EE3301 Electrical Network Analysis Dr. Jeanne Pitz

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)

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.

Course Logistics: Schedule

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

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.

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) - 10-6 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

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 + -

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)

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

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 1.6022 X 10 -19 coulombs

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) -

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

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

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 −2000 + 𝑒 −800 −800𝑡 625𝜇 0

Solution in book

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

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, 2.112.13, 2.1, 2.18 2.19, 2.22, 2.28 2.34, 2.35, 2.36, 2.37

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.

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

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

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

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

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

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.

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

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

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

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

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 : 

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

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 - -

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).

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

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

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) -

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

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

Example

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

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 3603.75 Dep src -2V 750 1500 1.8K resis 1012.5 8V 8 6000 power 6150

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

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)

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

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

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

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.

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= 12.8+7.2 30=20 30=12Ω 𝑣=5∗12=60𝑉 𝑃=𝑣∗𝑖=5∗60=300 𝑊 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑

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= 60 30 + 60 20 =2𝐴+3𝐴 60−𝑣1=7.2∗3=21.6 𝑣1=60−21.6=38.4𝑉 i10= V1 16 = 38.4 16 =2.4𝐴 V1

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

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

Similarly for V2: 𝑣2=𝑖𝑅2= 𝑅2 𝑅1+𝑅2

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

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

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= 480 16 =30Ω

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

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=1200+480=1680V 𝑃=20𝐴∗1680𝑉=33,600𝑊

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

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

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

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

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

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 = 1000 150 100 𝑅 𝑥 =1500Ω

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) 2500+250 =227.3Ω 𝐼 𝑇 = 5 227.3 =22𝑚𝐴 𝑖1= 5 250 =20𝑚𝐴 𝑖 2 = 5 2500 =2𝑚𝐴

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

Skip delta and wye configurations

Backup

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

Steps to check E =250 −𝑒 −3200𝑡 3200 + 2 𝑒 −2000𝑡 2000 + −𝑒 800 −800𝑡 625𝜇 0 E =250 −𝑒 −2 3200 + 2 𝑒 −1.25 2000 − 𝑒 800 −0.5 −250 −𝑒 0 3200 + 2 𝑒 0 2000 − 𝑒 800 0 E = −250𝑒 −2 3200 + 500 𝑒 −1.25 2000 − 250𝑒 800 −0.5 + 250 3200 − 500 2000 + 250 800 E = −.078125𝑒 −2 +0.25 𝑒 −1.25 − 0.3125𝑒 −0.5 + 0.078125−0.25+0.3125 E = .078125(1−𝑒 −2 )+0.25 𝑒 −1.25 −1 + 0.3125(1−𝑒 −0.5 E =0.078125 .86466 +0.25 −0.713495 +0.325 .39347 𝑚𝐽 E = 0.06755−0.178373+0.1279 =.017078𝑚𝑊=17.078𝜇𝐽

Steps to check E =250 −𝑒 −3200𝑡 3200 + 2 𝑒 −2000𝑡 2000 + −𝑒 800 −800𝑡 625𝜇 0 E =250 −𝑒 −3200∗625𝜇 3200 + 2 𝑒 −2000∗625𝜇 2000 − 𝑒 800 −800∗625𝜇 −250 −𝑒 0 3200 + 2 𝑒 0 2000 − 𝑒 800 0 E =250 −𝑒 −2 3200 + 2 𝑒 −1,25 2000 − 𝑒 800 −0.5 +250 1 3200 − 2 2000 + 1 800 E =250 −𝑒 −3200∗625𝜇 3200 + 2 𝑒 −2000∗625𝜇 2000 − 𝑒 800 −800∗625𝜇 + 250 3200 − 500 2000 + 250 800 E = −0.078125𝑒 −3200∗625𝜇 +0.25 𝑒 −2000∗625𝜇 −0.325 𝑒 −800∗625𝜇 + 0.078125−0.25+0.3125 0.078125−0.25+0.3125 E = −0.078125𝑒 −2 +0.25 𝑒 −1.25 −0.325 𝑒 −0.5 +.0140625 E = −0.01057+0.0716−0.197 +.0140625=