Kirchoff’s Laws Direct:

KCL, KVL, Ohm’s Law ⇒ V IR VG I= ⇒ =G R= −1

Ohm’s Law: V I I1 1 23 6= ⋅ = ⋅V I2 312= ⋅

(always get 1 equation/Resistor)

KCL:A: - +

B:

5 0

5 01 2 3

1 2 3

I I I

I I I

+ − =− − + =

eq. are dependent

(in general, get n-1 indep. for nodes)

KVL: − + − =V V1 224 0 write 1 loop equation for each loop with a voltagenot in the current set of equations.

⇒ Eliminate either V1 or I using Ohm’s Law

eq: A: − + + − =53 6 12

01 1 2V V V

⇒− −

=

+

−1

3

1

6

1

12 1

21 1

5

24

V

V

KVL: − + − =V V1 224 0

Cramer’s Rule: A X B[ ] ⋅ =v v

XDet A B A

Det Ai

i i N

= [ ][ ]

− +0 1 1.. ..

524V

+

-6Ω3ΩV1

V2+I1 I2

- +

12Ω I3A

B

We can always write in terms of only V, or I variables using ohm’s law:

KCL: A: − + + =I I I1 2 3 0

KVL: M1: − + + =12 2 4 01 2I I

M2 : − + − =4 6 8 02 3 1I I I

−

− −

=

1 1 1

2 4 0

8 4 6

0

12

0

1

2

3

I

I

I

Note:

if V IR=R G= −1

⇔V I G

R

+

B

2Ω 6Ω

4Ω

12V

V1 V3

8 1I

A

M1 M2

V2

+ +

+-

-

+

I3

I2

I1

CurrentDependentVoltage Source:

+

Rules for nodes:

1) Convert all voltage to current sources

1) Determine a reference node and identify unknown relative voltages

2) Use KCL at each unknown node:

at node A: G V G V G V IAA A AB B AN N A− − − =....B: -G V G V G V IAB A BB B BN N B+ − − =....N: -G V G V G V IAN A BN B NN N N+ − − =....

where: Gii = Σ all conductances connected to node iGji = 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.

2 2 22

1 1 1

R

V

10Ω

5V

1

1R

V

R+-

⇔ ⇔I

12 A 10 1Ω = .eg:

⇒4 Ω

2 Ω 5 Ω

4 Ω10 Ω

20 Ω+

-

9A

12V

Eq:

3A

14

Ref

110 1

20

14

12

15

9A

A B C

A: 0 25 5 25 0 5 0 25 3 9. . . . .+ +( ) − − = −V V VA B C (A)

B: −0 5. VA + + +( ) − =0 5 0 1 0 2 02 0. . . .V VB C

C: −.25VA − + + +( ) =0 2 0 2 0 25 0 05 9. . . .V VB C

⇒1 0 0 5 2 5

5 8 2

25 2 5

6

0

9

. . .

. . .

. . .

− −− −− −

=−

V

V

V

A

B

C

⇒ =V VA 4 V VB = 8 33. V VC = 23 3.

One can also solve for N independent currents in N meshes

Consider:

We know that KVL ⇒voltages around a loop = 0.

Current I1 flows in a simple mesh.

I I R I R I R V V

I I R I R I R V VB A

C B

1 1 1 1 2 2 2

2 2 3 2 2 1 2

0

0

:

:

⋅ + ⋅ − ⋅ + − =⋅ + ⋅ − ⋅ + − =

We can as usual re-arrange to a form that can be written by inspection:

1

21 2 1 2 2

2 1 2 3 2

1 2 2

2 2 3

1

2

:

:

R R I R I V V

R I R R I V V

R R R

R R R

I

I

V V

V VA B

B C

A B

B C

+( ) − = −− + +( ) = −

⇒+ −

− +

=−−

* 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.

R1 R3R3A B C

VA VC

VB

R2

I2I1

+

-

+

-

+

-

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 traverseeach loop in same direction, ie, clockwise.

3) Use KVL for each loop in terms of the mesh current variable.

iff no dependent sources:

1: R I R I R I VN N11 1 12 2 1 1− − − =...2: − − − − =R I R I R I VN N12 1 22 2 2 2...N: − − − − =R I R I R I VN N NN N N1 1 2 2 ...

Rii = sum of all resistance in mesh IRij = sum of all common resistance to meshes I,JVI = sum of voltage rises in mesh I, in direction of current I1

Eq: A Wheatstone Bridge

⇒ 12 8 20 8 20 241 2 3+ +( ) − − =I I I− + + +( ) − =8 8 4 6 6 01 2 3I I I− − + + +( ) =20 6 20 6 10 01 2 3I I I

40 8 20

8 18 6

20 6 36

24

0

0

1

2

3

− −− + −− − +

⋅

=

I

I

I

I1

I2

I3

12 Ω

4 Ω

10 Ω20 Ω

8 Ω

24V

+

-

How to solve this system?⇒ Gaussian Elimination to Triangular form:

1

2

3

20 4 10

4 9 3

10 3 18

12

0

0

:

:

:

− −− −− −

=

(divide both sides by 2)

3 10 059

809

53⋅ = +[ ] [ ]− −

1

2

3

0

4 9 3

10 3 18

12

0

0

1309

173'

−

− −− −

=

3 3 016

106⋅ = [ ] [ ]− =-1

2

1

2

3

0

0

10 3 18

12

0

0

1309

173

175

172

'

'

−

−

− −

=

2 2217

173

23

349' '⋅ ⋅ = ⋅ = [ ]− 017

3

1

2

3

0 0

0

10 3 18

12

0

0

969

173

172

'

" −

− −

=

⇒ 969 1 1

5412⋅ = =I V I A

Then 2 173

54

172 2 2

34' = ⋅ = ⋅ ⇒ =+ I I A

3' ⇒ + ⋅ + ⋅ = ⋅ ⇒ =10 3 1854

34 3

34I A I3

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 currentselsewhere in the circuit:

Eq:

if we choose B as the reference node ⇒

A: . . .8 2 2 5+( ) − =V VA C

C: −( ) + +( ) = − +. . .2 2 3 5 4 1V V IA C

now I V VA C= ⋅ −( )0 2.

⇒1 2

1 1 3

−−

=−

.

.

V

V

s

sA

C

I14 1I

.8v

.2v .3v

A B

C

5A

Superposition for Circuits

Consider:

We could solve this by either node or mesh analysis, but there may be a simplerapproach:

If we suppress source #1 (i.e. make a short circuit) we can find I I1 2' ', .

Similarly, I I1 2, could be written without source #2.

Total currents and voltages superpose ⇒ suppress one at a time and thensuperpose the results:

Suppress 1:

⇒ 24 12 6 3 5V I= + +( )( ) ⋅ '

So I A VA' '= ⋅ = ⇒ = +7108

24149

563

24 volt

I1

12818

' = −I2

12818

' =

6Ω 3Ω

12Ω5Ω

12Ω

A B

24

I1

A B

2:

I2I1

+

-+

-

C

1: 30

24

6Ω 3Ω

5Ω

12Ω

Now: Suppress I2 , run I1

30 6 12 8 1V I= + ( )( ) ⋅ ' '

6Ω

12Ω

3Ω

5Ω12Ω

I1' ' I2

' '

+

-30

Exponential Excitation of circuits: Admittance & Impedance

Idea: Exponential function is easy to analyze ⇒simple to add/multiply/integrates, etc.Also is a common case for circuit excitation

Consider:

where V Aest= for complex s, A, real t.

We know all currents are same in circuit, V V V VR C L+ + = KVL

V i t RR = ( ) ⋅ VC

i t dtC = ⋅ ( )∫1V L

di t

dtL = ( )

or i tV t

RRR( ) = ( )

i t CdV t

dtLC( ) = ( )

i tL

V t dtL C( ) = ( )∫1

for our circuit V t Aest( ) = ⇒ V t A eRst( ) = ' , V t A eC

st( ) = ' ' …

where: Ae A A A est st= + +( )' ' ' ' ' ' KVL

we have i tA

ReR

st( ) = 'i t C A s eC

st( ) = ⋅ ⋅' ' iA e

sLL

st

= ' ' '

+

-

R C

V VL

i

L

VRVC

Note: for this kind of excitation, i t C A s eLst( ) = ⋅ ⋅' '

= ⋅ ( )sC V tC

i ts

V t

i tR

V t

L L L

R R

( ) = ( )

( ) = ( )

1

1

So: sC has same units/behavior as 1R

: conductance.

Eg. 2:

Parallel ⇒ admittances ⇒

11

= +Vsv

⇒ =+

VI

s1

for i t e t( ) = −10 2 s A= − =2 10,

V e est t=−

= − = − −101 2

10 10 2 volts

i 1Ω

iR iC

1F

i

V

ref

1s11

Sinewave (Sinusodial) excitation:

We know: e o i oio = +cos sin

⇒ cosoe eio io

= + −

2sin o

e e

i

io io

= − −

2

We wish to study circuits with excitation: v t V wt( ) = +( )cos φ

= +( ) = ++ − − −Ve e v e v eiwt i iwt i

iiwt iwt

2 2ϕ ϕ

vV

ei1 2

= ϕ vV

e i2 2

= − ϕ

note V is complex.

We define:

for s iw=

R C L

sC

R sL

iwC

R iwL

sL

sC

R iwL

iwC

Admittance

Impedance

sinusoid case

1R

1

1

1 1

1

( )

Admittances compose like conductances, Impedances compose like resistances.

So for our circuit: v t R i t L i t dtdi tdt C( ) = ⋅ ( ) + ⋅ + ( )( ) ∫1

for i t Iest( ) =

⇒ ( ) = ⋅ + ⋅ ⋅ +( )v t R I L I s eIsC

st

or IV

R Ls

V

z ssC

=+ +

=1 ( )

we write impedances as Z s( ) , admittance as Y(s)

so i t IeV

R Lsest

sC

st( ) = =+ +

⋅1

eq: if V e t= ⋅ −10 2 (10 0V t@ = → decreasing)

R = 1Ω L H= 12 C F= 1

2

i t e es

s

st t( ) =+ +

= − −101

102

2

2 (Amps)

Basic Trick: Extend the circuit techniques for node and mesh analysis to alsohandle Impedence and Admittances

⇒ generalize circuits which can be analyzed.

note: we can apply superposition to solve this:

suppress v1 or v2 and solve for other.

IR iwL

V

iwC1

2

1=

+ +I

R iwL

V

iwC2

2

1=

− + −

if we write I I1 2, in exponential complex form:

I I ei1 1

1= σ I I ei2 2

2= σ

we get: I IR wL

V

wC

1 22

2 1 2= =

+ −( )o

wL

RwC

11

1

2= − −

= −−tan σ

i t I e ei iwt1 1

1( ) = σ ....

i t i i I e ei wt i wt1 1 2 1

1 1( ) = + = +( )+( ) − −( )σ σ

=+ −( )

⋅ +( )V

R wLwt

wC2 1 2 1cos σ

I, V are called Phasors

R L

CV t

V wt

( ) =cos

+

-

+

-+

-

v iwte2

v iwte2−

v1

v2