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6
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002 Circuits and Electronics, Spring 2007
...
mit
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Downloaded on [DD Month YYYY]
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002 Fall 2000

Lecture 2

Review
Lumped Matter Discipline LMD:

Constraints we impose on ourselves to simplify
our analysis

∂φ B
=0
∂t
∂q
=0
∂t

Outside elements
Inside elements
wires resistors sources

Allows us to create the lumped circuit
abstraction

Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6
...
MIT
OpenCourseWare (http://ocw
...
edu/), Massachusetts Institute of Technology
...


6
...
002 Circuits and Electronics, Spring 2007
...
mit
...
Downloaded on [DD Month YYYY]
...
002 Fall 2000

Lecture 2

Review
Review
Maxwell’s equations simplify to
algebraic KVL and KCL under LMD!
KVL:

∑ jν j = 0
loop

KCL:

∑jij = 0
node

Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6
...
MIT
OpenCourseWare (http://ocw
...
edu/), Massachusetts Institute of Technology
...


6
...
002 Circuits and Electronics, Spring 2007
...
mit
...
Downloaded on [DD Month YYYY]
...
002 Fall 2000

Lecture 2

Method 1: Basic KVL, KCL method of
Circuit analysis
Goal: Find all element v’s and i’s
1
...
write KCL for all nodes
3
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002 Circuits and Electronics, Spring 2007
...
mit
...
Downloaded on [DD Month YYYY]
...
002 Fall 2000

Lecture 2

Method 1: Basic KVL, KCL method of
Circuit analysis

Element Relationships
For R,

V = IR

For voltage source, V = V0
For current source, I = I 0

R
+–

V0

Io
3 lumped circuit elements

Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6
...
MIT
OpenCourseWare (http://ocw
...
edu/), Massachusetts Institute of Technology
...


6
...
002 Circuits and Electronics, Spring 2007
...
mit
...
Downloaded on [DD Month YYYY]
...
002 Fall 2000

Lecture 2

Associated variables discipline
i

+
ν

Element e

Current is taken to be positive going
into the positive voltage terminal

Then power consumed
by element e

= νi is positive

Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6
...
MIT
OpenCourseWare (http://ocw
...
edu/), Massachusetts Institute of Technology
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6
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002 Circuits and Electronics, Spring 2007
...
mit
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Downloaded on [DD Month YYYY]
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002 Fall 2000

Lecture 2

Analyze
ν 0 …ν 5 ,ι0 …ι5
1
...
KCL at the nodes
a: i0 + i1 + i4 = 0
3 independent
b: i2 + i3 − i1 = 0
equations
d: i5 − i3 − i4 = 0
e: − i0 − i2 − i5 = 0 redundant
3
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002 Circuits and Electronics, Spring 2007
...
mit
...
Downloaded on [DD Month YYYY]
...
002 Fall 2000

Lecture 2

Other Analysis Methods
Method 2— Apply element combination rules

A
B

C

R1

R2 R3

G1

G2

V1

GN

V2

+–

…

+–

⇔

I2

⇔

R1 + R2 +

G1 + G2

1
Gi =
Ri

⇔

D
I1

RN

⇔

+ RN

+ GN

V1 + V2
+–

I1 + I 2

Surprisingly, these rules (along with superposition, which
you will learn about later) can solve the circuit on page 8
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6
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MIT
OpenCourseWare (http://ocw
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edu/), Massachusetts Institute of Technology
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6
...
002 Circuits and Electronics, Spring 2007
...
mit
...
Downloaded on [DD Month YYYY]
...
002 Fall 2000

Lecture 2

Method 3—Node analysis
Particular application of KVL, KCL method
1
...

2
...

These are the primary unknowns
...
Write KCL for all but the ground
node, substituting device laws and
KVL
...
Solve for node voltages
...
Back solve for branch voltages and
currents (i
...
, the secondary unknowns)

Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6
...
MIT
OpenCourseWare (http://ocw
...
edu/), Massachusetts Institute of Technology
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6
...
002 Circuits and Electronics, Spring 2007
...
mit
...
Downloaded on [DD Month YYYY]
...
002 Fall 2000

Lecture 2

Example: Old Faithful
plus current source

V0
R1 R
3

+ V e1
– 0
R2

R4
e2

R5

for
I1 convenience,
write
1
Gi =
Ri

KCL at e1
(e1 − V0 )G1 + (e1 − e2 )G3 + (e1 )G2 = 0

KCL at e2
(e2 − e1 )G3 + (e2 − V0 )G4 + (e2 )G5 − I1 = 0
Step 3

Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6
...
MIT
OpenCourseWare (http://ocw
...
edu/), Massachusetts Institute of Technology
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6
...
002 Circuits and Electronics, Spring 2007
...
mit
...
Downloaded on [DD Month YYYY]
...
002 Fall 2000

Lecture 2

In matrix form:
− G3
⎡ G1V0 ⎤
⎡G1 + G2 + G3
⎤ ⎡ e1 ⎤
= ⎢
⎥
⎢
G3 + G4 + G5 ⎥ ⎢e2 ⎥
− G3
⎣G4V0 + I1 ⎦
⎣
⎦ ⎣ ⎦

conductivity
matrix

sources

unknown
node
voltages

Solve
G3
⎡G3 + G4 + G5
⎤ ⎡ G1V0 ⎤
G3
G1 + G2 + G3 ⎥ ⎢G4V0 + I1 ⎥
⎡ e1 ⎤ ⎢
⎣
⎦ ⎣
⎦
2
⎢e ⎥ =
(G1 + G2 + G3 )(G3 + G4 + G5 ) − G3
⎣ 2⎦

(

)(

) ( )(

)

G +G +G G V + G G V + I
3
4
5 1 0
3 4 0 1
e =
1 G G +G G +G G +G G +G G +G G +G 2 +G G +G G
1 3
1 4
1 5
2 3
2 4
2 5
3
3 4
3 5
e2 =

(G3 )(G1V0 ) + (G1 + G2 + G3 )(G4V0 + I 1 )
G1G3 + G1G4 + G1G5 + G2G3 + G2G4 + G2 G5 + G3 + G3G4 + G3G5
2

(same denominator)

Notice: linear in V0 , I1 , no negatives
in denominator
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6
...
MIT
OpenCourseWare (http://ocw
...
edu/), Massachusetts Institute of Technology
...


6
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2 K

G2 ⎫
1
⎬=
G4 ⎭ 3
...
5 K

I1 = 0

(

)(

)

G G V + G +G +G G V + I
e = 3 10 1 2 3 40 1
2 G + G + G + G + G + G −G 2
1 2 3
3 4 5 3
1
1
1
G +G +G =
+
+
=1
1
2
3 8
...
9 1
...
5 3
...
2

1
1
1
×
+ 1×
3
...
2 1
...
5

Check out the
DEMO

e2 = 0
...
8V0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6
...
MIT
OpenCourseWare (http://ocw
...
edu/), Massachusetts Institute of Technology
...


6

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