assignments due mutual inductance mutual inductance ...
ASSIGNMENTS DUE • Today (Monday):
ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 29
• Homework #10 Due • Activities 29-1, 29-2 (In Class)
• Tuesday/Wednesday: • Activity 30-1 (In Class)
• Thursday: • Activities 31-1, 31-2 (In Class) • End of Unit III Material
MUTUAL INDUCTANCE Note Direction of i 2
M
i1
•
L2
v1 L1 −
+
M>0
−
M M → − M Some Inductors May Be Negative!
1
MUTUAL INDUCTANCE When Mutual Inductance happens in a Circuit: => Replace Coupled Inductors with Tee Model => Tee Model has 3 Uncoupled Inductors => Negative Inductances are not Uncommon => Look for "Transformer-like" Configuration => Label Nodes Carefully
Then do regular AC Steady State Circuit Analysis
• Most Commercial Power to Large Users is 3 Phase: • Most Residential Power is 1 Phase • Brief Overview of 3 Phase Circuits: • Electric Power Engineering Courses if Want More Detail:
3 PHASE SOURCES
3 PHASE POWER
• Wye Generator:
• Why Use 3 Phase Power?: • Easy to Generate • Provides Constant Instantaneous Power to Load • Uses Fewest Wires to Deliver Constant Instantaneous Power
3 PHASE SOURCES
• Produces “phase voltages” va(t), vb(t), vc(t) • These create “line voltages, vab(t), vbc(t), vca(t)
• Delta Generator: • Produces “line voltages” vab(t), vbc(t), vca(t) • These create “phase voltages, va(t), vb(t), vc(t)
3 PHASE SOURCES Frequency Domain:
Time Domain: 3 Phase + v Generator ab − +
Wye Or Delta
3 PHASE AC POWER
−
v a , v b , v c = Phase Voltages v ab , v bc , v ca = Line Voltages
b
v ca
v bc −
a
+
c
+
+ va
+ vc
vb
−
−
−
3 Phase + V Generator ab − +
Wye Or Delta
b
V ca
V bc −
a
−
+
V a , V b , V c = Phase Voltages V ab , V bc , V ca = Line Voltages
c
+
+ Va
+ Vc
Vb
−
−
−
2
3 PHASE SOURCES Phase Voltages: • v a (t) = Voltage at "a" Relative to Ground
3 PHASE SOURCES
• v a (t) = 2Vφ cos(ω t + φ ) Volts;
Vca
2Vl
• Vφ = Phase Voltage (RMS Value)
• v b (t) = Voltage at "b" Relative to Ground • v c (t) = Voltage at "c" Relative to Ground • v c (t) = 2Vφ cos(ω t + φ + 1200 ) Volts
3 PHASE SOURCES 2Vφ
0 600 30
2Vl
Vb
Va
2Vφ
2Vl
2Vφ
Vbc
Va
2Vφ
• v b (t) = 2Vφ cos(ω t + φ − 1200 ) Volts
2Vφ
Let V a = 2Vφ / 00
Vc
2Vl
Vab
Vb
3 PHASE SOURCES
V ab
Vca
2Vl
2Vl / 2 3 cos 30 = = 2 2Vφ
Let V a = 2Vφ / 00
Vc
300
2Vφ
2Vφ
Vbc
0
2Vl 0
3 PHASE SOURCES Line Voltages: • v ab (t) = Voltage Between Line "a" and Line "b" • v ab (t) = 2Vl cos(ω t + φ + 30 ) Volts 0
• Vl = Line Voltage (RMS Value) = 3Vφ
Va
300
2Vφ
=> Vl = 3Vφ
/V ab = / V a + 30
2Vl
Vab
Vb
Vl = 3Vφ
/V ab = / V a + 300
3 PHASE SOURCES • Phase Voltages in Phasor Form: Va = 2Vφ /φ o Volts V b = 2Vφ /φ − 120o Volts Vφ =
Vl = Phase Voltage (RMS) 3
• v bc (t) = Voltage Between Line "b" and Line "c"
Vc = 2Vφ /φ + 120o Volts
• v bc (t) = 2Vl cos(ω t + φ − 900 ) Volts
• Line Voltages in Phasor Form:
• v ca (t) = Voltage Between Line "c" and Line "a"
Vab = 2Vl /φ + 30o Volts
• v ca (t) = 2Vl cos(ω t + φ + 1500 ) Volts
V bc = 2Vl /φ − 90o Volts Vl = 3 Vφ = Line Voltage (RMS) Vca = 2Vl /φ + 150o Volts
3
3 PHASE SOURCES
3 PHASE LOADS
• Line Voltages Always “lead” Phase Voltages by 30o
a
Ia
b
Ib
c
Ic
Vl = 3Vφ /V ab = / V a + 300
• If know any one of phase or line voltages => can write down all others by inspection
Ia , Ib , Ic = Line Currents
3 PHASE LOADS Balanced Delta Load
a
Ia
b
Ib
Z∆
c
Ic
Z∆
Z∆
Ia , Ib , Ic = Line Currents
n
RMS Voltage Across ZY = Vφ RMS Current Thru ZY = Il
• Phase Voltage across each ZY • Line Current through each ZY
• Delta: • Line Voltage across each ZDelta • Line Current Does NOT Flow through each ZDelta
3 PHASE LOADS • Can Convert any Delta Load to an Equivalent Wye Load:
a'
V Ia = a ZY
ZY
• Wye or Delta: • Wye:
RMS Voltage Across Z∆ = Vl
Ia
Va
ZY • n' Balanced Wye Load
3 PHASE LOADS
Equivalent 1 Phase Model a
ZY
• ZYeq = ZDelta / 3 • Easier to analyze circuit using Wye loads
ZY or Z∆ / 3
n'
• Equivalent 1 Phase Model for Wye: • • • •
See Figure Phase Voltage across ZY; Line Current thru ZY Ia = Va / ZY Write down all other I’s, V’s by Inspection
4
s-DOMAIN CIRCUIT ANALYSIS • Now Want to Go Back to Laplace Transform Analysis of Circuits: • More on Network Functions, H(s)
• More on Poles and Zeros • Stability of Circuits • Then We will be at the End of Unit III
CIRCUIT ELEMENTS IN S-DOMAIN +
IR
+
+
IL sL Ω
VR
RΩ
VL
VC
Li L (0+ ) −
−
IC 1 Ω sC
• Replace i(t), v(t) => I(s), V(s) • Replace R, L, C => s-domain Models • Use V (s) = Z (s) I (s)
• Use All Techniques from Unit I: • Series/Parallel; Voltage/Current Divider • Thevenin/Norton; Source Conversions • Node Equations, Mesh Equations
WITH NO INITIAL STORED ENERGY
ZR
RΩ
ZL
sL Ω ZC
1 Ω sC
−
For a 2nd Order Circuit, Differential Equation is: d2 y dy d2x dx + a1 + a 0 y = b 2 2 + b1 + b0 x 2 dt dt dt dt
Take Laplace Transform with NO IC's:
(a 2s + a1s + a 0 )Y(s) = (b 2s + b1s + b 0 )X(s) 2
• Transform Circuit from Time Domain to s-Domain:
v C (0+ ) s
NETWORK FUNCTION
a2
s-DOMAIN CIRCUIT ANALYSIS
2
NETWORK FUNCTION Define H(s) =
Y(s) with No Initial Stored Energy X(s)
H(s) =
a 2s 2 + a1s + a 0 b 2s 2 + b1s + b 0
(a 2 , a1 , a 0 ) come from LHS of Diff Eq (b 2 , b1 , b 0 ) come from RHS of Diff Eq If Find H(s) → Write Down Differential Equation
5
NETWORK FUNCTIONS H(s) =
Y(s) Output with No Initial Stored Energy = X(s) Input
b b 2 2 b1 ( s + s+ 0 ) ∆ ∆ a2 b2 b2 H(s) = ; 2α = a1/a 2 ; ω0 2 = a 0 / a 2 a0 a1 2 (s + s + ) a2 a2
NETWORK FUNCTIONS • Punch Line: • • • • •
Can find H(s) directly from Circuit Use s-domain Circuit Analysis to find H(s) Use s-domain models for R, L, C with no IC’s Use Generalized Impedance and Ohm’s Law All Techniques from Unit I
Denominator = s 2 + 2α s + ω02 = Characteristic Polynomial Roots of Characteristic Equation determine form of y N (t)
NETWORK FUNCTIONS
ACTIVITY 29-2
• Find H(s) from Circuit Analysis => H(s) Contains A Lot of Useful Information: • • • •
Can Find form for yN(t) Can Find Differential Equation for y(t) Can Determine Stability; Poles and Zeros Can Sketch AC Steady State Behavior (Will do this in Unit IV)
+ vL −
R
1H
i
1 F 400
v(t)
• Let’s Practice with Activity 29-2: Find H in (s) =
I(s) V(s)
+ vC
.1v L
−
Find H(s) =
VC (s) V(s)
ACTIVITY 29-2 • Input = v(t) : • Will first find Hin(s) = I (s) / V (s) • Will also find H (s) = VC (s) / V (s) • Practice s-domain circuit analysis
• Time Domain => s-Domain: • v(t), i(t) => V(s), I(s) • R,L,C => R, sL, 1 / sC • Use techniques from Unit I • Mesh equations
6
ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 29
• Homework #10 Due • Activities 29-1, 29-2 (In Class)
• Tuesday/Wednesday: • Activity 30-1 (In Class)
• Thursday: • Activities 31-1, 31-2 (In Class) • End of Unit III Material
MUTUAL INDUCTANCE Note Direction of i 2
M
i1
•
L2
v1 L1 −
+
M>0
−
M M → − M Some Inductors May Be Negative!
1
MUTUAL INDUCTANCE When Mutual Inductance happens in a Circuit: => Replace Coupled Inductors with Tee Model => Tee Model has 3 Uncoupled Inductors => Negative Inductances are not Uncommon => Look for "Transformer-like" Configuration => Label Nodes Carefully
Then do regular AC Steady State Circuit Analysis
• Most Commercial Power to Large Users is 3 Phase: • Most Residential Power is 1 Phase • Brief Overview of 3 Phase Circuits: • Electric Power Engineering Courses if Want More Detail:
3 PHASE SOURCES
3 PHASE POWER
• Wye Generator:
• Why Use 3 Phase Power?: • Easy to Generate • Provides Constant Instantaneous Power to Load • Uses Fewest Wires to Deliver Constant Instantaneous Power
3 PHASE SOURCES
• Produces “phase voltages” va(t), vb(t), vc(t) • These create “line voltages, vab(t), vbc(t), vca(t)
• Delta Generator: • Produces “line voltages” vab(t), vbc(t), vca(t) • These create “phase voltages, va(t), vb(t), vc(t)
3 PHASE SOURCES Frequency Domain:
Time Domain: 3 Phase + v Generator ab − +
Wye Or Delta
3 PHASE AC POWER
−
v a , v b , v c = Phase Voltages v ab , v bc , v ca = Line Voltages
b
v ca
v bc −
a
+
c
+
+ va
+ vc
vb
−
−
−
3 Phase + V Generator ab − +
Wye Or Delta
b
V ca
V bc −
a
−
+
V a , V b , V c = Phase Voltages V ab , V bc , V ca = Line Voltages
c
+
+ Va
+ Vc
Vb
−
−
−
2
3 PHASE SOURCES Phase Voltages: • v a (t) = Voltage at "a" Relative to Ground
3 PHASE SOURCES
• v a (t) = 2Vφ cos(ω t + φ ) Volts;
Vca
2Vl
• Vφ = Phase Voltage (RMS Value)
• v b (t) = Voltage at "b" Relative to Ground • v c (t) = Voltage at "c" Relative to Ground • v c (t) = 2Vφ cos(ω t + φ + 1200 ) Volts
3 PHASE SOURCES 2Vφ
0 600 30
2Vl
Vb
Va
2Vφ
2Vl
2Vφ
Vbc
Va
2Vφ
• v b (t) = 2Vφ cos(ω t + φ − 1200 ) Volts
2Vφ
Let V a = 2Vφ / 00
Vc
2Vl
Vab
Vb
3 PHASE SOURCES
V ab
Vca
2Vl
2Vl / 2 3 cos 30 = = 2 2Vφ
Let V a = 2Vφ / 00
Vc
300
2Vφ
2Vφ
Vbc
0
2Vl 0
3 PHASE SOURCES Line Voltages: • v ab (t) = Voltage Between Line "a" and Line "b" • v ab (t) = 2Vl cos(ω t + φ + 30 ) Volts 0
• Vl = Line Voltage (RMS Value) = 3Vφ
Va
300
2Vφ
=> Vl = 3Vφ
/V ab = / V a + 30
2Vl
Vab
Vb
Vl = 3Vφ
/V ab = / V a + 300
3 PHASE SOURCES • Phase Voltages in Phasor Form: Va = 2Vφ /φ o Volts V b = 2Vφ /φ − 120o Volts Vφ =
Vl = Phase Voltage (RMS) 3
• v bc (t) = Voltage Between Line "b" and Line "c"
Vc = 2Vφ /φ + 120o Volts
• v bc (t) = 2Vl cos(ω t + φ − 900 ) Volts
• Line Voltages in Phasor Form:
• v ca (t) = Voltage Between Line "c" and Line "a"
Vab = 2Vl /φ + 30o Volts
• v ca (t) = 2Vl cos(ω t + φ + 1500 ) Volts
V bc = 2Vl /φ − 90o Volts Vl = 3 Vφ = Line Voltage (RMS) Vca = 2Vl /φ + 150o Volts
3
3 PHASE SOURCES
3 PHASE LOADS
• Line Voltages Always “lead” Phase Voltages by 30o
a
Ia
b
Ib
c
Ic
Vl = 3Vφ /V ab = / V a + 300
• If know any one of phase or line voltages => can write down all others by inspection
Ia , Ib , Ic = Line Currents
3 PHASE LOADS Balanced Delta Load
a
Ia
b
Ib
Z∆
c
Ic
Z∆
Z∆
Ia , Ib , Ic = Line Currents
n
RMS Voltage Across ZY = Vφ RMS Current Thru ZY = Il
• Phase Voltage across each ZY • Line Current through each ZY
• Delta: • Line Voltage across each ZDelta • Line Current Does NOT Flow through each ZDelta
3 PHASE LOADS • Can Convert any Delta Load to an Equivalent Wye Load:
a'
V Ia = a ZY
ZY
• Wye or Delta: • Wye:
RMS Voltage Across Z∆ = Vl
Ia
Va
ZY • n' Balanced Wye Load
3 PHASE LOADS
Equivalent 1 Phase Model a
ZY
• ZYeq = ZDelta / 3 • Easier to analyze circuit using Wye loads
ZY or Z∆ / 3
n'
• Equivalent 1 Phase Model for Wye: • • • •
See Figure Phase Voltage across ZY; Line Current thru ZY Ia = Va / ZY Write down all other I’s, V’s by Inspection
4
s-DOMAIN CIRCUIT ANALYSIS • Now Want to Go Back to Laplace Transform Analysis of Circuits: • More on Network Functions, H(s)
• More on Poles and Zeros • Stability of Circuits • Then We will be at the End of Unit III
CIRCUIT ELEMENTS IN S-DOMAIN +
IR
+
+
IL sL Ω
VR
RΩ
VL
VC
Li L (0+ ) −
−
IC 1 Ω sC
• Replace i(t), v(t) => I(s), V(s) • Replace R, L, C => s-domain Models • Use V (s) = Z (s) I (s)
• Use All Techniques from Unit I: • Series/Parallel; Voltage/Current Divider • Thevenin/Norton; Source Conversions • Node Equations, Mesh Equations
WITH NO INITIAL STORED ENERGY
ZR
RΩ
ZL
sL Ω ZC
1 Ω sC
−
For a 2nd Order Circuit, Differential Equation is: d2 y dy d2x dx + a1 + a 0 y = b 2 2 + b1 + b0 x 2 dt dt dt dt
Take Laplace Transform with NO IC's:
(a 2s + a1s + a 0 )Y(s) = (b 2s + b1s + b 0 )X(s) 2
• Transform Circuit from Time Domain to s-Domain:
v C (0+ ) s
NETWORK FUNCTION
a2
s-DOMAIN CIRCUIT ANALYSIS
2
NETWORK FUNCTION Define H(s) =
Y(s) with No Initial Stored Energy X(s)
H(s) =
a 2s 2 + a1s + a 0 b 2s 2 + b1s + b 0
(a 2 , a1 , a 0 ) come from LHS of Diff Eq (b 2 , b1 , b 0 ) come from RHS of Diff Eq If Find H(s) → Write Down Differential Equation
5
NETWORK FUNCTIONS H(s) =
Y(s) Output with No Initial Stored Energy = X(s) Input
b b 2 2 b1 ( s + s+ 0 ) ∆ ∆ a2 b2 b2 H(s) = ; 2α = a1/a 2 ; ω0 2 = a 0 / a 2 a0 a1 2 (s + s + ) a2 a2
NETWORK FUNCTIONS • Punch Line: • • • • •
Can find H(s) directly from Circuit Use s-domain Circuit Analysis to find H(s) Use s-domain models for R, L, C with no IC’s Use Generalized Impedance and Ohm’s Law All Techniques from Unit I
Denominator = s 2 + 2α s + ω02 = Characteristic Polynomial Roots of Characteristic Equation determine form of y N (t)
NETWORK FUNCTIONS
ACTIVITY 29-2
• Find H(s) from Circuit Analysis => H(s) Contains A Lot of Useful Information: • • • •
Can Find form for yN(t) Can Find Differential Equation for y(t) Can Determine Stability; Poles and Zeros Can Sketch AC Steady State Behavior (Will do this in Unit IV)
+ vL −
R
1H
i
1 F 400
v(t)
• Let’s Practice with Activity 29-2: Find H in (s) =
I(s) V(s)
+ vC
.1v L
−
Find H(s) =
VC (s) V(s)
ACTIVITY 29-2 • Input = v(t) : • Will first find Hin(s) = I (s) / V (s) • Will also find H (s) = VC (s) / V (s) • Practice s-domain circuit analysis
• Time Domain => s-Domain: • v(t), i(t) => V(s), I(s) • R,L,C => R, sL, 1 / sC • Use techniques from Unit I • Mesh equations
6
