1. EE3301 Electrical Network Analysis
Dr. Jeanne Pitz

2. Electrical Network Analysis
Analysis and design of RC, RL, and RLC electrical networks
Sinusoidal steady state analysis of passive networks using phasor representation
mesh and nodal analyses
Introduction to the concept of impulse response and frequency analysis using the Laplace transform.
Prerequisites: MATH 2420 ( integral & differential eq)
PHYS 2326 (electricity & magnetism) .
Corequisite: EE 3101 (ENA lab)

3. Course Logistics The grade will be based on 4 tests
Homework will be assigned but NOT collected.
There will be a quiz every Wed. closed book and notes.
It will cover the material that has already covered and the homework. No quiz on test days.
3 Tests are scheduled during the semester.
They are NOT comprehensive.
The date will not be changed but what’s covered may be modified.
The grade will be based on 4 tests
The 4th will be the percentage based on the sum of the quizzes 10 best quizzes.

4. Course Logistics: Schedule

5. Chapter 1 Overview International units Voltage and Current Definitions
“Ideal” circuit element
Power and energy

6. homework Read chapter 1 esp. pages 10-18 Work problems: 1.14 a-d
1.18 a-c
1-25 a-e
1-26
Answers are in the back of the book.

7. Circuit language Voltage is measured across two points
Current is measured through an element
Scale factor prefix smaller than 1
Milli (m) – 10-3
Micro (m)
Nano (n) – 10-9
Pico (p) – 10-12
Scale factor prefix larger than 1
Kilo( K) – 103
Mega( M) – 106
Giga (G)– 109
Terra (T) – 1012

8. Definitions in circuit theory
Voltage is the energy per unit charge of separating a pos and neg charge, that is the potential difference between two points
Current is the rate of change of charge
Current and voltage sign passive convention + -

9. Conventions in circuit theory
Voltage is joules per coulomb, the energy required to move a positive charge of 1colomb through the element.
Power is i(t)*v(t)

10. definitions Power is the time rate of delivering or absorbing energy
In terms of voltage and current:

11. Other facts
Current and voltage are VECTOR quantities having both magnitude and direction
If you compute power with the passive sign convention
Positive power is absorbed
Negative power is supplied
Energy is neither created or destroyed; it must be conserved.
The charge on an electron is X coulombs

12. Power sign conventions
P is pos = vi if the current is entering the positive terminal.
See fig 1.6 for other cases
v(t) +
i(t) -

13. Problem 1-14 pg 20 a, b + a. V=125V I= 10 A v(t) =1250 W B->A
Supplied by B
Absorbed by A
v(t)
i(t) A B - b.
V=5V
I= -240A
W= 1200W
A->B.
Supplied by A
Absorbed by B

14. Prob 1-18 a.
𝑣 𝑡 =50 𝑒 −1600𝑡 −50 𝑒 −400𝑡 𝑉
𝑖(𝑡)=5𝑒 −1600𝑡 −5 𝑒 −400𝑡 𝑚𝐴
Find the power at 625 s =0.625ms
Work out the exponents:
1600*0.625ms=1000ms = 1.00s
400*0.625 ms = 0.25s
1000*0.625 ms = 625ms = .625 s
𝑣=50 𝑒 −1 −50 𝑒 −0.25 𝑉=18.39−38.94 𝑉=−20.5𝑉
𝑖=5𝑒 −1 −5 𝑒 −0.25 𝑚𝐴=−2.05 𝑚𝐴
𝑝=𝑣∗𝑖=−20.5𝑉∗−2.05 𝑚𝐴=42.2 𝑚𝑊

15. Prob 1-18 b
𝑣 𝑡 =50 𝑒 −1600𝑡 −50 𝑒 −400𝑡 𝑉 𝑖 𝑡 =5 𝑒 −1600𝑡 −5 𝑒 −400𝑡 𝑚𝐴 𝑝 𝑡 =𝑣 𝑡 ∗𝑖 𝑡 𝑒𝑛𝑒𝑟𝑔𝑦= 0 625𝜇 𝑝 𝑡 𝑑𝑡
𝑝(𝑡)=250 𝑒 −3200𝑡 −2 𝑒 −2000𝑡 + 𝑒 −800𝑡
𝑒 𝑎𝑡 𝑑𝑡= 𝑒 𝑎𝑡 𝑎
E =250 𝑒 −3200𝑡 −3200 − 2 𝑒 −2000𝑡 − 𝑒 −800 −800𝑡 𝜇 0

16. Solution in book

17. Chapter 2 Voltage and current sources Electrical resistance Ohm’s law
Independent, dependent
Electrical resistance
Ohm’s law
Resistive Circuits
Circuit models
Kirchhoff's law
Analysis of independent vs. dependent current sources

18. homework Read pages 24-48 Make sure you understand the examples
Look it the following homework problems
2.2, 2.4, 2.5, 2.7, , 2.1, , 2.22, , 2.35, 2.36, 2.37

19. Ideal Sources
Ideal Voltage sources: can supply as much current as the circuit requires.
Ideal Current sources : supplies current regardless of the voltage across it.
Independent Sources: don’t rely on any other circuit parameters
Dependent Sources : depend on some parameter in the circuit.

20. Sources: voltage + - DC Independent Voltage source: +
Constant value regardless of what is attached. The sign indicates it’s polarity (positive on top) + + - 5V -

21. Sources: current DC Current source : independent
arrow show it’s direction
Constant value regardless of what is attached.
Current flows around the loop.
10A

22. Dependent Voltage Sources
Dependent DC Voltage source : dependent on some parameter in the circuit. + + -
5 V  -

23. Dependent current Sources
Dependent DC Current source : dependent on some parameter in the circuit. +
25a A a -

24. Assessment problem 2.1 𝑣𝑔= 𝑖𝑏 4 𝑖𝑏=−8𝐴 𝑣𝑔= -2 V P8A =−2 ∗8=−16 W
a. What value of ib makes this circuit valid:
b. What power is associated with the 8A source: a, b,
𝑣𝑔= 𝑖𝑏 4 𝑖𝑏=−8𝐴
𝑖𝑏 4 ≜−2𝑉
𝑣𝑔= -2 V
P8A =−2 ∗8=−16 W
16 W delivered
Current entering + node

25. Networks in Electrical Engineering
A network is an interconnection of components such as sources, resistors inductors and capacitors.
In studying first, DC (direct current) circuits composed only of resistors and sources, we can learn the basics of electricity, that will be useful with more complex circuits.
In Direct current circuits the voltage or current is assumed to be constant overtime.
For example the 4V supply never varies with time. Similarly for the current sources they are considered 2A and 4A for all time.

26. Resistors and Ohm’s Law
Resistance is the property of a component to impede the flow of current.
Such a component is called a “resistor”.
The symbol for a resistor with resistance R is shown below.
Ohms law relates the current through the resistor to the voltage across it. V=I*R R

27. Circuit Elements + - Resistors v i resist the flow of current
Obey ohm’s law
Voltage and current relationships
V=ir ohms law.
R is usually considered a constant.
but can be temperature dependent + - v i

28. Resistors: Current and Voltage conventions
The conventions for signs are as follows. Ir Vr + - R
Vr= Ir * R

29. Single loop analysis
Ohm’s law : the voltage across a resistor is directly proportional to the current flowing through it.
For simple resistors R is constant so
v=i*R
In DC analysis the voltage (or current ) is constant so on a plot of I vs. V, the slope is 1/R i
1/R v

30. Conductance
If we use the reciprocal of ohms we get Conductance which is measured in “siemens” S.
some literature uses “mho”
G= 1/R
Mho : 

31. Prob 2.11 v=i*r =>r=v/i
Slope m = Δy/Δx; on this graph y is current in mA and x is V
Find slope from 2 points: (54,2m) (108,4m): m=(4-2)mA/(108-54)V=1/R
R=26K

32. Adding resistors series vs parallel
Series share the same current
RT = R1 + R2
Parallel share the same voltage
1/RT= 1/R1 +1/R2
RT= R1*R2/(R1 + R2) R1 R2 I V + R1 R2 - -

33. Circuit Models Some terminology
Branch – a portion of the circuit containing a single element and the nodes at either end of the element.
Node – a point connecting two or more elements.
Loop – a closed path in which no node is encountered more than once (except the starting point).

34. Loops, nodes, branches node 6V - 8V + Vx 4A - + * - 2A branch + 4V - -
* - 2A
branch + 4V - - 6V + - 2V + 6A
Closed loop
node

35. Kirchhoff’s laws Current law Voltage law
Sum of currents entering or leaving a node is 0, taking direction into account
Voltage law
Sum of voltages around a closed loop is 0 taking polarity into account

36. Circuit convention rules
Sum of voltages around a closed loop is 0.
Sum of currents at a node is 0.
Power passive sign convention:
Arrange I and v so magnitude are positive then if the sign of power is positive it being absorbed; if the sign of power is negative it being supplied. +
v(t)
i(t) -

37. Problem 1-22 Vx 2A 6V 4A
- 8V +
- +
+ - + 4V - - 6V + - 2V + 6A
proceed in a loop; Add the voltage if you reach the + terminal first; subtract if you reach the – terminal first.
0 = -8V + Vx - 6V - 4V
Vx= 18V

38. Analysis with dependent sources
Write the equations using the “parameter” specified.
Then another equation setting the parameter
Solve the equations simultaneously

39. Example

40. Example assessment 2.9
Solve for current i1 write a kvl loop equation around the outside loop.
Voltage v: kvl around the outside loop, solve for v
0=−5V+54Ki1−1V+𝐯−1.8𝐾 30𝑖1 +8𝑉 0=−5V+54Ki1−1V+𝐯−54𝐾𝑖1+8𝑉 0=−5V−1V+𝐯+8𝑉 6−8=𝒗=−2 𝑉
5𝑉=54𝑘∗𝑖1−1𝑉+6𝑘∗31𝑖1 6= 54𝑘+186𝑘 𝑖1 𝑖1= 6 240k =25𝜇𝐴

41. Power calculations: assessment 2.9 c,d
component
Voltage(v)
Current(A)
Power(W)
absorbed
delivered
5V supply 5V
25
125
54K resis
1.35V
33.7
33.75
1V supply 1V
6K resis
4.65V
775 
Dep src
-2V
750
1500
1.8K resis
1012.5 8V 8
6000
power
6150

42. Voltage and current division
Voltage divides across resistors in series in proportion to their values.
Vt = V1 + V2 = I R1 + I R2
Current divides through resistors in proportion to their values.
It = I1 + I2 =V/R1 + V/R2

43. Chapter 3 Resistors in series Resistors in parallel
Voltage divider circuits
Current divider circuits
Measuring voltage
Measuring current
Measuring resistance
Wheatstone bridges
(skip section 3.7)

44. Adding Resistors in Series
Two resistors connected at a single node
Write Kirchhoff's law around the loop:
𝑣𝑖=𝑅1𝑖+𝑅2𝑖+𝑅3𝑖+𝑅4𝑖+𝑅5𝑖+𝑅6𝑖+𝑅7𝑖 𝑣𝑖=𝑖 𝑅1+𝑅2+𝑅3+𝑅4+𝑅5+𝑅6+𝑅7 𝑣𝑖=𝑖𝑅𝑒𝑞 𝑅𝑒𝑞= 𝑅1+𝑅2+𝑅3+𝑅4+𝑅5+𝑅6+𝑅7

45. Adding Resistors in Parallel
Parrallel connected elements have the same voltage
Write Kirchhoff's current law at node a.
𝑖𝑠= 𝑖1+𝑖2+𝑖3+𝑖4
𝑖𝑠= 𝑣𝑎𝑏 𝑅1 + 𝑣𝑎𝑏 𝑅2 + 𝑣𝑎𝑏 𝑅3 + 𝑣𝑎𝑏 𝑅4 =𝑣𝑎𝑏 1 𝑅1 + 1 𝑅2 + 1 𝑅3 + 1 𝑅4

46. Adding Resistors in Parallel
𝑖𝑠=𝑣𝑎𝑏 1 𝑅1 + 1 𝑅2 + 1 𝑅3 + 1 𝑅4 = 𝑣𝑎𝑏 𝑅𝑒𝑞
1 Req = 1 𝑅1 + 1 𝑅2 + 1 𝑅3 + 1 𝑅4

47. Example Assessment Problem 3.1
Find the voltage v
Find the power delivered by the current source
Find the power dissipated by the 10  resistor.

48. Solution 𝑅𝑒𝑞=((16| 64)+7.2) |30Ω 𝑅𝑒𝑞= 12.8+7.2 30=20 30=12Ω 𝑣=5∗12=60𝑉
a. solve for Req then v = 5A*Req
b. solve for power delivered by the 5A source
𝑅𝑒𝑞=((16| 64)+7.2) |30Ω 𝑅𝑒𝑞= =20 30=12Ω 𝑣=5∗12=60𝑉
𝑃=𝑣∗𝑖=5∗60=300 𝑊 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑

49. Solution Find the power dissipated by the 10  resistor
To find V1 across the 10 + 6 resistors we must first determine the current in 30 and 7.2 then find current in the 10 branch
𝑅1=16||64+7.2=20Ω 5A= =2𝐴+3𝐴 60−𝑣1=7.2∗3=21.6 𝑣1=60−21.6=38.4𝑉 i10= V1 16 = =2.4𝐴 V1

50. Solution 𝑉10=10Ω∗2.4𝐴=24𝑉 𝑃10=24∗2.4=57.6𝑊
Now we have the current in the branch we can find voltage in the 10  and power is v*i.
𝑉10=10Ω∗2.4𝐴=24𝑉 𝑃10=24∗2.4=57.6𝑊

51. Voltage Divider
Resistors can used to develop voltage level from the source voltage.
𝑣𝑠=𝑖𝑅1+𝑖𝑅2=𝑖 𝑅1+𝑅2
𝑖= 𝑣𝑠 𝑅1+𝑅2
𝑣1=𝑖𝑅1
𝑣1= 𝑅1 𝑅1+𝑅2

52. Similarly for V2:
𝑣2=𝑖𝑅2= 𝑅2 𝑅1+𝑅2

53. Current Divider The current source is is divided at the node:
Is=i1 + i2
𝑣=𝑖1𝑅1=𝑖2𝑅2
𝑅𝑒𝑞= 𝑅1𝑅2 𝑅1+𝑅2 ⇒𝑣=𝑖𝑠𝑅𝑒𝑞
𝑖1=𝑖𝑠 𝑅2 𝑅1+𝑅2

54. Assessment problem 3.3 pg. 64
Find the value of R that will cause 4A of current to flow through the 80.
How much power will R dissipate?
How much power will the Current source supply

55. Solution to Assessment 3.3 a
Find the value of R that will cause 4A of current to flow through the 80.
Write an equation that includes R showing how the current divides between the two branches:
𝑖80= 20∗𝑅 𝑅+120 =4𝐴
20∗𝑅=4 𝑅+120
20R−4R=480
𝑅 20−4 =480
𝑅= =30Ω

56. Solution: Assessment 3.3 b
How much power will R dissipate?
Find the current in the branch iR
Find voltage at VR then power is vR*IR
𝑃R=𝑣∗𝑖𝑅
𝑖𝑅=20−4 𝐴=16𝐴
𝑣𝑅=30∗16=480𝑉
𝑃=16∗480=7680𝑊

57. Solution Assessment 3.3 c
How much power will the Current source supply?
Find the voltage Vs then total power is Vs*20A Vs VR
𝑉60=20𝐴∗60Ω=1200𝑉
𝑉𝑠−480𝑉=1200𝑉
𝑉𝑠= =1680V
𝑃=20𝐴∗1680𝑉=33,600𝑊

58. Solving circuits using voltage division
Generalizing voltage and current division to help solve and analyze circuits
Series connections
Same “i” goes through all the resistors
i= v 𝑅1+𝑅2+…𝑅𝑛−1+𝑅𝑛 = 𝑣 𝑅𝑒𝑞
𝑣𝑗=𝑅𝑗𝑖= 𝑣𝑅𝑗 𝑅𝑒𝑞

59. Solving circuits using current division
Next consider the circuit shown. Parallel connections:
All resistors have the same voltage drop
v =i(R1||R2||…Rn−1||Rn)=i∗Req
𝑖 𝑗 = 𝑣 𝑅 𝑗 = 𝑖 𝑅 𝑒𝑞 𝑅 𝑗

60. Resistance Measurements Wheatstone bridge
A Wheatstone bridge is a “balanced circuit used to measure the resistance of Rx by varying R3 until no current is measured in the meter ig
Then the unknown resistance is: 𝑅 𝑥 = 𝑅 2 𝑅 1 𝑅 3

61. Wheatstone bridge cont’d
If ig =0 then:
Points a and b are at the same potential since no current is flowing in the meter (ig=0)
𝑖 1 + 𝑖 2 = 𝑖 3 + 𝑖 𝑥 𝑖 1 = 𝑖 3 𝑎𝑛𝑑 𝑖 2 = 𝑖 𝑥
𝑣=𝑖1𝑅1+𝑖3𝑅3 𝑣= 𝑖 2 𝑅 2 + 𝑖 𝑥 𝑅 𝑥

62. Wheatstone bridge cont’d
Now divide equation 1 by equation 2.
𝑖3𝑅3= 𝑖 𝑥 𝑅 𝑥 (1)
𝑖1 𝑅 1 = 𝑖 2 𝑅 2
𝑖3 𝑅 1 = 𝑖 𝑥 𝑅 2 (2)
𝑅3 𝑅 1 = 𝑅 𝑥 𝑅 2
𝑅 𝑥 = 𝑅 2 𝑅3 𝑅 1

63. Assessment problem 3.7 𝑅 𝑥 = 𝑅 2 𝑅3 𝑅 1 𝑅 𝑥 = 1000 150 100 𝑅 𝑥 =1500Ω
What is Rx?
𝑅 𝑥 = 𝑅 2 𝑅3 𝑅 1 𝑅 𝑥 = 𝑅 𝑥 =1500Ω

64. Assessment problem 3.7 b How much power is dissipated by he circuit
Now the power in each resistor is R*i2
𝑅𝑇= (2500∗250) =227.3Ω 𝐼 𝑇 = =22𝑚𝐴 𝑖1= =20𝑚𝐴 𝑖 2 = =2𝑚𝐴

65. Power Calculations P100= 100*(20mA)2=40mW P150= 150* (20mA)2=60mW
So yes all resistors are well within the 250mW max limits

66. Skip delta and wye configurations

67. Backup

68. Content chapter 1 Outline: introduction Components: Parameters
Resistors.
Inductors
Capacitors
Parameters
Voltage
current
Passive sign convention
Power and energy ts

69. Steps to check
E =250 −𝑒 −3200𝑡 𝑒 −2000𝑡 −𝑒 800 −800𝑡 𝜇 0
E =250 −𝑒 − 𝑒 − − 𝑒 800 −0.5 −250 −𝑒 𝑒 − 𝑒
E = −250𝑒 − 𝑒 − − 250𝑒 800 − −
E = − 𝑒 − 𝑒 −1.25 − 𝑒 − −
E = (1−𝑒 −2 ) 𝑒 −1.25 − (1−𝑒 −0.5
E = − 𝑚𝐽
E = − = 𝑚𝑊=17.078𝜇𝐽

70. Steps to check
E =250 −𝑒 −3200𝑡 𝑒 −2000𝑡 −𝑒 800 −800𝑡 𝜇 0
E =250 −𝑒 −3200∗625𝜇 𝑒 −2000∗625𝜇 − 𝑒 800 −800∗625𝜇 −250 −𝑒 𝑒 − 𝑒
E =250 −𝑒 − 𝑒 −1, − 𝑒 800 − −
E =250 −𝑒 −3200∗625𝜇 𝑒 −2000∗625𝜇 − 𝑒 800 −800∗625𝜇 −
E = − 𝑒 −3200∗625𝜇 𝑒 −2000∗625𝜇 −0.325 𝑒 −800∗625𝜇 − −
E = − 𝑒 − 𝑒 −1.25 −0.325 𝑒 −
E = − − =