Chapter 31
Alternating Current
Circuits
Alternating Current Circuits
• Alternating Current - Generator
• Wave Nomenclature & RMS
• AC Circuits: Resistor; Inductor; Capacitor
• Transformers - not the movie
• LC and RLC Circuits - No generator
• Driven RLC Circuits - Series
• Impedance and Power
• RC and RL Circuits - Low & High Frequency
• RLC Circuit - Solution via Complex Numbers
• RLC Circuit - Example
• Resonance
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Generators
By turning the coils in the magnetic field an emf is
generated in the coils thus turning mechanical energy into
alternating (AC) power.
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Generators
Rotating the Coil in a Magnetic Field Generates an Emf
• Examples: Gasoline generator
• Manually turning the crank
• Hydroelectric power
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Generators
φm = NBAcosθ
θ = ωt
φm = NBAcosωt
d
= - φm = NBAωsinωt
dt
=
sinωt;
= NBAω
ε
ε ε
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peak
Chap31-AC Circuits-Revised: 6/24/2012
ε
peak
5
Wave Nomenclature and RMS
Values
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Wave Nomenclature
Apeak-peak = Ap-p = 2Apeak = 2Ap; Ap = Ap-p /2
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Shifting Trig Functions
 sin 
x=A 
  ωt - ϕ 
 cos 
x= A
The minus sign means that the
phase is shifted to the right.
sin  t

2π
ϕ

cos  T
{ }
A plus sign indicated the phase
is shifted to the left
x = A sin  ωt - π2 
x = A ( sinωt cos π2 - sin π2 cosωt )
x = A ( sinωt (0) - (1)cosωt )
x = -Acosωt
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Shifting Trig Functions

π
sin  ωt -  = 0
2

Shifted Trig Functions
1.50
π
ωt - = 0
2
π
ωt =
2
π 1
1
T
t=
;
=
ω 2π
2ω
π T
T
t=
=
2 2π 4
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sin(ωt)
1.00
sin(ωt-δ)
0.50
-3.00
-2.00
-1.00
0.00
0.00
1.00
2.00
3.00
-0.50
-1.00
-1.50
Time
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Root Mean Squared
The root mean squared (rms) method of averaging is
used when a variable will average to zero but its effect
will not average to zero.
Procedure
• Square it (make the negative values positive)
• Take the average (mean)
• Take the square root (undo the squaring operation)
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Root Mean Squared Average
Sine Functions
1.5
1.0
0.5
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
-0.5
SIN(Theta)
SIN2(Theta)
RMS Value
-1.0
-1.5
Angle (Radians)
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Average of a Periodic Function
T
V = Vavg
Vavg
1
= ∫ V(t)dt; V(t) = V p sinωt
T o
T
Vp
1
= ∫ V p sinωtdt =
T o
ωT
Vavg = -
Vp
ωT
ωT
∫ sinxdx = 0
Vp
ωT
cos(ωT)
∫
d(cosx)
cos(0)
(1 - 1) = 0
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Root Mean Squared
T
V
( )
2
= V
( )
V2
( )
V2
avg
avg
=
=
avg
V p2
T
T
∫
sin 2 ωtdt =
o
V p2
ωT
π =
V p2
2
V p2
VRMS ≡
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1
= ∫ V 2 (t)dt; V(t) = V p sinωt
T o
2
2
(V )
2
avg
=
1
2
V p = 0.707V p
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Root Mean Squared
Root
VRMS ≡
Square
(V )
2
avg
1
=
V p = 0.707V p
2
Mean
The RMS voltage (VRMS )is the DC voltage that has the same
effect as the actual AC voltage.
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RMS Power
Pavg
1
= Vp I p
2
since
Pavg
Pavg
VRMS =
Vp
1
=
2 VRMS
2
= VRMS I RMS
(
2
)(
and I RMS =
2 I RMS
Ip
2
)
The average AC power is the product of the DC equivalent
voltage and current.
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Resistor in an AC Circuit
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Resistor in an AC Circuit
For the case of a resistor in an AC circuit the VR across the
resistor is in phase with the current I through the resistor.
In phase means that both waveforms peak at the same time.
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Resistor in an AC Circuit
P(t) = I (t)R = ( I cosωt ) R
2
2
p
2
P(t) = I p2 Rcosωt
The instantaneous power is a function of time. However, the average
power per cycle is of more interest.
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Inductors in an AC Circuit
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Coils & Caps in an AC Circuit
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Inductors in an AC Circuit
For the case of an inductor in an
AC circuit the VL across the
inductor is 900 ahead of the current
I through the inductor.
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Inductors in an AC Circuit
I = I p sinωt =
Ip =
VL peak
ωL
X L =ωL
=
VL peak
ωL
(
cos ωt - π 2
)
VL peak
XL
XL is the inductive reactance
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Average Power - Inductors
P(t) = V I = (V
cosωt )( I sinωt )
L
L peak
p
P(t) = VL peak I p cosωt sinωt
T
Pavg
1
= ∫ VL peak I p cosωt sinωtdt
T0
Pavg =
Pavg =
VL peak I p
T
VL peak I p
2T
T
∫ cosωt sinωtdt
0
T
∫ sin2ωtdt = 0
0
Inductors don’t dissipate energy, they store energy.
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Average Power - Inductors
Inductors don’t dissipate energy, they
store energy.
The voltage and the current are out of
phase by 90o.
As we saw with Work, energy
changed only when a portion of the
force was in the direction of the
displacement.
In electrical circuits energy is
dissipated only if a portion of the
voltage is in phase with the current.
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Capacitors in an AC Circuit
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Capacitors in an AC Circuit
VC =
ε cosωt = V
p
Cp
cosωt
Q = VC C = VC p Ccosωt = Q p cosωt
dQ
I=
= -ωQ p sinωt = -I p sinωt
dt
I = -ωQ p sinωt = I p cos ωt + π 2
(
)
For the case of a capacitor in an
AC circuit the VC across the
capacitor is 900 behind the current
I on the capacitor.
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Capacitors in an AC Circuit
VCp
VCp
I p =ωQ p = ωCV Cp =
=
1
ωC X C
1
XC =
ωC
XC is the capacitive reactance.
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Electrical Transformers
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Electrical Transformers
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Electrical Transformers
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Electrical Transformers
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Electrical Transformers
Both coils see the same magnetic flux and the cross sectional
areas are the same
B = µ 0nI
µ0 n1 I 1 = µ0 n2 I 2
n1 I 1 = n2 I 2
n1
I2 =
I1
n2
N2
I 1 n2
N2
L
=
=
=
N1
I 2 n1
N1
L
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Electrical Transformers
Conservation of Energy
Primary Power = Secondary Power
Vin I 1 = Vout I 2
Vout
I1 N 2
= =
Vin
I2
N1
N2
Vout =
Vin
N1
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Induced voltage/loop
More loops => more voltage
Voltage steps up but the current
steps down.
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LC and RLC Circuits Without a
Generator
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LC Circuit - No Generator
To start this circuit some energy must be placed in it since
there is no battery to drive the circuit. We will do that by
placing a charge on the capacitor
Since there is no resistor in the circuit and the resistance of
the coil is assumed to be zero there will not be any losses.
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LC Circuit - No Generator
Apply Kirchhoff’s rule
dI Q
L + =0
dt C
Since I = dQ
dt
d 2Q Q
L 2 + =0
dt
C
d 2Q
1
=Q
2
dt
LC
1
ωR =
LC
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This is the harmonic
oscillator equation
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LC Circuit - No Generator
Q(t) = Q p cosωt
dQ
= -ωQ p sinωt
dt
I(t) = -ωQ p cos ωt + π 2
I(t) =
(
)
The circuit will oscillate at the frequency
ωR. Energy will flow back and forth
from the capacitor (electric energy) to
the inductor (magnetic energy).
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RLC Circuit - No Generator
Like the LC circuit some energy must initially be placed in
this circuit since there is no battery to drive the circuit. Again
we will do this by placing a charge on the capacitor
Since there is a resistor in the circuit now there will be losses
as the energy passes through the resistor.
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RLC Circuit - No Generator
Apply Kirchhoff’s rule
dI
Q
dQ
L + IR + = 0 ; I =
dt
C
dt
d 2Q
dQ 1
L 2 +R
+ Q=0
dt
dt C
Restoring force “kx”
Damping term - friction
“ma” term
MFMcGraw-PHY 2426
The damping term causes a damping of the
natural oscillations of the circuit.
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RLC Circuit - No Generator
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RLC Circuit - No Generator
dI
Q
Multiply by I
L + RI + = 0
dt
C
dI
QI
2
LI
+I R+
=0
dt
C
2


d 1 2
d
1
Q
2
=0
 LI  + I R + 
dt  2
dt  2 C 

d  1 2 1 Q2 
2
LI
+
=
-I
R


dt  2
2 C 
The rate of change of the
=
stored energy
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-
Power dissipated in the
resistor
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Series RLC Circuit with Generator
We have already examined the components in this circuit to
understand the phase relations of the voltage and current of
each component
Now we will examine the power relationships
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Series RLC Circuit with Generator
ε(t) = ε
peak
sinωt =
ε sinωt
p
Apply Kirchhoff’s Loop rule to the circuit
dI(t) Q(t)
RI(t) + L
+
= (t)
dt
C
ε
t
dQ
= I ⇒ Q(t) = Qo + ∫ I(t')dt'
0
dt
dI(t) 1 t
RI(t) + L
+ ∫ I(t')dt' = (t); with Q0 = 0
dt
C 0
ε
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Series RLC Circuit with Generator
dI(t) 1 t
RI(t) + L
+ ∫ I(t')dt' = (t)
dt
C 0
ε
Steady state ⇒ I(t) = I p sinωt
dI(t)
= ω I p cosωt;
dt
∫
t
0
I(t')dt' =
∫
t
0
I p sinωt'dt' = -
1
RI p sinωt + ωLI p cosωt I p cosωt =
ωC
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Ip
ω
cosωt
ε sinωt
p
45
Series RLC Circuit with Generator
1
RI p sinωt + ωLI p cosωt I p cosωt =
ωC
ε sinωt
p
Change all “cos” to “sin” by shifting the angle
1
π
RI p sinωt + ωLI p sin ωt + 2 +
I p sin ωt - π 2 = ε sinωt
p
ωC
(
The inductive voltage is
90o ahead of the current
MFMcGraw-PHY 2426
)
(
)
The capacitive voltage is
90o behind of the current
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46
Impedance in a Series RLC Circuit
1
π
I p sin ωt - π 2 = ε sinωt
RI p sinωt + ωLI p sin ωt + 2 +
p
ωC
(
)
(
)
The coefficients are voltages
R
L
C
RI p
ωLI p
I p ωC
X LI p
XCI p
X L = ωL
X C = 1 ωC
The R and XL and XC values are
called impedances. That is a
generlized term for resistance
since they all have units of ohms.
XL is the inductive reactance.
XC is the capacitive reactance.
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Power in a Series RLC Circuit
Now we go back to the original equation and multiply by
I(t) = I p sinωt and integrate over one cycle: 0 => T
1
RI p sinωt + ωLI p cosωt I p cosωt =
ωC
T
2
RI p2 ∫ sinωtdt
+ ωLI
T
2
p
∫sinωtcosωtdt -
o
o
∫
T
0
2
sin ωtdt = π
MFMcGraw-PHY 2426
∫
T
0
I p2
ωC
ε sinωt
p
T
∫sinωtcosωt dt =
o
T
ε I ∫ sinωtdt
2
p p
o
sinωt cosωtdt = 0
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Power in a Series RLC Circuit
RI p2 = E p I p
Power in resistor
Power out of
battery
• Power is only dissipated in the resistor.
• The inductor stores energy in its magnetic field.
• The capacitor stores its energy in its electric field.
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Series RLC Circuit with Generator
We have used this equation to demonstrate the behavior
of the three types of components: R, L and C, butWe still haven’t solved the equation
dI(t) 1 t
RI(t) + L
+ ∫ I(t')dt' = (t); with Q0 = 0
dt
C 0
ε
Before we actually solve it
we need to introduce
complex variables that will
be used in the solution.
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The RC Circuit
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The RC Circuit - Low Freq
1
RI p sinωt + ωLI p cosωt I p cosωt =
ωC
ε sinωt
p
Let L => 0
1
RI p sinωt I p cosωt =
ωC
ε sinωt
p
For ωt 0; cosωt 1; sinωt 0
Low ω
1
I p = p (0)
ωC
I p = 0 ⇒ Open circuit
ε
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The RC Circuit - High Freq
1
RI p sinωt I p cosωt =
ωC
ε sinωt
p
For ω 1 CR
1
p
I p sinωt I p cosωt =
sinωt
ωCR
R
ε
I p sinωt =
εp
High ω
sinωt
R
Ip =
εp
R
At
Athigh
highfrequency
frequencythe
the
cap
capacts
actsas
asaashort
shortcircuit.
circuit.
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The RL Circuit
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RI p
The RL Circuit - Low Freq
1
sinωt + ωLI cosωt I cosωt = ε sinωt
ωC
p
p
p
Let C => 0
RI p sinωt + ωLI p cosωt =
ε sinωt
p
For ωL 1
RI p sinωt =
RI p =
ε
ε sinωt
Low ω
p
p
At
Atlow
lowfrequency
frequencyLLacts
acts
as
asaashort
shortcircuit.
circuit.
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The RL Circuit - High Freq
RI p sinωt + ωLI p cosωt =
ε sinωt
p
For ωL 1
ωLI p cosωt =
ε sinωt
p
multiply by cosωt and average
T
1
ωLI p ∫ cos 2 ωtdt =
T0
ε
T
1
sinωtcosωtdt
p
∫
T0
High ω
ωLI p π = 0
I p = 0 ⇒ Open circuit
At
Athigh
highfrequency
frequencyLLacts
actsas
asan
anopen
opencircuit.
circuit.
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Coils & Caps in an AC Circuit
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Complex Numbers for AC Circuits
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Complex Numbers for AC Circuits
The basic complex (imaginary) number is “i.”
To avoid confusion we replace “i” with “j”
j = -1
2
j = jj = -1 -1 = -1
j 3 = jj 2 = j ( -1) = -j
4
2
2
j = j j = ( -1)( -1) = +1
j 5 = jj 4 = j (+1) = j
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Complex Numbers for AC Circuits
Let a and b be real numbers
Then z is a complex number and z* is the complex
conjugate
y
z = a + bj
z* = a - bj
z
(a,b)
θ
x
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Complex Numbers for AC Circuits
The magnitude of z
Magn ( z ) = z = ( z * ) z =
( a - bj )( a + bj )
z = a 2 + abj - abj - j 2b 2
z = a 2 + b2
y
( )
tanθ = b a ; θ = tan -1 b a
z = z ( cosθ + jsinθ ) = z e jθ
The exponential representation of a
complex number will prove useful in
solving the RLC differenial eqn.
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z
(a,b)
θ
x
61
Complex Numbers for AC Circuits
y
z = z ( cosθ + jsinθ ) = z e jθ
z
e jθ can be viewed as a rotation
operator in a complex space
e
jπ 2
(a,b)
θ
= j
x
jπ
e = -1
e
j 3π 2
= -j
e j2π = e0 = +1
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Why Complex Numbers ?
y
z = z ( cosθ + jsinθ ) = z e jθ
z
(a,b)
θ
x
Complex numbers simplify the solution of the integraldifferential equations encountered in series RLC AC circuits.
The use of complex numbers simplifies the lead-lag nature of
the voltage and current in AC circuits.
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Phasor Notation
This diagram depicts a series RLC
circuit driven at a frequency that
causes the inductive voltage to be
greater than the capacitive voltage.
This gives the circuit an overall
inductive nature - the current (in
phase with VR) is lagging the applied
voltage Vapp.
All of these voltage vectors (phasors) have a common time
component (ejωt) and so they all rotate at this common frequency.
By suppressing this common rotation the concepts are easier to
understand.
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RLC Circuit Solution
dI(t) 1 t
RI(t) + L
+ ∫ I(t')dt' = (t)
dt
C 0
ε
I(t) = I p e jωt
The solution of a differential equation
begins with the selection of a trial solution
dI
= jωI p e jωt = jωI ;
dt
I
∫ I(t)dt = jω
MFMcGraw-PHY 2426
I
RI + jωLI +
=E
jωC

j 
 R + jωL I = E
ωC 


1  E
R + j  ωL  =
ωC  I

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RLC Circuit Solution

1 
E
R + j  ωL  =
ωC 
I


1 
E
Z = R + j  ωL  =
ωC 
I

E
Z=
I
The quantity Z is called the impedance
and it is a complex variable
These are complex
variables
E = I Z is a complex version of
Ohm’s Law
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Complex Impedance

1 
Z = R + j  ωL 
ωC


z =

1 
z * z = R +  ωL 
ωC


( )
2
2
1
ωL X L - XC
ωC
tanθ =
=
R
R
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Relative Voltage Phases - Inductive
Impedance Space
Voltage Space
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Phases in an Inductive AC Circuit
RLC series
L only series
Earlier time
Later time
All vectors rotating at a
common frequency ω
In an inductive circuit the voltage peaks first and the
current peaks later.
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Phases in a Capacitive AC Circuit
RLC series
C only series
Earlier time
Later time
All vectors rotating at a
common frequency ω
In capacitive circuit the current peaks first and the
voltage peaks later.
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RLC Series AC Circuit
Example
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R= 250Ω, L = 1.20mH,
C = 1.80µF, Vp = 120v, f = 60Hz
Determine the following:
(a.) XL - Inductive reactance
(b.) XC - Capacitive reactance
(c.) Z - Impedance
(d.) θ - Phase angle
(e.) Ip - Peak current
(f.) IRMS - RMS current
(g.) ωR - Resonance frequency
(h.) Pavg - Average Power
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R= 250Ω, L = 1.20mH,
C = 1.80µF, Vp = 120v, f = 60Hz
Determine the following:
(a.) XL - Inductive reactance
(b.) XC - Capacitive reactance
(c.) Z - Impedance
First calc: ω = 2πf = 2(3.14)60 = 377 rad/s
XL = ωL= 377(1.20x10-3) = 0.452Ω
XC = 1/ωC= 1/((377)(1.80x10-6)) = 1474Ω
Z = R 2 + (X L - X C )2 = 250 2 + (0.452 - 1474)2
Z = 1495 Ω
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R= 250Ω, L = 1.20mH,
C = 1.80µF, Vp = 120v, f = 60Hz
Determine the following:
(d.) θ - Phase angle
X L - XC 

-1  0.452 - 1474 
-1  -1474 
0
θ = tan 
=
tan
=
tan
=
-80.4

R
250



 250 
-1
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R= 250Ω, L = 1.20mH,
C = 1.80µF, Vp = 120v, f = 60Hz
Determine the following:
(e.) Ip - Peak current
(f.) IRMS - RMS current
+jωt
V p +j(ωt-θ)
V Vpe
+j(ωt-θ)
I= =
=
e
=
I
e
p
Z
Z p e+jθ
Zp
Vp
120
Ip =
=
= 80.3mA
Z
1495
1st minus from the division.
I = I p e+j(ωt-θ) = 80.3mAe+j(ωt+80.4)
I RMS =
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Ip
2
= 0.707I p = 56.7mA
Chap31-AC Circuits-Revised: 6/24/2012
2nd minus from
the angle.
75
R= 250Ω, L = 1.20mH,
C = 1.80µF, Vp = 120v, f = 60Hz
Determine the following:
(g.) ωR - Resonance frequency
ωR =
1
=
LC
1
(1.20x10 )(1.80x10 )
-3
-6
= 21.5krad/s
ω
21.5x10 3
f =
=
= 3.42kHz
2π
6.28
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R= 250Ω, L = 1.20mH,
C = 1.80µF, Vp = 120v, f = 60Hz
Determine the following:
(h.) Pavg - Average Power
1
S = VI * V = V p e+jωt I = I p e+j(ωt+80.4)
2
1
1
+jωt
-j(ωt+80.4)
S = (V p e )( I p e
= V p I p e+j(ωt-ωt-80.4)
)
2
2
V p I p -j(80.4)
1
-j(80.4)
S = Vp I pe
=
e
= VRMS I RMS e -j(80.4)
2
2 2
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Determine the following: (h.) Pavg - Average Power
V p I p -j(80.4)
1
-j(80.4)
S = Vp I pe
=
e
= VRMS I RMS e -j(80.4)
2
2 2
S = VRMS I RMS ( cos ( 80.4 ) - jsin ( 80.4 ) )
S = VRMS I RMS cos ( 80.4 ) - jVRMS I RMS sin ( 80.4 )
Pavg = VRMS I RMS cos ( 80.4 )
Power Factor
Pavg = 84.3 ( 56.7x10 -3 ) ( 0.167 ) = 0.803W
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Voltages
VR = I RMS R = ( 56.7x10 -3 ) ( 250 ) = 14.2V
VL = I RMS X L = ( 56.7x10 -3 ) ( 377 ) ( 1.20x10 -3 ) = 0.0265V
(
)
VC = I RMS X C = ( 56.7x10 -3 ) / 377 ( 1.80x10 -6 ) = 83.6V
2
2
2
V = VR + (VL - VC ) = ( 14.2 ) + (0.0256 - 83.6)2
V = 84.8 = VRMS
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Resonance in a Series RLC Circuit
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Resonance in a Series RLC
Circuit
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Power Transfer and Resonance
d 2Q
dQ
Q
L 2 +R
+ + = V p cosωt
dt
dt
C
Z = R + ( X L - XC )
2
Pavg = I
2
RMS
R=
V
2
p
Z2
R
I = I p cos ( ωt - δ ) =

1 
Z = R +  ωL 
ωC


V p2
2
Pavg =
2
Vp
Z
cos ( ωt - δ )
2
2
R
2
2
2
2
 ωL   ω - ω0 
1+ 

 
2
R


  ω
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Q Factor = Measure of Stored Energy
V p2
Pavg =
Q-Factor
R
2
2
2
2


 ωL  ω - ω0
1+ 



2

 R   ω
ω0
f0
E
Q = 2π
=
=
∆E ∆ω ∆f
E = Total Energy and
∆E is the dissipated energy
∆ω = FWHM
FWHM = Full Width at Half
Maximum
As an approximation
ω0 L
Q
R
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Resonance in a Series RLC Circuit
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RLC Parallel Circuit
We’re not covering this type of circuit
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Extra Slides
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