements
involves derivatives
I(t) = C
d
Vc (t)
dt
VL (t) = L
d
I(t)
dt
Voltage and current are not simply proportional for reactive
elements. Ohm’s law does not apply.
Three categories of time behavior
1. Direct Current (DC) Voltages and currents are constants in time.
Example: batteries - circuits driven by batteries
2. Transients Voltages and currents change in time after a switch is
opened or closed. Changes diminish in time and stop if you wait
long enough.
S
R
VL(t)
VL (t) = V0 exp[- tR / L]
V0
L
t
3. Alternating Current (AC). The voltages and currents continually
change sinusoidally in time.
frequency
V(t) = V0 cos[wt + q]
amplitude
phase
Examples: our power grid when it is on. f=60 Hz, V=110 V (RMS)
audio signals
communication signals
Power in microwave ovens
Power in MRI machines
Real Life voltages involve DC, AC and Transients
http://www.tvtower.com/images/rf-spectrum.jpg
Power Supply: converts AC
to DC
Present inside almost all
home electronics
Inverter: converts DC to AC
Plugs into cigarette lighter,
charges laptop.
Don’t run a hair straighter on
one of these while driving in
your car.
AC - Circuits
First Rule of AC - Circuits - everything oscillates at the same
frequency
If a circuit is driven by a source with frequency w, and you
wait for all transients to die out, the circuit will reach a state where
every voltage and current is oscillating at the same frequency w.
Often this is called a “steady state” even though every thing is
oscillating.
The problem then becomes: Find the amplitude and phase of each
voltage and current.
Cc
Vs (t) = V0 cos[wt + q]
Cs
Ls
Lc/p
Rc/p
Lc/m
Rc/m
Lc/m
Rc/m
Lc/n
Rc/n
Complicated circuit:
Rs, Ls, and Cs
Every voltage will be in the form
Vn (t) = V0n cos[wt + qn ]
Every current will be in the form
Im (t) = I 0m cos[wt + qm ]
Problem is to find the amplitudes and phases
Cc
Some general comments
about circuits driven by a
source with frequency w.
Cs
Ls
Lc/p
Rc/p
Lc/m
Rc/m
Lc/m
Rc/m
Lc/n
Rc/n
Vs (t) = V0 cos[wt + q]
1. All voltages and currents oscillate at the same frequency w.
2. Amplitudes and phases of voltages and currents depend on
source and Rs, Cs, Ls, and w.
3. Amplitudes of voltages and currents are proportional to
source voltage.
4. Phases of voltages and currents do not depend on amplitude
of source voltage.
5. Shifting the phase of the source shifts the phase of all
voltages and currents by the same amount.
Let’s do a “simple” example
Vs(t)
Vs (t) = V0 cos[wt + qs ]
R
I(t)
L
Can take: qs = 0
Current I, flows through both R and L
I(t) = I 0 cos[wt + qI ]
Resistor Voltage VR = RI(t) = RI 0 cos[wt + qI ]
Inductor Voltage
d
p
VL = L I(t) = - wLI 0 sin[wt + qI ] = wLI 0 cos[wt + qI + ]
dt
2
Inductor voltage is 90 degrees out of phase with resistor voltage and
current
I(t)
When I(t) is maximum
I(t) = I 0 cos[wt + qI ]
t
VL(t)
VL(t) is zero
and
decreasing
VL = - wLI 0 sin[wt + qI ]
p
= wLI 0 cos[wt + qI + ]
2
VL leads I by p/2
B
C
Which could be true?
3
a. Red is the voltage
across an inductor,
black is the current
through that inductor
b. Black is the voltage
across an inductor,
red is the current
through that inductor
c. neither of the above
Voltage or Current
2
1
0
-1
-2
-3
0
2
4
6
8
time
10
12
14
B
C
Which could be true?
3
a. Red is the voltage
across a capacitor,
black is the current
through that capacitor
b. Black is the voltage
across a capacitor, red
is the current through
that capacitor
c. neither of the above
Voltage or Current
2
1
0
-1
-2
-3
0
2
4
6
8
time
10
12
14
Let’s do a “simple” example
Vs(t)
Vs (t) = V0 cos[wt + qs ]
R
I(t)
L
Can take: qs = 0
Current I, flows through both R and L
I(t) = I 0 cos[wt + qI ]
Resistor Voltage VR = RI(t) = RI 0 cos[wt + qI ]
Inductor Voltage
d
p
VL = L I(t) = - wLI 0 sin[wt + qI ] = wLI 0 cos[wt + qI + ]
dt
2
Inductor voltage is 90 degrees out of phase with resistor voltage and
current
I(t)
When I(t) is maximum
I(t) = I 0 cos[wt + qI ]
t
VL(t)
VL(t) is zero
and
decreasing
VL = - wLI 0 sin[wt + qI ]
p
= wLI 0 cos[wt + qI + ]
2
VL leads I by p/2
Kirchoff’s Voltage Law: sum of voltages around loop=0 for all t
Vs (t) = VR (t) + VL (t)
V0 cos[wt] = RI 0 cos[wt + qI ]- wLI 0 sin[wt + qI ]
Find
How to solve:
1. Use trigonometric identities
cos[wt + qI ] = cos[wt]cos[qI ]- sin[wt]sin[qI ]
sin[wt + qI ] = sin[wt]cos[qI ]+ cos[wt]sin[qI ]
2. Collect terms multiplying
sin[wt] and cos[wt]
After regrouping
“Reactance” XL = wL
V0 cos[wt] = cos[wt]( RcosqI - XL sinqI ) I 0
- sin[wt]( RsinqI + XL cosqI ) I 0
Can only be satisfied for all t if coefficients of cos and sin are
separately equal.
V0 = ( RcosqI - XL sinqI ) I 0
0 = ( RsinqI + XL cosqI )
Solution:
tanqI = - XL / R
I 0 = V0 / R 2 + X L2
Solution: I 0 = V0 / R 2 + X L2
Inductor voltage
Resistor Voltage
VL = V0
tanqI = - XL / R
XL = wL
p
cos[wt + qI + ]
2
2
2
R + XL
VR = V0
XL
R
2
R + X
2
L
cos[wt + qI ]
R, XL
XL = wL
R
Resistance and Reactance equal
w = wc
wc = R / L = 1 / t
w
Crossover network
VR , VL
V0
V0 / 2
VR
VL
w = wc
w
wc = R / L = 1 / t
Low frequency
Inductor is short
All voltage appears
across resistor
Vs(t)
R
I(t)
L
High frequency
Inductor is open
All voltage appears
across inductor
Recall for a moment when life was simple - DC circuits.
I
V0
Wouldn’t you do
anything to get back
to that simple way of
analyzing circuits?
R1
V0
I=
R1 + R2
R2
A. Yes
B. No
C. What do you mean
by anything?
Phasors: sinusoidal signals can be represented as vectors rotating
in a plane. Later we will see that this is the complex plane
Think of the time dependent voltage as the projection of the
rotating vector on to the horizontal axis
VL = V0
XL
R2 + XL2
cos[wt + qI +
p
]
2
What are the phasors for the
voltages in our circuit?
VS (t) = V0 cos[wt]
tanqI = - XL / R
wt
qI
VR = V0
R
2
R + X
2
L
cos[wt + qI ]
VR(t) and VL(t) form two sides of a right triangle, the
hypotenuse is Vs(t)
The magnitude of
the instantaneous
value of the emf
represented by this
phasor is
A. constant.
B. increasing.
C. decreasing.
D. It’s not possible to tell without knowing t.
Bottom Line
Everything you learned about DC circuits can be applied to AC
circuits provided you do the following:
1. Replace all voltages and currents by their complex phasor
amplitudes. In practice this means putting a hat on each letter.
2. Treat inductors as resistors with “resistance” jwL
3. Treat capacitors as resistors with “resistance” 1/(jwC)
j= - 1
Phasors - a way of representing complex numbers
Imaginary number
Complex number
j= - 1
Z = X + jY
Engineers use j
Physicists and mathematicians
use i
X is the real part
Y is the imaginary part
Complex numbers follow the same rules of algebra as regular numbers
Z1 = X1 + jY1
Addition:
Z2 = X2 + jY2
Z1 + Z2 = (X1 + X2 ) + j(Y1 + Y2 )
Multiplication:
-1
Z1Z2 = (X1 + jY1 )(X2 + jY2 ) = X1 X2 + j 2Y1Y2 + j(X1Y2 + X2Y1 )
A complex number is specified by two real numbers
Y
Z = X + jY
Instead of real and imaginary parts can give
magnitude and phase
Z
Z =
q
X2 + Y 2
tan q = Y / X
X
Multiplying complex numbers - part 2
Magnitudes multiply
Phases add
Z 3 = Z1Z2
Z3 = Z1 Z2
Z2
Z1
q3 = q1 + q2
Exponential
ex
function
Plot for real x
But, what if x is imaginary?
e0=1
1
x
e jq = cosq + j sinq
X = cosq, Y = sinq
Z = cosq + j sinq
Let
sin q
Then you can show:
So:
dZ
= jZ
dq
Z(q) = Z(0)e jq = e jq
Z(0) = 1+ j0 = 1
1
q
cos q
Phasors
Suppose I have an oscillating voltage
V(t) = V0 cos[wt + q]
I can write this as the real part of a complex number.
V(t) = Re È
(ÍÎ V0e jq )e jwt ×Ý
Þ
Call this Vˆ0 a complex amplitude or “phasor”
In this class, hat means a complex #
Vˆ0 = V0 e jq
Vˆ0
q
Vˆ0 = V0
Magnitude of phasor gives peak
amplitude of signal.
Angle give phase of signal.
ˆ e jwt ×
V (t) = Re È
V
ÍÎ 0
Ý
Þ
Vˆ0e jwt
Multiplying Vˆ0 by e
rotates the angle of the product
by wt
jwt
Vˆ0
wt
q
Remember:
Z3 = Z1 Z2
q3 = q1 + q2
V0 cos(wt + q)
How to use in circuits:
1. Every voltage and current is written in phasor form:
ˆ e jwt ×
Vs (t) = Re È
V
ÍÎ 0
Ý
Þ
ˆ jwt ×
I(t) = Re È
Ie
ÍÎ
Ý
Þ
ˆ e jwt ×
VL (t) = Re È
V
ÍÎ L Ý
Þ
2. Write every circuit law in terms of phasors:
Example: Ohm’s Law VR(t) = R I(t)
ˆ e jwt ×= R Re ÈIe
ˆ jwt ×= Re[RIe
ˆ jwt ]
VR (t) = Re È
V
ÍÎ R Ý
ÍÎ
Ý
Þ
Þ
3. Drop the Real.
Real parts are equal and lets say imaginary parts are equal too. Why not?
ˆ jwt
VˆRe jwt = RIe
jwt
e
4. Cancel
VˆR = RIˆ
5. The result is the same Ohm’s law we love, but with phasors!
What about Inductors?
VL (t) = L
d
I(t)
dt
Substitute in phasors
Only t dependence
È d ˆ jwt ×
d
jwt ×
jwt ×
ˆ
ˆ
ˆ jwt ×
È
È
VL (t) = Re ÍVL e Ý= L Re ÍIe Ý= Re ÍL Ie Ý= Re È
jwL
Ie
ÍÎ
Ý
Î
Þ
Þ
Þ
ÍÎ dt
Ý
dt Î
Þ
3. Drop the Real
jwt
4. Cancel e
ˆ jwt
VˆL e jwt = jwLIe
VˆL = jwLIˆ = jXL Iˆ
5. The result is the same Ohm’s law we love, but with resistance
replaced by
jX L
Back to our circuit
Vs(t)
Vˆ0 = VˆR + VˆL
KVL:
R
I(t)
VˆR = RIˆ
VˆL = jXL Iˆ
L
Result:
Recall DC circuit result:
ˆ
V
0
Iˆ =
R + jXL
V0
I=
R1 + R2
Bottom Line
Everything you learned about DC circuits can be applied to AC
circuits provided you do the following:
1. Replace all voltages and currents by their phasor amplitudes.
In practice this means putting a hat on each letter.
2. Treat inductors as resistors with “resistance” jwL
3. Treat capacitors as resistors with “resistance” 1/(jwC)
KVL
RLC Circuit
Vs(t)
Vˆ0 = VˆR + VˆL + VˆC
R
VˆC = 1 / ( jwC)Iˆ
VˆR = RIˆ
VˆL = jwLIˆ
I(t)
C
Current phasor
Complex Impedance
L
Iˆ =
Vˆ0
Vˆ0
=
R + j[wL - 1 / (wC)] Z
Z = R + j[wL - 1 / (wC)]
Magnitude of Impedance Z =
Phase of Impedance
R 2 + [wL - 1 / (wC)]2
tanf = [wL - 1 / (wC)] / R
Resonance: At what frequency is the amplitude of the
current maximum?
Complex Amplitude
Iˆ = Iˆ e jq =
Vˆ0
Vˆ0
=
R + j[wL - 1 / (wC)] Z
Current Amplitude
Iˆ =
Z
=
Vˆ0
R2 + [wL - 1 / (wC)]2
Current is largest when this term is zero
At resonance:
Iˆ =
Vˆ0
Vˆ0
R
w = w0 = 1/ LC
Resonant frequency
How narrrow is the Resonance?
I max
I max / 2
Iˆ =
w0 / Q
Q=
Vˆ0
Z
Vˆ0
=
R2 + [wL - 1 / (wC)]2
L
/R
C
Width of resonance determined by
when these two are equal
Quality Factor
1.5
Decaying transient
Quality factor determines rate
of decay of transient
1
envelope
V(t)
0.5
envelope = e-
0
-0.5
wo t /(2Q )
Power dissipated in R
w0
=
Energy stored in L & C
Q
-1
0
20
40
60
wt
0
80
100
Vs(t)
I(t)
L
R
C
A
Z = R + j[wL - 1 / (wC)]
B
Z = R- j[wL - 1 / (wC)]
C
- 1
- 1
×
Z = 1/ È
R
+
jwL
+
jwC
(
)
ÍÎ
Ý
Þ
What is the impedance
of the parallel
combination of an R, L,
and C?
Phasors for R-L circuit
Write currents and voltages in phasor form
Vs(t) I(t) R
KVL
L
ˆ e jwt × V (t) = Re ÈVˆ e jwt ×
Vs (t) = Re È
V
ÍÎ 0
Ý
ÍÎ L Ý
Þ L
Þ
ˆ jwt × V (t) = Re ÈVˆ e jwt ×
I(t) = Re È
Ie
ÍÎ
Ý
ÍÎ R Ý
Þ R
Þ
Write circuit equations for phasor amplitudes
KVL:
Result:
ˆ
V
Iˆ = 0
Z
Impedance Z = R + jXL
0 = VˆL + VˆR - Vˆ0
VˆL = j(wL)Iˆ = jXL Iˆ
VˆR = RIˆ
Result:
Impedance has a magnitude and phase
ˆ
V
Iˆ = 0
Z
Z = Z e jf Z
XL
Impedance Z = R + jXL
Z
Z =
fZ
R
Resistor Voltage
R
R
VˆR = RIˆ = Vˆ0 = Vˆ0 eZ
Z
p
j( - f Z )
jX
X
VˆL = jXL Iˆ = Vˆ0 L = Vˆ0 L e 2
Z
Z
Note:
j= e
tanf Z = XL / R
jf Z
Inductor Voltage
p
j
2
R 2 + X L2
VˆL
Vˆ0
fZ
VˆR
Vˆ0 = VˆL + VˆR
Power Dissipated in Resistor
Current
I(t) = I R cos[wt]
Instantaneous Power
p(t) = RI 2 = RI R2 cos2 [wt]
Average over time is 1/2
Average Power
1 2
P = RI R
2
Root Mean Square (RMS) Voltage and Current
Current
I(t) = I R cos[wt]
Average Power
1 2
P = RI R
2
Peak current
What would be the equivalent DC current as far as average
power is concerned?
I RMS
Average Power
IR
=
2
2
P = RI RMS
No pesky 2
What is the peak voltage for 110 V-AC- RMS?
A: 156 V
Power Delivered to a Capacitor
Voltage V(t) = VC cos[wt]
Current
I(t) = CdV(t) / dt
I(t) = - wCVC sin[wt]
Instantaneous Power
p(t) = IV = - wCVC2 cos[wt]sin[wt]
wCVC2
p(t) = sin[2wt]
2
Average Power
P= 0
Rank in order, from largest to smallest, the cross-over
frequencies
of these four circuits.
A series RLC circuit has VC = 5.0 V, VR
= 7.0 V, and VL = 9.0 V. Is the
frequency above, below or equal to the
resonance frequency?
A. Above the resonance frequency
B. Below the resonance frequency
C. Equal to the resonance frequency
The emf and the current
in a series RLC circuit
oscillate as shown. Which
of the following would
increase the rate at which
energy is supplied to the
circuit?
A. Decrease ε0
B. Increase C
C. Increase L
D. Decrease L