Kirchoff’s Laws Direct:
KCL, KVL, Ohm’s Law ⇒
V = IR ⇒ VG = I
G = R −1
V1 = 3 ⋅ I1 = 6 ⋅ I2
V2 = 12 ⋅ I3
Ohm’s Law:
(always get 1 equation/Resistor)
5
I2
I1
+
V2
3Ω
V1
I3
12Ω
A
+
6Ω
+
-
24V
B
KCL:
A:
- 5 + I1 + I2 − I3 = 0
B:
5 − I1 − I2 + I3 = 0
eq. are dependent
(in general, get n-1 indep. for nodes)
KVL: −V1 + 24 − V2 = 0
write 1 loop equation for each loop with a voltage
not in the current set of equations.
⇒ Eliminate either V1 or I using Ohm’s Law
eq:
A:
−5 +
V1 V1 V2
+ −
=0
3 6 12
KVL: −V1 + 24 − V2 = 0
Cramer’s Rule:
v
 13 + 16 
⇒
 −1


−1
−1
12
 V1   5 
V  = 24 
 2  
v
[ A] ⋅ X = B
Xi =
[
Det A0..i −1 B Ai +1.. N
Det[ A]
]
We can always write in terms of only V, or I variables using ohm’s law:
2Ω
+
V1
I1
12V
6Ω
A
I2
+
+
I3
V3
4Ω
V2
-
+
-
8 I1
+
M1
M2
Current
Dependent
Voltage Source:
B
KCL: A:
− I1 + I2 + I3 = 0
KVL: M1 :
−12 + 2 I1 + 4 I2 = 0
M2 :
−4 I2 + 6 I3 − 8 I1 = 0
 −1 1 1   I1   0 
2
4 0   I2  = 12 
 


−8 −4 6   I3   0 
R
Note:
+
V
if
V = IR
R = G −1
⇔
I
G
Rules for nodes:
1) Convert all voltage to current sources
1
R
1
⇔
1
R
1
V
R
+
V
-
10Ω
eg:
1
⇔
1
2
10Ω = .1
A
I 5V
2
2
2
2
1) Determine a reference node and identify unknown relative voltages
2) Use KCL at each unknown node:
at node A:
B:
N:
GAAVA − GABVB − .... − GAN VN = I A
-GABVA + GBBVB − .... − GBN VN = I B
-GAN VA + GBN VB − .... − GNN VN = I N
where: Gii = Σ all conductances connected to node i
G ji = Σ all conductances between node i and node j
Ii = Σ all current sources connected to node i
⇒ Define N equations in N unknowns
all other voltages and currents by Ohm’s Law from V and S.
Eq:
9A
9A
4Ω
2Ω
4Ω
+
12V
1
⇒
5Ω
1
A
1
10 Ω
20 Ω
4
2
4
1
B
1
10
3A
Ref
5
C
1
20
A: (0.25 + .5 + .25)VA − 0.5VB − 0.25VC = 3 − 9 (A)
B: −0.5VA +(0.5 + 0.1 + 0.2)VB − .02VC = 0
C: −.25VA −0.2VB + (0.2 + 0.25 + 0.05)VC = 9
 1.0 −0.5 −2.5 VA  −6
⇒  −.5
.8
−.2  VB  =  0 
 


−.25 −.2
.5  VC   9 
⇒ VA = 4V
VB = 8.33V
VC = 23.3V
One can also solve for N independent currents in N meshes
Consider:
R1
A
R3
B
+
R2
VA
-
I1
I2
C
+
VC
-
+
VB
-
We know that KVL ⇒
voltages around a loop = 0.
Current I1 flows in a simple mesh.
I1: I1 ⋅ R1 + I1 ⋅ R2 − I2 ⋅ R2 + VB − VA = 0
I2 : I2 ⋅ R3 + I2 ⋅ R2 − I1 ⋅ R2 + VC − VB = 0
We can as usual re-arrange to a form that can be written by inspection:
1: ( R1 + R2 ) I1 − R2 I2 = VA − VB
 R1 + R2
⇒
2: − R2 I1 + ( R2 + R3 ) I2 = VB − VC
 − R2
− R2 
R2 + R3 
 I1  VA − VB 
 I  = V − V 
C
 2  B
* Note that KCL is always solved implicitly since at each node we have a sum of currents in meshes:
each mesh enters and leaves with the same current.
In general:
1) Connect each current source with parallel res. to voltage source with series R.
2)
Select a current variable and mesh for each simple loop (usually we traverse
each loop in same direction, ie, clockwise.
3)
Use KVL for each loop in terms of the mesh current variable.
iff no dependent sources:
1:
2:
N:
R11 I1 − R12 I2 − ... − R1 N I N = V1
− R12 I1 − R22 I2 − ... − R2 N I N = V2
− R1 N I1 − R2 N I2 − ... − RNN I N = VN
Rii = sum of all resistance in mesh I
Rij = sum of all common resistance to meshes I,J
VI = sum of voltage rises in mesh I, in direction of current I1
Eq: A Wheatstone Bridge
12 Ω
8Ω
+
24V
I2
4Ω
I1
20 Ω
⇒
(12 + 8 + 20) I1 − 8 I2 − 20 I3 = 24
−8 I1 + (8 + 4 + 6) I2 − 6 I3 = 0
−20 I1 − 6 I2 + (20 + 6 + 10) I3 = 0
 40 −8 −20   I1  24 
 −8 +18 −6  ⋅  I  =  0 

  2  
−20 −6 +36  I3   0 
I3
10 Ω
How to solve this system?
⇒ Gaussian Elimination to Triangular form:
1:
2:
3:
 20 −4 −10  12 
 −4 9
−3  =  0 

  
−10 −3 18   0 
3 ⋅ 5 9 = [ −980
1'
2
3
+ 10] [0]
−17
0  12 
 130
9
3
 −4 9 −3 =  0 

  
−10 −3 18   0 
3 ⋅ 16 = [ −610
1'
2'
3
−5
3
-1
2
3] = [0]
−17
0  12 
 130
9
3
 −17
17
0  =  0 
2
 5
 
−10 −3 18  0 
2' ⋅ 172 ⋅ 173 = 2' ⋅ 23 = [ −934
1'
2"
3
(divide both sides by 2)
17
3
0]
0 0  12 
 969
 −17 17 0  =  0 
2
  
 3
−10 −3 18  0 
⇒
96
9
Then
2' =
3' ⇒
+10 ⋅ 5 4 + 3 ⋅ 3 4 = 18 ⋅ I3
⋅ I1 = 12V I1 = 5 4 A
+17
3
⋅ 45 = 172 ⋅ I2 ⇒ I2 = 3 4 A
⇒ I3 = 3 4 A
Dependent Voltage & Current Sources:
We model the activity of many active components by use of a “programmable”
voltage or current source whose strength is a function of the voltage or currents
elsewhere in the circuit:
Eq:
.8v
A
I1
.2v
.3v
5A
C
if we choose B as the reference node ⇒
A:
C:
(.8 + .2)VA − .2VC = 5
−(.2)VA + (.2 + .3)VC = −5 + 4 I1
now I = 0.2 ⋅ (VA − VC )
⇒
 1 −.2  VA   s 
−1 1.3  V  = − s 

  C  
B
4 I1
Superposition for Circuits
Consider:
6Ω
3Ω
A
B
12Ω
+
1: 30
-
I1
+
2:
5Ω
I2
24
-
C
We could solve this by either node or mesh analysis, but there may be a simpler
approach:
If we suppress source #1 (i.e. make a short circuit) we can find I1' , I2' .
Similarly, I1 , I2 could be written without source #2.
Total currents and voltages superpose ⇒ suppress one at a time and then
superpose the results:
Suppress 1:
6Ω
3Ω
A
12Ω
12Ω
I1
24
⇒ 24V = (12 + 6 (3 + 5)) ⋅ I'
So I' =
7
14
56
⋅ 24 =
A ⇒ VA' =
+ 24 volt
108
9
3
I1' =
−128
18
I2' =
128
18
B
5Ω
Now: Suppress I2 , run I1
6Ω
+
30
-
I1''
30V = (6 + 12 (8)) ⋅ I1' '
3Ω
12Ω
12Ω
I2''
5Ω
Exponential Excitation of circuits:
Idea:
Admittance & Impedance
Exponential function is easy to analyze ⇒
simple to add/multiply/integrates, etc.
Also is a common case for circuit excitation
Consider:
i
VR
VC
R
C
+
V
-
L
VL
where V = Ae st for complex s, A, real t.
We know all currents are same in circuit, VR + VC + VL = V
VR = i(t ) ⋅ R
or
iR (t ) =
VR (t )
R
VC =
1
⋅ i(t )dt
C ∫
VL = L
di(t )
dt
dVC (t )
dt
iL (t ) =
1
VC (t )dt
L∫
iL (t ) = C
for our circuit V (t ) = Ae st ⇒ VR (t ) = A' e st , VC (t ) = A' ' e st …
where: Ae st = ( A' + A' ' + A' ' ' )e st
we have iR (t ) =
KVL
A' st
e iC (t ) = C ⋅ A' ' ⋅s e st
R
A' ' ' e st
iL =
sL
KVL
Note: for this kind of excitation, iL (t ) = C ⋅ A' ' ⋅s e st
= sC ⋅ VC (t )
So: sC has same units/behavior as
iL (t ) =
1
VL (t )
sL
iR (t ) =
1
VR (t )
R
1
: conductance.
R
Eg. 2:
iR
1Ω
i
iC
1F
Parallel ⇒ admittances ⇒
V
1
i
ref
1=
⇒V =
for
V
+ sv
1
I
1+ s
i(t ) = 10e −2 t
V=
s = −2, A = 10
10
= −10e st = −10e −2 t volts
1− 2
1s
Sinewave (Sinusodial) excitation:
eio = cos o + i sin o
We know:
⇒
coso =
eio + e − io
2
sin o =
eio − e − io
2i
We wish to study circuits with excitation: v(t ) = V cos( wt + φ )
=
V iwt + iϕ
(e + e−iwt − iϕ ) = vieiwt + v2e−iwt
2
v1 =
V iϕ
e
2
v2 =
V − iϕ
e
2
C
sC
L
note V is complex.
R
We define:
for s = iw
Admittance
Impedance
(sinusoid case)
1
R
R
1
R
R
1
sC
iwC
1
iwC
1
sL
sL
1
iwL
iwL
Admittances compose like conductances, Impedances compose like resistances.
So for our circuit: v(t ) = R ⋅ i(t ) + L ⋅
for
di ( t )
dt
+
1
C
∫ i(t )dt
i(t ) = Ie st
⇒ v( t ) = ( R ⋅ I + L ⋅ I ⋅ s +
or I =
I
sC
)e st
V
V
=
R + Ls + 1 sC z( s)
we write impedances as Z (s) , admittance as Y(s)
so
i(t ) = Ie st =
eq: if V = 10 ⋅ e −2 t
R = 1Ω
V
⋅ e st
1
R + Ls + sC
( 10V @t = 0 → decreasing )
L = 12 H
i(t ) =
C = 12 F
10
e st = −10e −2 t (Amps)
s
2
1+ 2 + s
Basic Trick: Extend the circuit techniques for node and mesh analysis to also
handle Impedence and Admittances
⇒ generalize circuits which can be analyzed.
R
+
L
V (t ) =
C
V cos wt
-
+
v1
+
v2
-
v
2
eiwt
v
2
e − iwt
note: we can apply superposition to solve this:
suppress v1 or v2 and solve for other.
I1 =
V
R + iwL +
1
V
I2 =
2
2
R − iwL + 1 − iwC
iwC
if we write I1 , I2 in exponential complex form:
I1 = I1eiσ 1
we get: I1 = I2 =
I2 = I2 eiσ 2
V
o1 = − tan −1 

2
R2 + ( wL − 1 wC )
2
i1 (t ) = I1eiσ 1 eiwt ....
(
i1 (t ) = i1 + i2 = I1 e (
=
i wt + σ 1 )
+e
V
R2 + ( wL − 1 wC )
2
− i ( wt − σ 1 )
)
⋅ cos( wt + σ 1 )
I, V are called Phasors
wL − 1 wC 
= −σ 2
R 