Node and Mesh Equations

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Node and Mesh Equations

• NETWORKS 1 ECE 09.201.02
• 10/22/07 Lecture 11
• ROWAN UNIVERSITY
• College of Engineering
• Prof. John Colton
• DEPARTMENT OF ELECTRICAL COMPUTER ENGINEERING
• Fall 2007 - Quarter One

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Some Administrative Items

• Test 2 Results Mean 74.8, SD 19.1
• Test 3 10/23 800 AM 915 AM RH 131
• Problem Set 7 due today
• Problem Set 7 Solutions on website
• Today 1215 200 Review
• 200 315 Lab

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Some Administrative Items

• Capacitors
• Energy Storage in a capacitor
• Series and parallel capacitor combinations
• Inductors
• Energy Storage in an inductor
• Series and parallel inductor combinations
• Circuits with energy storage elements

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Capacitor Energy Storage
Parallel plate capacitor
C Ae/d
e is the dielectric constant f/m A is the area
of the plates m d is the plate separation m
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Capacitor Energy Storage
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Capacitor types
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Capacitor current and voltage
i dq/dt C dv/dt Assume C 1000 µF Assume
v(0) 0 Since charge cannot change
instantaneously. an instantaneous change of
capacitor voltage is not possible
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Capacitor current and voltage
i dq/dt C dv/dt Assume C 1000 µF Assume
v(0) 0 Since charge cannot change
instantaneously. an instantaneous change of
capacitor voltage is not possible, but an
instantaneous change in current is possible
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Capacitor voltage and current
t v(t) (1/C)?i(t)dt
-8 Assume C 1000 µF Assume v(0)
0 What if v(0) ?0 ? 0
t v(t)
(1/C)?i(t)dt (1/C)?i(t)dt
- 8 0
t vc
(0) (1/C)?i(t)dt
0
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Capacitor voltage and current
t v(t) (1/C)?i(t)dt
-8 Assume C 1000 µF Assume v(0)
0 What if v(0) ?0 ? 0
t v(t)
(1/C)?i(t)dt (1/C)?i(t)dt
- 8 0
t vc
(0) (1/C)?i(t)dt
0
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Capacitor voltage and current
t v(t) (1/C)?i(t)dt
-8 Assume C 1000 µF Assume v(0)
0 What if v(0) ?0 ? 0
t v(t)
(1/C)?i(t)dt (1/C)?i(t)dt
- 8 0
t vc
(0) (1/C)?i(t)dt
0
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Energy Storage in a capacitor
i(t) C dv(t)/dt
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Energy Storage in a capacitor
i(t) C dv(t)/dt

t
t v(t) wc
?v(t) i(t)dt ? v(t) C dv/dt dt C?v dv
-8 -8
v (-8) wc ½ C v 2(t)
Joules (1/2C) q 2(t) since q C v for
example, if C 1 µF and v 100 volts, then wc
½ C v 2(t) ½ ( 0.1)(100) 2 500 J

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Example Capacitor energy storage
Suppose the switch is closed for a long period of
time and is opened at t 0 What is vc (0)
? Since charge flows continuously until the
capacitor was fully charged, vc (0 -) 10
volts Since voltage cannot change
instantaneously across a capacitor, vc (0)
10 volts
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Example Capacitor energy storage
Suppose the switch is closed for a long period of
time and is opened at t 0 What is vc (0)
? Since charge flows continuously until the
capacitor was fully charged, vc (0 -) 10 volts
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Example Capacitor energy storage
Suppose the switch is closed for a long period of
time and is opened at t 0 What is vc (0)
? Since charge flows continuously until the
capacitor was fully charged, vc (0 -) 10
volts Since voltage cannot change
instantaneously across a capacitor, vc (0)
10 volts
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Example Capacitor power and energy
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Example Capacitor power and energy
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Example Capacitor power and energy
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Example Capacitor power and energy
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Combining capacitors in parallel
Since the same voltage v appears across each
capacitor i C1 dv/dt C2 dv/dt
CN dv/dt Cp dv/dt
Cp C1 C2 CN So the equivalent
capacitance for N capacitors in parallel is the
sum of their capacitances
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Combining capacitors in series
Since the same current i goes through each
capacitor v(t) v1(t) v2(t) vN(t)
t
t t
t v(t) (1/C1)?i(t)dt
(1/C2)?i(t)dt (1/CN)?i(t)dt t
(1/Cs)?i(t)dt -8
-8 -8
-8
1/Cs 1/C1 1/C2 1/CN So the inverse
of the equivalent capacitance for N capacitors in
series is the sum of their inverse capacitances
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Example Combining capacitances
Suppose C1 C2 C3 2 mF and at t 0, v1 (0)
10v and v2 (0) v3 (0) 20v
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Example Combining capacitances
Suppose C1 C2 C3 2 mF and at t 0, v1 (0)
10v and v2 (0) v3 (0) 20v
Replace two parallel capacitors with Cp Cp C1
C2 vCp ( 0) 20v
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Example Combining capacitances
Suppose C1 C2 C3 2 mF and at t 0, v1 (0)
10v and v2 (0) v3 (0) 20v
Replace two parallel capacitors with Cp Cp C1
C2 vCp ( 0) 20v
Replace two series capacitors with Cs Cs C1Cp
/(C Cp ) vCs ( 0) 10v 20v 30v
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Inductor Energy Storage
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Inductor Energy Storage
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Inductor types
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Inductor voltage and current
v L di/dt Assume L 0.1H Assume i(0)
0 Since magnetic flux cannot change
instantaneously, an instantaneous change of
inductor current is not possible, but an
instantaneous change in voltage is possible
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Inductor voltage and current
v L di/dt Assume L 0.1H Assume i(0)
0 Since magnetic flux cannot change
instantaneously, an instantaneous change of
inductor current is not possible, but an
instantaneous change in voltage is possible
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Inductor current and voltage
t i(t) (1/L)?v(t)dt
-8 Assume L 0.1H Assume i(0)
2A What if i(0) ?0 ? 0
t v(t)
(1/C)?i(t)dt (1/C)?i(t)dt
- 8 0
t vc
(0) (1/C)?i(t)dt
0
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Inductor current and voltage
t i(t) (1/L)?v(t)dt
-8 Assume L 0.1H Assume i(0) 2A
t i(t)
i(0) (10)?v(t)dt
-8 2 20t
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Energy Storage in an inductor
v(t) L di(t)/dt
t
t i(t) wL
?i(t) v(t)dt ? i(t) L di/dt dt L? i di
-8 -8
i (-8) wc ½ L i 2(t)
Joules The inductor is said to have memory and
to store energy

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Combining inductors in series
Since the same current i goes through each
inductor v(t) v1(t) v2(t) vN(t) v(t)
L1 di/dt L2 di/dt LN di/dt Ls di/dt
Ls L1 L2 LN
So the equivalent inductance for N inductors in
series is the sum of their inductances
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Combining inductors in parallel
Since the same voltage v is across each
inductor i(t) i1(t) i2(t) iN(t)
t t
t
t i(t) (1/L1)?v(t)dt
(1/L2)?v(t)dt (1/LN)?v(t)dt t
(1/Lp)?v(t)dt -8
-8 -8
-8
1/Ls 1/L1 1/L2 1/LN So the inverse
of the equivalent inductance for N inductors in
parallel is the sum of their inverse inductances
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Example Combining inductances
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Combining Energy Storage Elements
Leq L1L2 /(L1 L2)
Leq (L1 L2)
Ceq (C1 C2)
Ceq C1C2 /(C1 C2 )