Basic circuits analysis method lecture note

```6.002
CIRCUITS AND
ELECTRONICS
Basic Circuit Analysis Method
(KVL and KCL method)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 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 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Review
LMD allows us to create the
lumped circuit abstraction
i
+
v
Lumped circuit element
power consumed by element = vi
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 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 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Review
a
R1
+
–
b
R4
R3
R2
d
R5
c
DEMO
vca + vab + vbc = 0
KVL
ica + ida + iba = 0
KCL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Method 1: Basic KVL, KCL method of
Circuit analysis
Goal: Find all element v’s and i’s
1. write element v-i relationships
(from lumped circuit abstraction)
2. write KCL for all nodes
3. write KVL for all loops
lots of unknowns
lots of equations
lots of fun
solve
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Method 1: Basic KVL, KCL method of
Circuit analysis
Element Relationships
For R,
V = IR
For voltage source, V = V0
R
+–
V0
For current source, I = I 0
J
Io
3 lumped circuit elements
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
KVL, KCL Example
a
+
ν1
+
ν 0 = V0
–
R3
b
+
ν2
–
ν4
R1
–
+
–
+
+ν 3 –
R2
–
R4
d
+
ν5
–
R5
c
The Demo Circuit
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 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 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
KVL, KCL Example
a
+
+
ν 0 = V0
–
ν1
i0
L1
+
–
–
+
ν2
–
i4
i1 L 2
+
R1
ν 4 R4
–
R3
b i3
d
+ν 3 –
i2
i5
+
R2
ν 5 R5
L3
–
c
The Demo Circuit
L4
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Analyze
ν 0 …ν 5 ,ι0 …ι5
1. Element relationships (v, i )
given v3 = i3 R3
v0 = V0
v4 = i4 R4
v1 = i1 R1
v5 = i5 R5
v2 = i2 R2
12 unknowns
6 equations
2. 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. KVL for loops
L1: − v0 + v1 + v2 = 0
3 independent
equations
L2: v1 + v3 − v4 = 0
s
L3: v3 + v5 − v2 = 0
n
o
L4: − v0 + v4 + v5 = 0 redundant ati
s
n
w
u
no
k
eq
n
u
2
2
1
1
/
ugh @#!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Other Analysis Methods
Method 2— Apply element combination rules
A
B
C
R1
R2 R3
G1
G2
V1
V2
+–
+–
…
RN
GN
⇔
+ RN
+ GN
V1 + V2
+–
J
J
J
I2
G1 + G2
1
Gi =
Ri
⇔
D
I1
⇔
⇔
R1 + R2 +
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 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Other Analysis Methods
Method 2— Apply element combination rules
I =?
Example
R1
V +
–
R3
R2
I
I
V +
–
R1
V +
–
R2 R3
R2 + R3
R = R1 +
R
R2 R3
R2 + R3
V
I=
R
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Method 3—Node analysis
Particular application of KVL, KCL method
1. Select reference node ( ground)
from which voltages are measured.
2. Label voltages of remaining nodes
with respect to ground.
These are the primary unknowns.
3. Write KCL for all but the ground
node, substituting device laws and
KVL.
4. Solve for node voltages.
5. Back solve for branch voltages and
currents (i.e., the secondary unknowns)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Example: Old Faithful
plus current source
V0
R2
R5
J
e2
+ V e1
– 0
Step 1
R4
R1 R
3
I1
Step 2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Example: Old Faithful
plus current source
V0
+ V e1
– 0
R2
R4
e2
R5
J
R1 R
3
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 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Example: Old Faithful
plus current source
V0
e2
R2
R5
+ V e1
– 0
J
R4
R1 R
3
I1
1
Gi =
Ri
KCL at e1
(e1 − V0 )G1 + (e1 − e2 )G3 + (e1 )G2 = 0
KCL at l2
(e2 − e1 )G3 + (e2 − V0 )G4 + (e2 )G5 − I1 = 0
move constant terms to RHS & collect unknowns
e1 (G1 + G2 + G3 ) + e2 (−G3 ) = V0 (G1 )
e1 (−G3 ) + e2 (G3 + G4 + G5 ) = V0 (G4 ) + I1
2 equations, 2 unknowns
(compare units)
Solve for e’s
Step 4
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
In matrix form:
− G3
⎡ G1V0 ⎤
⎡G1 + G2 + G3
⎤ ⎡ e1 ⎤
=
⎢G V + I ⎥
⎢
G3 + G4 + G5 ⎥⎦ ⎢⎣e2 ⎥⎦
− G3
⎣ 4 0 1⎦
⎣
conductivity
matrix
sources
unknown
node
voltages
Solve
G3
⎡G3 + G4 + G5
⎤ ⎡ G1V0 ⎤
G3
G1 + G2 + G3 ⎥⎦ ⎢⎣G4V0 + I1 ⎥⎦
⎡ e1 ⎤ ⎢⎣
⎢e ⎥ =
(G1 + G2 + G3 )(G3 + G4 + G5 ) − G3 2
⎣ 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 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
Solve, given
G1 ⎫
1
=
⎬
G5 ⎭ 8.2 K
G2 ⎫
1
⎬=
G4 ⎭ 3.9 K
1
G3 =
1.5 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 8.2 3.9 1.5
(
G3 + G4 + G5 =
)(
)
1
1
1
+
+
=1
1.5 3.9 8.2
1
1
1
×
+ 1×
3.9 V
e2 = 8.2 1.5
0
1
1− 2
1.5
Check out the
DEMO
e2 = 0.6V0
If V0 = 3V , then e2 = 1.8V0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
6.002 Fall 2000
Lecture 2
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