Connections• Terminal – point where an element connects to other elements

• Branch – portion of a circuit with only 2 terminals

• Series elements - same current in a branch

• Node – connection point of branches

• Parallel elements – share same two nodes

• Loop – a closed connection of branches

• Mesh – a loop that does not contain other loops

Kirchoff’s Current Law• Equal amounts of current enter and exit a node

Kirchoff’s Voltage Law• Sum of the voltages around a closed path is zero

Ohm’s Law: V=IR

1

in to n o d e

0N

nn

I

1(voltage rises in

clock-wise direction)

0N

nn

V

RV

I

Voltage Divider Current Divider

Power: If P>0 then the device consumes powerIf P<0 then the device supplies power

Energy

Maximum Power Transfer:

Superposition• Choose one current or voltage source and remove all others

• Replace voltage source by short circuit• Replace current source by open circuit

• Solve for voltages and/or currents of interest.

WP VI

Watts

V

I

0

0

ft

f

t

E P d P t t VI t

11

1 2

RV V

R R

2

1

1 2

RI I

R R

L thR R

2

max4

th

th

VP

R

Repeat forEVERYsource in thecircuit

Resistors in Series: Resistors in Parallel:

Procedure to find Thevenin Equivalent Circuit• A. Find Equivalent Resistance

• Remove Load (resistor or sub-circuit)• Remove Sources

• V source -> short circuit• I source -> open circuit

• Find R

• B. Find Open Circuit Voltage• Remove Load• Solve Circuit (node voltage, or mesh current)• Find voltage at load terminals (still without the load)

Procedure to find Norton Equivalent Circuit• B. Find Short Circuit Current

• Replace Load with a Short• Solve Circuit (node voltage, or mesh current)• Find current at the load terminals (still with the short)

eq iseries

R R 1

1 1N

neq nR R

a

b

RT

a

b

RTVT IT

SourceTransformation

T N N

TN

T

V R I

VI

R

Node Voltage Method – Systematic Application of KCL• Label all n of the Nodes and Select a Reference Node

• Decide if the remaining n-1 Node Voltages are dependent orindependent. A connected voltage source will make a nodedependent. Count the m dependent nodes.

• Write KCL equations at each of the n-1-m independent nodes. Write mequations to relate the dependent node voltages to the sourcevoltages.

• Solve n-1 equations.

Mesh Current Method - Systematic application of KVL• Label all n of the mesh currents in a clockwise direction.

• Decide if the remaining n-1 mesh currents are dependent orindependent. A connected current source will make a meshdependent. Count the m dependent meshes.

• Write KVL equations at each of the n-m independent meshes. Write mequations to relate the dependent mesh currents to the sourcecurrents. Define a supermesh for the special case of a shared currentsource

• Solve n equations

ECE 307 4

Preliminary - how to write the currentat a node using node voltages

Extend this to any resistor in the circuit

ECE 307 5

Adjacent Node of Interest

Adjacent Node of Interest

0R

R

V V V

V V V

ReferenceNode

R

+

-

Node ofInterest

Vnode of interestVadjacent

AdjacentNode

KVL at “Node of Interest”

+

Can now apply thisto all nodes and resistorsin the circuit withoutthinking about the signconvention+

-VR

Adjacent Node of Interest

Into Node of InterestR

V VVI

R R

Ohm’s Law for “Resistor of Interest”

Into Node of InterestI

ECE 307 6

1. Label all meshes

2. Identify dependent meshes(current sources)

3. Write n-m=2 KVL eqns.

n=2

m=0

1 1 1 2 1 2 2rises inCW directionin mesh 1

2 2 3 3 2 4 2 1 2rises inCW directionin mesh 2

( ) 0

( ) 0

n

n

V V i R V i i R

V V i R V i R i i R

Form for resistors in shared branch

Form for resistors in “outside” branch

General Form in a Mesh of Interest:

(imesh of interest-iadjacent)R

i1 i2

zeros 40 2 8 , 8 6 6 , 6 4 20 , , ,x x y y x y y z z y z x y z

ECE 307 7

( )( ) C

C

dv ti t C

dt

( )( ) L

L

di tv t L

dt

1

1 1N

iEQ iL L

1

N

EQ ii

L L

1

1 1N

iEQ iC C

1

N

EQ ii

C C

Series Connected: Parallel Connected:

cosx t A t Sinusoids 1 2

Hertz or Hz 2 rad/s secondsf f TT T

supplied to device stored or dissipated 0 0

t t

p t v t i t W t p t dt v t i t dt

2

00

tt

dissipated

v t v tW v t dt dt

R R

2 2 2stored

00

1 1if 0 0

2 2

tt di t

W L i t dt Li t Li t idt

2 2 2stored 0

0

1 1if (0) 0

2 2

tt dv t

W v t C dt Cv t Cv t vdt

2 maxmax0

10.707

2

T

rms

VV v t dt V

T max cosv t V t

Complex numberscos sinje j

Euler’s Identity

Rectangular Form x A jB

180 180 orjx Ce C Polar Form

2 2 1tan /C A B B A cos sinA C B C

1j

Capacitors and Inductors

RMS

ECE 307 8

( ) cos sinjj Ae A jA A V( ) cos( )v t A t Phasor Transformation

Impedance

)()()( jIjZjV ( )

( )( )

V jZ j

I j

00 Re jRZ R R

290j

LZ j L L Le

21 1 1

90j

CZ ej C C C

1

N

EQ ii

Z Z

Series Impedances

1

1 1N

iEQ iZ Z

Parallel Impedances

Transform circuit to frequency domain thenproceed similar to DC Analysis

Series and Parallel Reduction forImpedancesMesh Current Method for AC circuitsNode Voltage Method for AC circuitsThevenin Equivalent for AC circuitsSuperposition for AC circuits

Transform back to find time domain signal

Frequency Domain or Phasor Domain Circuit Analysis

2 2rms rms

V IV IV I

2

22

1cos cos cos cos

2 2

1cos cos

2

av

VVI VP

Z Z

Z I Z

V I

I

jjZ Z e R jX

j

V

I

0

1cos

2

T

av

VIP p t dt

T

2

1

f

Tf

cosavP

pfV I

0 1pf

*S VI

cos sin

av

S V I jV I

P jQ

cos

sin

avP V I

Q V I

|S| - apparent power

- units volt-amperes (VA)

Re

Im

s Q - reactive power- volt-amperes reactive (VAR)

Pav - real power (absorbed by load resistance)- watts (W)

2 2

avP R Q X I I

* 2*/ /

av

S Z Z

P jQ

V V V

2

2

1if

1if

avZ R PR

Z jX QX

V

V

Complex Power Equations

Power factor correction

2

2

sin

cosavP V I I R

Q V I I X

load needingpf correction

Shunt reactanceto give pf correction

avP

2A AI X Q

2B BI X Q

S

Power factor Triangle

Condition for correction:

Make: B AQ Q

1 1if ifA B A BX L X X X L

C C

avS P

1V 2V

1I2I

1 2:n n

1 : Nor2

1

nN

n

12 1 2

IV NV I

N

* * *21 2 21 21 2S N S

N

VV I I V I

SV

SZ

LZ

* or andL S L S L SZ Z R R X X

2

max 4

S

av

S

PR

V

SV

SZ

LZ

a

b

1V

2V

SZ

1 2:n n2

1ab LZ Z

N

* *

2 2 2

1 L LS S S ab L

R XZ R jX Z Z j

N N N

Max Power

2 2andL S L SR N R X N X

inductor

1capacitor

S

L

X

C

Max Power to load

Power thru Ideal Transformer

Ideal Transformer Equations

2

2

10182

Octal (base 8)

Hexadecimal (base 16)

2 10000 0

2 10001 1

2 10010 2

2 10011 3

2 10100 4

2 10101 5

2 10110 6

2 10111 7

2 101000 8

2 101001 9

2 101010 10

2 101011 11

2 101100 12

2 101101 13

2 101110 14

2 101111 15

Common Binary to Decimal

101101 1 0 1 1 0 1

10110110 10 110 110

Binary

Binary

10110110 1011 0110 B6 B6H

Binary

Boolean Logic Operations, Truth Tables, and Logic Expressions

X Y Z

0 0 0

0 1 1

1 0 1

1 1 1

X Y Z

0 0 0

0 1 0

1 0 0

1 1 1

X Y

X Y XY

X X

0 1

1 0

5 4 3 2 1 02

10

2 2 2 2101101 1 0 1 1 0 1

1 0 1 1

2 2

32 16 8 4 2 1

45

0 1

2

8

2 1 0

10

10110110 10 110 110

266

8 8 8

128 48

2

1

6

5

6

6

2

Binary-to-decimal

Binary-to-Octal

2 1610110110 1011 0110 B6 B6H

Binary-to-Hexadecimal

Boolean Logic Operations, Truth Tables, and Logic Expressions

X

1

0

18210

5 4 3 2 1 0101101 1 0 1 1 0 12 2

32 16 8 4 2 1