een 100 electrical circuits i - kau. een 100... · 2019. 6. 30. · an electric circuit is an...

EEN 100

Electrical Circuits I

Prof. Dr. Youssef A. Mobarak

1

Course Specifications

Course TitleEnglish

Code /No

ARABIC

code/no.

contact hours /week c.U.

Th. Pr. Tr. TCU

Electrical Circuits I EEN 100 100هن ك 2 2 2 3

Pre-requisites PHYS 202

Electric quantities and circuit elements; Kirchhoff’s laws; Mesh and node analyses;

Network theorem and transformations; Sinusoidal steady-state analysis using

phasors; Power Triangle and power factor correction.

Course Objectives:

By completion of the course, the students should be able to:

1. identify the fundamental electric quantities.2. analyse and evaluate responses of circuits containing

resistance, capacitance and inductance elements according tofundamental circuit laws.

3. identify the node voltages in resistive circuits containingcurrent sources and voltage sources using nodal analysis

4. analyse the mesh currents and branch currents in resistivecircuits containing voltage sources and current sources usingmesh analysis

5. apply theorems to simplify a resistive circuit.

Course Objectives:

6. apply the source transformation and Y-Δ transformation to

simplify circuits.

7. express an AC steady-state circuit to a phasor circuit

8. recognize the concepts of power factor, complex power,

and conservation of power.

9. work with a small team to carry out experiments in electric

circuits and prepare reports that present lab work.

Contents:

1- Circuit elements, ohm's law & Kirchhoff's laws.

2- Simple resistive circuits.

3- Circuit analysis

4- Network theorems

5- Inductors and capacitors

6- Analysis of sinusoidal steady-state circuits

7- Power triangle and power factor correction

Weighting of assessments

Quizzes 20 %

Lab. 20 %

Mid-term exam 20 %

Final exam 40 %

Total 100 %

Text book:

C. K. Alexander & M. N. Sadiqu, Fundamentals of

Electric Circuits, 4th ed., M.C.-Graw-Hill, 2009

Supplementary references

James Nilson & Susan Riedel, Electric Circuits,

8th ed., Pearson Prentice Hall, 2008.

12 June 2019 7

Student Learning Outcomes

(a): An ability to apply knowledge of mathematics,

science, and engineering.

(b): An ability to design and conduct experiments,

as well as to analyze and interpret data.

(e): An ability to identify, formulate, and solve

engineering problems

12 June 2019 8

Ch 1: 1Basic Concepts

Spring 2018

EEN 100Electrical Circuits I

Chapter 1

Basic Concepts


Spring 2018

An electric circuit is an interconnectionof electrical elements.

What are the electrical elements?

Introduction to electric circuits


Spring 2018

Charge and current

• the most basic quantity in an electric circuit is the electric charge.

• Charge is an electrical property of the atomic particles ofwhich matter consists, measured in coulombs (C).

• Positive charge is called proton, negative charge is called electron.

• The charge of an electron (e) = -1.602 ×10-19 C.

E#2 How many electrons in one coulomb of charge?

E#1 How much charge represented by 4600 electrons?

4600 × -1.602 ×10-19 = -7.369×10-16 C

1/1.602 ×10-19 = 6.24×1018 electrons

Charge


Spring 2018

Current

• When a conducting wire (consisting of several atoms) isconnected to a battery (a source of electromotive force), thecharges are compelled to move; positive charges move in onedirection while negative charges move in the oppositedirection.

• This motion of charges creates electric current. • It is conventional to take the current flow as the movement of

positive charges. That is, opposite to the flow of negativecharges

• Electric current is the time rate of change of charge,measured in amperes (A).


Spring 2018

DC currentCurrent remains constant with time

Types of current

Also, charge can be obtained as:

AC currentCurrent varies sinusoidally with time


Spring 2018

E#3 10 C of charges flow past a point in a wire in 2 s. What is thecurrent in A?

𝑖 =∆𝑄

∆𝑡=10

2= 5 𝐴

E#4 The total charge entering a terminal is given by𝑞 = 20𝑡 sin 𝜋𝑡 mC. Calculate the current at t= 0.5 s.

𝑖 =𝑑𝑞

𝑑𝑡= 20 sin 𝜋𝑡 +20𝜋𝑡 cos 𝜋𝑡 𝑚𝐴

at t= 0.5 s

𝑖 = 20 sinπ

2+ 20

π

2cos

𝜋

2= 20 𝑚𝐴


Spring 2018

E#5 The charge entering a certain element is shown in thefollowing figure. Calculate the current at a) t= 1 ms, b) 6 ms, c)10 ms.

𝑞 = ቐ40𝑡 𝑚𝐶 0 < 𝑡 < 280 𝑚𝐶 2 < 𝑡 < 8

−20𝑡 + 240 𝑚𝐶 8 < 𝑡 < 12

a) At t=1 ms, 𝑖 =𝑑𝑞

𝑑𝑡=

40 𝑚𝐴b) At t=6ms, 𝑖 = 0c) At t=10 ms, 𝑖 = −20 𝑚𝐴


Spring 2018

E#6 Determine the total charge entering a terminal between t=1s and t=2 s if the current is given by 𝑖 = 3𝑡2 − 𝑡 A.

E#7 Determine the total charge entering an element within thefirst 5s, if the current through the element is given by 𝑖

= ቊ5 0 < 𝑡 < 2𝑡2 + 1 𝑡 > 2

Answer 5.5 C

Answer 52 C


Spring 2018

E#8 The current flowing past a point in a device is shown in Thefollowing figure. Calculate the total charge through the point.


Spring 2018

Voltage (or potential difference) between point a, point b is theenergy (or work done) required to move a unit charge from a tob in an electric circuit, measured in volts (V).

𝑣𝑎𝑏 =𝑑𝑊

𝑑𝑞

Using polarity, there are two equivalent representations:(a) point a is 9 V higher than point b(b) point b is -9 V above point a, i.e. point b is 9 V lower than point a, i.e. point a is 9 V higher than point b

𝑣𝑎𝑏 = 𝑣𝑎 − 𝑣𝑏

Voltage


Spring 2018

Power is the time rate of expending or absorbing energy, measured in watts (W).

𝑝 =𝑑𝑊

𝑑𝑡

In terms of voltage and current:

𝑝 =𝑑𝑊

𝑑𝑡=𝑑𝑊

𝑑𝑞∙𝑑𝑞

𝑑𝑡= 𝑣 𝑖

Absorbing power Supplying power

Power can be absorbed or supplied (note sign):

Power and Energy


Spring 2018

E#9 Find the power for the following elements. Also, determineif the power absorbed or supplied.

a) b) c) d)

Answer (a) 12 W, (b) 12 W, (c) -12 W, (d) -12 W


Spring 2018

E#10 Find the power for the following elements. Also,determine if the power absorbed or supplied.

Answer (a) 40 W, (b) -24 W, (c) 15 W

a) b) c)


Spring 2018

E#11 Find the power delivered to an element at t=3 ms if thecurrent entering its positive terminal is 𝑖 = 40 cos 𝜋𝑡, and 𝑣 = 3𝑖


Spring 2018

Energy

Energy is the capacity to do work, measured in joules (J).

From definition for power:

𝑝 =𝑑𝑊

𝑑𝑡𝑑𝑊 = 𝑝 𝑑𝑡 𝑊 = න

𝑡𝑜

𝑡

𝑝 𝑑𝑡 = න

𝑡𝑜

𝑡

𝑣 𝑖 𝑑𝑡

Electric utility companies use watt-hours (Wh) to calculate energy.

E#12 How much energy does a 100-W electric bulb consume intwo hours.

𝑊 = 𝑝 𝑡 = 100 × 2 × 60 × 60 = 720000 𝐽 = 720 𝐾𝐽

𝑊 = 𝑝 𝑡 = 100 × 2 = 200 𝑊𝐻

Or


Spring 2018

E#13 A stove element draws 15 A when connected to a 240 V line. How longdoes it take to consume 60 kJ?

𝑡 =𝑊

𝑝=𝑊

𝑣 𝑖=

60000

240 × 15= 16.67 𝑠

E#14 The following figure shows a circuit with five elements. If P1=-205 W,P2=60 W, P4=45 W, P5=30 W, calculate the power received or delivered byelement 3.

𝑃 = 0

𝑃1 + 𝑃2 + 𝑃3 + 𝑃4 + 𝑃5 = 0

𝑃3 = 205 − 60 − 45 − 30 = 70𝑊 (𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑)

Law of conservation of energy: “The algebraic sum of power in a circuit at any timemust be zero”. So, total power supplied (delivered) = total power absorbed (received)


Spring 2018

E#15 How much energy does a 60 W bulb consume in 8 hoursfor a month (30 days)? Determine the cost for the consumption if1 kWh = 21 cents.


Spring 2018

Circuit elements

We have five ideal circuit elements (components): Voltage source, Currentsource, Resistor, Inductor, Capacitor

There are two types of elements:(1) Active elements which supply energy (voltage and current sources).(2) Passive elements which absorb energy (resistor, capacitor, inductor).


Spring 2018

Electrical sources

Electrical source: is a devise that is able to convert non-electric energy into electric energy and vise versa.

Voltage and current sources can be divided into two types:

1) Independent sourceProvides a specified amount (of voltage or current) independentfrom other circuit elements.

Independent current source (arrow

indicates direction of current)

Time-varying independent

voltage source

Constant-voltage independent

voltage source


Spring 2018

2) Dependent sourceSource quantity is controlled by another (voltage or current) sourceThere are four types of dependent sources:1. Voltage-Controlled Voltage Source (VCVS) 2. Current-Controlled Voltage Source (CCVS) 3. Voltage-Controlled Current Source (VCCS) 4. Current-Controlled Current Source (CCCS)

dependent current source

dependent voltage source


Spring 2018

E#16 Calculate the power supplied or absorbed by each elementin the following circuit.

Answer -100 W, 60 W, 48 W, -8 W


Spring 2018

E#17 Calculate the power supplied or absorbed by each elementin the following circuit.

Answer -40 W, 16 W, 9 W, 15 W


Spring 2018

E#18 Find the current I in the following network.

Answer 3 A

Ch 02:

Basic Laws

Spring 2018


Chapter 2

Basic Laws

1

Ch 02:

Basic Laws

Spring 2018

Ohm’s law states that the voltage v across a resistor is directlyproportional to the current i flowing through the resistor.

Ohm’s law

𝑣 𝛼 𝑖

𝑣 = 𝑖 𝑅

The resistance R of an element denotes its ability to resist theflow of electric current; it is measured in ohms (Ω ).

𝑅 =𝑣

𝑖

Resistance can be calculated fromphysical properties of a material as:

𝑅 =𝜌 𝑙

𝐴

2

Ch 02:

Basic Laws

Spring 2018

Resistivity of common materials

• Good conductors such as copper and aluminum have lowresistivity while insulators such as mica and glass have highresistivity.

3

Ch 02:

Basic Laws

Spring 2018

Types of resistors

1) Fixed resistors

2) Variable resistors

Representation

Representation

Actual

Actual

4

Ch 02:

Basic Laws

Spring 2018

Conductance

𝐺 =1

𝑅=𝑖

𝑣

Conductance is the ability of an element to conduct electriccurrent; it is measured in mhos (Ʊ) or Siemens (S).Mathematically, conductance is the reciprocal of resistance.

Extreme cases of resistance

Short circuit Open circuit

5

Ch 02:

Basic Laws

Spring 2018

Calculation of power absorbed by resistor

𝑝 = 𝑣 𝑖

𝑣 = 𝑖 𝑅

𝑝 = 𝑖2 𝑅 =𝑣2

𝑅

Since

Therefore

Or in terms of conductance G 𝑝 =𝑖2

𝐺= 𝑣2 𝐺

E#1 In the following circuit, calculate the current i, conductanceG, and the power p.

𝑖 =𝑣

𝑅=

30

5000= 6𝑚𝐴

𝐺 =1

𝑅=

1

5000= 0.2 𝑚𝑆

𝑝 = 𝑣 𝑖 = 30 × 6 × 10−3 = 180 𝑚𝑊

and

6

Ch 02:

Basic Laws

Spring 2018

E#2 In the following circuit, calculate the voltage v, and thepower p.

E#3 A voltage source of 20 sin 𝜋𝑡 V is connected across a 5 kΩresistor. Find the current through the resistor and the power dissipated.

𝑖 =𝑣

𝑅=20 sin𝜋𝑡

5000= 4 sin 𝜋𝑡 𝑚𝐴

𝑝 = 𝑣 𝑖 = 20 sin𝜋𝑡 × 0.004 sin𝜋𝑡 = 80 𝑠𝑖𝑛2 𝜋𝑡𝑚𝑊

𝑣 = 𝑖 𝑅 = 2 × 10−3 × 10 × 103 = 20 𝑉

𝑝 = 𝑣 𝑖 = 20 × 2 × 10−3 = 40 𝑚𝑊

7

Ch 02:

Basic Laws

Spring 2018

E#5 Calculate the current i when: a) the switch in position 1, andb) the switch in position 2.

E#4 A bar of silicon is 4 cm long with a circular cross section. Ifthe resistance of the bar is 240 Ω at room temperature, what isthe cross-sectional radius of the bar.

𝑎) 𝑖 =𝑣

𝑅=

15

100= 0.15 𝐴 = 150 𝑚𝐴

𝑏) 𝑖 =𝑣

𝑅=

15

150= 0.1 𝐴 = 100 𝑚𝐴

𝑠𝑖𝑛𝑐𝑒 𝑅 =𝜌 𝑙

𝐴∴ 𝐴 =

𝜌 𝑙

𝑅=6.4 × 102 × 4 × 10−2

240= 0.1067 𝑚2

𝑠𝑖𝑛𝑐𝑒 𝐴 = 𝜋 𝑟2 ∴ 𝑟 =𝐴

𝜋=

0.1067

𝜋= 0.1843 𝑚 = 18.43 𝑐𝑚

8

Ch 02:

Basic Laws

Spring 2018

Nodes, Branches, and Loops

▪ A branch represents a single element such as a voltage sourceor a resistor.

▪ A node is the point of connection between two or morebranches.

▪ A loop is any closed path in a circuit. Formed by starting at anode, passing through a set of nodes and returning to thestarting node (without passing through any node more thanonce). There are two types of loops (independent, dependent)

A network with b branches, n nodes, and l loops

9

Ch 02:

Basic Laws

Spring 2018

This circuit has:▪ 5 branches (10 V source, 5 Ω resistor, 2 Ω resistor, 3 Ω resistor, 2 A

source)▪ 3 nodes (a, b, c)▪ 6 loops

E#6 Determine the number of nodes, branches, and loops in thefollowing circuit.

10

Ch 02:

Basic Laws

Spring 2018

E#7 Determine the number of nodes, branches, and loops in the following circuit.

𝑛 = 3, 𝑏 = 4, 𝑙 = 3

E#8 Find the number of nodes, branches, and loops in each of the followingcircuits.

𝑛 = 5, 𝑏 = 6, 𝑙 = 3 𝑛 = 5, 𝑏 = 7, 𝑙 = 6

11

Ch 02:

Basic Laws

Spring 2018

Kirchhoff’s Laws

Ohm’s law by itself is not sufficient to analyze circuits.However, when it is coupled with Kirchhoff’s two laws, we havea sufficient, powerful set of tools for analyzing a large variety ofelectric circuits. Kirchhoff’s laws were first introduced in 1847 bythe German physicist Gustav Robert Kirchhoff (1824–1887).These laws are formally known as Kirchhoff’s current law(KCL) and Kirchhoff’s voltage law (KVL).

Kirchhoff’s current law (KCL)

“The algebraic sum of all currents at any node is zero”

𝑛=1

𝑁

𝑖𝑛 = 0

▪ For simplicity: Current entering a node is positive, and leaving a node is negative.

▪ Or total currents entering = total currents leaving

12

Ch 02:

Basic Laws

Spring 2018

𝑠𝑖𝑛𝑐𝑒

𝑛=1

𝑁

𝑖𝑛 = 0 ∴ 𝑖1 − 𝑖2 + 𝑖3 + 𝑖4 − 𝑖5 = 0

𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑖1 + 𝑖3 + 𝑖4 = 𝑖2 + 𝑖5

13

Ch 02:

Basic Laws

Spring 2018

Kirchhoff’s voltage law (KVL)

“The algebraic sum of all voltages around a loop is zero”

𝑚=1

𝑀

𝑣𝑚 = 0

▪ For simplicity: Start at any branch (element) and go clockwise.▪ In drop case (+ -), vm is positive, while in rise case (- +), vm is

negative.▪ Or sum of voltage drops = sum of voltage rises.

−𝑣1 + 𝑣2 + 𝑣3 − 𝑣4 + 𝑣5 = 0

𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑣2 + 𝑣3 + 𝑣5 = 𝑣1 + 𝑣4

14

Ch 02:

Basic Laws

Spring 2018

E#9 For the following circuit, use KCL to find the branchcurrents I1 to I4.

Answer 12 A, -10 A, 5 A, -2 A

15

Ch 02:

Basic Laws

Spring 2018

E#10 For the following circuit, find the current io and voltage vo.

Applying KCL at point m

0.5 𝑖𝑜 + 3 − 𝑖𝑜 = 0

𝑖𝑜 = 6 𝐴

𝑣𝑜 = 6 × 4 = 24 𝑉

By ohm’s law

E#11 For the following circuit, find the current io and voltage vo.

Applying KCL at point m

6 = 𝑖𝑜 +𝑖𝑜4+ 𝑖 → (1)

𝑣𝑜 = 𝑖 × 8 = 𝑖𝑜 × 2 → (2)

From 1,2

𝑣𝑜 = 8 𝑉

𝑖𝑜 = 4 𝐴

i

m

m

16

Ch 02:

Basic Laws

Spring 2018

E#12 For the following circuit, use KVL to find the voltages V1 to V4.

Answer -8 V, 6 V, -11 V, 7 V

17

Ch 02:

Basic Laws

Spring 2018

E#13 For the following circuit, obtain the voltage V1.

Answer 2 V

18

Ch 02:

Basic Laws

Spring 2018

E#14 For the following circuit, find voltages v1 and v2.

E#15 For the following circuit, find voltages vx and vo.

i

𝑣1 = 2𝑖 → 1 𝑣2 = −3𝑖 → (2)

By ohm’s law

Applying KVL

−20 + 𝑣1 − 𝑣2 = 0 → (3)

From 1,2,3

𝑖 = 4 𝐴, 𝑣1 = 8 𝑉 𝑣2 = −12 𝑉

i

𝑣𝑥 = 10𝑖 → 1 𝑣𝑜 = −5𝑖 → (2)

By ohm’s law

Applying KVL

−35 + 𝑣𝑥 + 2𝑣𝑥 − 𝑣𝑜 = 0 → (3)

From 1,2,3

𝑖 = 1 𝐴, 𝑣𝑥 = 10 𝑉 𝑣2 = −5 𝑉

19

Ch 02:

Basic Laws

Spring 2018

E#16 Find the currents and voltages as shown in the followingcircuit.

Answer 1.5 A, 0.25 A, 1.25 A, 3 V, 2 V, 5 V

20

Ch 02:

Basic Laws

Spring 2018

Series connection

Resistive circuit

• When a circuit contains resistors, it is called “Resistive Circuit”• It is possible to convert any complicated Resistive Circuit into

a simpler equivalent circuit

𝑣 = 𝑣1 + 𝑣2

𝑣1 = 𝑖 𝑅1, 𝑣2 = 𝑖 𝑅2

𝑣 = 𝑖 (𝑅1 + 𝑅2)

∴ 𝑣 = 𝑖 𝑅𝑒𝑞

21

Ch 02:

Basic Laws

Spring 2018

Voltage division

𝑣1 =𝑅1

𝑅1 + 𝑅2𝑣

𝑣2 =𝑅2

𝑅1 + 𝑅2𝑣

𝑅𝑒𝑞 = 𝑅1 + 𝑅2

For N resistors in series

𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 +⋯+ 𝑅𝑁 =

𝑛=1

𝑁

𝑅𝑛

𝑣1𝑣=

𝑖 𝑅1𝑖 (𝑅1 + 𝑅2)

𝑣2𝑣=

𝑖 𝑅2𝑖 (𝑅1 + 𝑅2)

22

Ch 02:

Basic Laws

Spring 2018

Parallel connection

𝑣 = 𝑖1 𝑅1 = 𝑖2 𝑅2

𝑖 = 𝑖1 + 𝑖2 =𝑣

𝑅1+

𝑣

𝑅2= 𝑣

1

𝑅1+

1

𝑅2=

𝑣

𝑅𝑒𝑞

∴1

𝑅𝑒𝑞=

1

𝑅1+

1

𝑅2

∴ 𝑅𝑒𝑞 =𝑅1𝑅2

𝑅1 + 𝑅2

23

Ch 02:

Basic Laws

Spring 2018

Current division

𝑖1 =𝑅𝑒𝑞

𝑅1𝑖 =

𝑅2𝑅1 + 𝑅2

𝑖

𝑖2 =𝑅𝑒𝑞

𝑅2𝑖 =

𝑅1𝑅1 + 𝑅2

𝑖

𝑖1𝑖=

𝑣/𝑅1𝑣/𝑅𝑒𝑞

𝑖2𝑖=

𝑣/𝑅2𝑣/𝑅𝑒𝑞

For N resistors in parallel

1

𝑅𝑒𝑞=

1

𝑅1+

1

𝑅2+

1

𝑅3+⋯+

1

𝑅𝑁

Note that: the equivalent resistance is always smaller than the resistance ofthe smallest resistor in the parallel combination.

The equivalent conductance of resistors connected in parallel is the sum oftheir individual conductances.

𝐺𝑒𝑞 = 𝐺1 + 𝐺2 + 𝐺3 +⋯+ 𝐺𝑁

24

Ch 02:

Basic Laws

Spring 2018

E#17 Find the equivalent circuit of the following circuit.

Answer 14.4 Ω

25

Ch 02:

Basic Laws

Spring 2018

E#18 Find the equivalent circuit of the following circuit.

Answer 11.2 Ω

26

Ch 02:

Basic Laws

Spring 2018

E#19 For the following circuit, find v1, v2, i1, i2, and the powerdissipated in 40 Ω resistor.

𝑅𝑒𝑞 = 12ԡ6 + 10ԡ40 = 4 + 8 = 12 Ω

𝑖 =𝑣

𝑅𝑒𝑞=15

12= 1.25 𝐴

𝑖1 =6

18× 1.25 = 416.67 𝑚𝐴

𝑖2 =10

50× 1.25 = 250 𝑚𝐴

𝑣1 = 416.67 × 10−3 × 12 = 5 𝑉

𝑣2 = 250 × 10−3 × 40 = 10 𝑉

𝑃40 = 10 × 250 × 10−3 = 2.5 𝑊

27

Ch 02:

Basic Laws

Spring 2018

Wye-delta transformations

Δ-Y conversion

𝑅1 =𝑅𝑏𝑅𝑐

𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐

𝑅2 =𝑅𝑎𝑅𝑐


𝑅3 =𝑅𝑎𝑅𝑏


Y- Δ conversion

𝑅𝑎 =𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅3𝑅1

𝑅1

𝑅𝑏 =𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅3𝑅1

𝑅2

𝑅𝑐 =𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅3𝑅1

𝑅3

28

Ch 02:

Basic Laws

Spring 2018

E#20 For the following circuit, find the current i.

Answer 12.46 A

29

Ch 02:

Basic Laws

Spring 2018

E#21 For the following circuit, find the voltage V.

Answer 42.18 V

30

Ch 3: 1Methods Of Analysis

Fall 2018


Chapter 3

Methods Of Analysis


Fall 2018

Nodal analysis

Nodal analysis provides a general procedure for analyzing circuits usingnode voltages as the circuit variables. Choosing node voltages instead ofelement voltages as circuit variables is convenient and reduces the number ofequations one must solve simultaneously.

Steps to determine node voltages

1. Select a node as the reference node.2. Assign voltages v1, v2, …, vn-1 to the remaining nodes. The voltages are

referenced with respect to the reference node.3. Apply KCL to each of the n-1 non-reference nodes. Use Ohm’s law to

express the branch currents in terms of node voltages.4. Solve the resulting simultaneous equations to obtain the unknown node

voltages.


Fall 2018

Nodal analysis


1. Select a node as the reference node.2. Assign voltages v1, v2, …, vn-1 to the remaining nodes.

The voltages are referenced with respect to the reference

node.

3. Apply KCL to each of the n-1 non-reference nodes.Use Ohm’s law to express the branch currents interms of node voltages.


Fall 2018

Nodal analysis


4. Solve the resulting simultaneous equations toobtain the unknown node voltages.

Current flows from a higher potential to a lower

potential in a resistor.


Fall 2018

E#1 For the following circuit, obtain node voltages.

In this circuit, there are 3 nodes. One is taken as a reference and the nodes 1 and 2 are node voltages. Apply KCL at node voltages 1, and 2.

1 = 𝑖1 + 𝑖2

i2 i3i1

At node 1

𝑖1 = 𝑖3 + 4At node 2

Use ohm’s law to determine the branch currents in terms of node voltages

𝑖1 =𝑣1 − 𝑣2

6


Fall 2018

𝑖2 =𝑣1 − 0

2

𝑖3 =𝑣2 − 0

7

Substituting at node voltages equations

1 =𝑣1 − 𝑣2

6+𝑣12

6 = 4𝑣1 − 𝑣2 → (1)

𝑣1 − 𝑣26

=𝑣27+ 4 7𝑣1 − 13𝑣2 = 168 → (2)

From 1,2

𝑣1 = −2 𝑉

𝑣2 = −14 𝑉


Fall 2018

E#2 For the following circuit, obtain the node voltages.

Apply KCL at node voltages 1, 2, and 3

3 = 𝑖1 + 𝑖𝑥

𝑖𝑥 = 𝑖2 + 𝑖3

𝑖1 + 𝑖2 = 2𝑖𝑥


Fall 2018


𝑖1 =𝑣1 − 𝑣3

4𝑖𝑥 =

𝑣1 − 𝑣22

𝑖2 =𝑣2 − 𝑣3

8𝑖3 =

𝑣24

Substituting at node voltages equations

3 =𝑣1 − 𝑣3

4+𝑣1 − 𝑣2

23𝑣1 − 2𝑣2 − 𝑣3 = 12 → (1)

𝑣1 − 𝑣22

=𝑣2 − 𝑣3

8+𝑣24

−4𝑣1 + 7𝑣2 − 𝑣3 = 0 → (2)

𝑣1 − 𝑣34

+𝑣2 − 𝑣3

8= 𝑣1 − 𝑣2 2𝑣1 − 3𝑣2 + 𝑣3 = 0 → (3)

Use Cramer’s rule


Fall 2018


Fall 2018

Therefore


Fall 2018

Nodal analysis with voltage sources

CASE 1 If a voltage source is connected between the reference node and a non-reference node, we simply set the voltage at the non-reference node equal to thevoltage of the voltage source. For example, as shown in the following circuit, v1= 10 V.

CASE 2 If the voltage source (dependent or independent) is connected between twonon-reference nodes, the two non-reference nodes form a generalized node or super-node and any element connected in parallel with it.


Fall 2018

We analyze a circuit with super-nodes using the same steps mentioned in the previoussection except that the super--nodes are treated differently.

Apply KCL at super-node

𝑖1 + 𝑖4 = 𝑖2 + 𝑖3


𝑖1 =𝑣1 − 𝑣2

2𝑖2 =

𝑣28

𝑖3 =𝑣36

𝑖4 =𝑣1 − 𝑣3

4

𝑣1 − 𝑣22

+𝑣1 − 𝑣3

4=𝑣28+𝑣36


36𝑣1 − 30𝑣2 − 20𝑣3 = 0

Since v1=10 V

3𝑣2 + 2𝑣3 = 36 → (1)


Fall 2018

Apply KVL at super-node

−𝑣2 + 5 + 𝑣3 = 0 → (2)

From 1,2

𝑣2 = 9.2 𝑉

𝑣3 = 4.2 𝑉


Fall 2018

E#3 For the following circuit, obtain the node voltages.

In this circuit, there is a super-node. Apply KCL at super-node

2 = 𝑖1 + 𝑖2 + 7

2 =𝑣12+𝑣24+ 7

8 = 2𝑣1 + 𝑣2 + 28

∴ 2𝑣1 + 𝑣2 + 20 = 0 → (1)


Fall 2018

Apply KVL at super-node

−𝑣1 − 2 + 𝑣2 = 0 → (2)

From 1,2

𝑣1 = −7.33 𝑉

𝑣2 = −5.33 𝑉


Fall 2018

E#4 For the following circuit, using nodal analysis, obtain thevoltage v and the current i.


Fall 2018

Answer -0.6 V, 4.2 A


Fall 2018

Mesh analysis (Loop analysis)

▪ Mesh analysis provides another general procedure for analyzingcircuits using mesh currents (loop currents) as the circuit variables.

▪ Nodal analysis applies KCL to find unknown voltages in a givencircuit, while mesh analysis applies KVL to find unknown currents.

▪ A mesh is a loop which does not contain any other loops within it.

Steps to determine the mesh currents:▪ Assign mesh currents i1, i2, …, in to the n meshes.▪ Apply KVL to each of the n meshes. Use Ohm’s law to express the

voltages in terms of the mesh currents.▪ Solve the resulting n simultaneous equations to get the mesh

currents.


Fall 2018

E#5 For the following circuit, find the branch currents I1, I2, andI3 using Mesh analysis.

Apply KVL

−15 + 5 𝑖1 + 10 𝑖1− 𝑖2 + 10 = 0

−10 + 10 𝑖2− 𝑖1 + 6 𝑖2 + 4 𝑖2 = 0

For mesh 1

For mesh 2

∴ −1 + 3 𝑖1 − 2 𝑖2 = 0 → (1)

∴ −1 − 𝑖1+2 𝑖2 = 0 → (2)

From 1,2

∴ 𝑖1= 1 𝐴, 𝑖2 = 1 𝐴

Therefore the branch currents

∴ 𝐼1= 1 𝐴, 𝐼2 = 1 𝐴, 𝐼3 = 0


Fall 2018

E#6 For the following circuit, find the branch currents Io usingMesh analysis.

−24 + 10 𝑖1− 𝑖2 + 12 𝑖1− 𝑖3 = 0

24 𝑖2 + 4 𝑖2− 𝑖3 + 10 𝑖2− 𝑖1 = 0

For mesh 1

For mesh 2

22 𝑖1 − 10 𝑖2 − 12 𝑖3 = 24 → (1)

−10 𝑖1 + 38 𝑖2 − 4 𝑖3 = 0 → (2)


Fall 2018

4 𝑖3− 𝑖2 + 4𝐼𝑜 + 12 𝑖3− 𝑖1 = 0

For mesh 3

−8 𝑖1 − 8 𝑖2 + 16 𝑖3 = 0 → (3)

Since𝐼𝑜 = 𝑖1− 𝑖2

4 𝑖3− 𝑖2 + 4 𝑖1− 𝑖2 + 12 𝑖3− 𝑖1 = 0

From 1,2,3 using Cramer's rule, the mesh currents are

𝑖1= 2.25 𝐴, 𝑖2 = 0.75 𝐴, 𝑖3 = 1.5 𝐴

Therefore

𝐼𝑜 = 𝑖1− 𝑖2= 2.25 − 0.75 = 1.5 𝐴


Fall 2018

Mesh analysis with current source

CASE 1 When a current source exists only in one mesh, as shown in thefollowing circuit, the mesh current i2= -5 A.

Then, for mesh 1

−10 + 4 𝑖1 + 6 𝑖1− 𝑖2 = 0

−10 + 4 𝑖1 + 6 𝑖1+5 = 0

𝑖1= −2 𝐴

𝑖2= −5 𝐴Since


Fall 2018

CASE 2 When a current source exists between two meshes, as shown in the followingcircuit, a super-mesh is created by excluding the current source and any elementsconnected in series with it.

For super-mesh−20 + 6 𝑖1 + 10 𝑖2 + 4 𝑖2 = 0 → (1)

Apply KCL at a node in the branch where the two meshes intersect

𝑖2= 6+ 𝑖1 → (2)

𝑖1= −3.2 𝐴, 𝑖2 = 2.8 𝐴From 1,2


Fall 2018

E#7. For the following circuit, find the branch current Io usingMesh analysis

2𝑖1 + 6 + 4 𝑖1− 𝑖3 + 𝑖1− 𝑖2 = 0

5𝑖2 + 𝑖2− 𝑖1 + 4 𝑖3− 𝑖1 + 12 = 0

For mesh 1

For super mesh

7𝑖1 − 𝑖2 − 4𝑖3 + 6 = 0 → (1)

−5𝑖1 + 6𝑖2 + 4𝑖3 + 12 = 0 → (2)

i1

i3

i2


Fall 2018

Apply KCL at a node in the branch where the mesh 2 and mesh 3 intersect

𝑖3= 3 + 𝑖2 → (3)

𝑖1 = −1.333 𝐴, 𝑖2 = −3.067 𝐴, 𝑖3 = −0.067 𝐴

From 1,2,3 using Cramer’s rule

∴ 𝑖𝑜 = −1.734 𝐴

𝑖𝑜 = 𝑖1− 𝑖2Since


Fall 2018

E#8 For the following circuit, using Mesh analysis to determinethe currents i1, i2, and i3.


Fall 2018

Answer 3.474 A, 0.4737 A, 1.1052 A

Ch 4: 1Circuit Theorems

Spring 2016


Chapter 4

Circuit Theorems


Spring 2016

Superposition theorem

If a circuit has two or more independent sources, one way to determinethe value of a specific variable (voltage or current) is to use nodal or meshanalysis. Another way is to determine the contribution of each independentsource to the variable and then add them up. The latter approach is known asthe superposition.

Steps to apply superposition theorem

1. Turn off all independent sources except one source. Find the output(voltage or current) due to that active source using nodal or mesh analysis.2. Repeat step 1 for each of the other independent sources.3. Find the total contribution by adding algebraically all the contributions dueto the independent sources.


Spring 2016

Two things have to be keep in mind:

1. When we say turn off all other independent sources: Independentvoltage sources are replaced by 0 V (short circuit) and Independentcurrent sources are replaced by 0 A (open circuit).

2. Dependent sources are left intact because they are controlled bycircuit variables.


Spring 2016

E#1 For the following circuit, Use the superposition theorem to find v.

a) Turn off current source 3A

𝑣1 =4

4 + 8× 6 = 2𝑉

b) Turn off voltage source 6V

𝑖3 =8

4 + 8× 3 = 2𝐴

𝑣2 = 2 × 4 = 8𝑉

∴ 𝑣 = 𝑣1 + 𝑣2 = 2 + 8 = 10𝑉


Spring 2016

E#2 For the following circuit, determine the current io using superposition.

Note: In this circuit, there are three sources: two independent and onedependent. Left the dependent source and turn off the independent sources.


Spring 2016

𝑖1 = 4𝐴 → (1)For loop 1

3 𝑖2 − 4 + 2𝑖2 − 5𝑖𝑜′ + 𝑖2 − 𝑖3 = 0 → (2)

For loop 2

5 𝑖3 − 4 + 𝑖3 − 𝑖2 + 5𝑖𝑜′ + 4𝑖3 = 0 → (3)

For loop 3

At node 0 𝑖3 + 𝑖𝑜′ = 4 → (4)

From 2,3,4 𝑖𝑜′ =

52

17𝐴

a) Turn off voltage source 20V


Spring 2016

b) Turn off current source 6A

2𝑖4 − 5𝑖𝑜′′ + 𝑖4 − 𝑖5 + 3𝑖4 = 0 → (1)

For loop 4

5𝑖5 + 𝑖5 − 𝑖4 + 5𝑖𝑜′′ + 4𝑖5 − 20 = 0 → (2)

For loop 5

Since

𝑖𝑜′′ = −𝑖5 → (3)

From 1,2,3 𝑖𝑜′′ =

−60

17𝐴

∴ 𝑖𝑜 = 𝑖𝑜′ + 𝑖𝑜

′′ =52

17−60

17= −

8

17= −0.471𝐴


Spring 2016

E#3 Determine the current I using superposition.

Answer 0.75A


Spring 2016

Thevenin’s theorem

Thevenin’s theorem states that a linear two-terminal circuit can bereplaced by an equivalent circuit consisting of a voltage source VTh inseries with a resistor RTh.

Where VTh is the open-circuit voltage at the terminals and RTh is theinput or equivalent resistance at the terminals when the independentsources are turned off.


Spring 2016

E#4 Find the Thevenin equivalent circuit of the following circuit to the left ofthe terminals a-b. Then find the current through the load resistance when a)RL=6Ω, b) RL=16Ω, and c) RL=36Ω.

To get RTh, all independent sources are turned off.

∴ 𝑅𝑇ℎ = 1 + ԡ4 12 = 1 +4 × 12

16= 4Ω


Spring 2016

To get VTh

−32 + 4𝑖1 + 12(𝑖1 − 𝑖2) = 0, 𝑖2 = −2𝐴

Apply KVL on loop 1 and 2

Therefore𝑖1 = 0.5 𝐴

∴ 𝑉𝑇ℎ = 12 𝑖1 − 𝑖2 = 12 0.5 + 2 = 30𝑉


Spring 2016

a) When RL=6Ω

b) When RL=16Ω

c) When RL=36Ω

∴ 𝐼𝐿 =𝑉𝑇ℎ

𝑅𝐿 + 𝑅𝑇ℎ=30

10= 3𝐴

∴ 𝐼𝐿 =30

20= 1.5𝐴

∴ 𝐼𝐿 =30

40= 0.75𝐴

Thevenin’s equivalent circuit


Spring 2016

Norton’s theorem

Norton’s theorem states that a linear two-terminal circuit can be replaced byan equivalent circuit consisting of a current source IN in parallel with aresistor RN.

where IN is the short-circuit current throughthe terminals and RN is the input orequivalent resistance at the terminals whenthe independent sources are turned off.

Observe the close relationship between Norton’s and Thevenin’s theorems.

𝑅𝑁 = 𝑅𝑇ℎ 𝐼𝑁 =𝑉𝑇ℎ𝑅𝑇ℎ


Spring 2016

E#6 Find the Norton equivalent circuit at the terminals a-b.

𝑅𝑁 = 5ԡ(8 + 4 + 8) =5 × 20

25= 4Ω

To get RN, all independent sources are turned off.


Spring 2016

To get IN

𝑖1 = 2𝐴 → (1)

−12 + 4 𝑖2 − 𝑖1 + 8𝑖2 + 8𝑖2 = 0 → (2)

From 1,2 𝐼𝑁 = 𝑖𝑠𝑐 = 𝑖2 = 1𝐴

Norton’s equivalent circuit


Spring 2016

Maximum power transfer

Differentiate the power with respect to RL, the maximum power istransferred to the load when the load resistance equals the Theveninresistance (RL = RTh). Therefore:

𝑃𝑚𝑎𝑥 =𝑉𝑇ℎ2

4 𝑅𝑇ℎ

𝑃 = 𝑖2𝑅𝐿 =𝑉𝑇ℎ

𝑅𝑇ℎ + 𝑅𝐿

2

𝑅𝐿


Spring 2016

E#7 Determine the value of RL for maximum power transfer in the followingcircuit. Then find the maximum power.

To get RTh

𝑅𝑇ℎ = 5 + 6ԡ12 = 5 +6 × 12

18= 9Ω

maximum power is transferred when RL = RTh= 9Ω


Spring 2016

To get VTh

Using mesh analysis

For loop 1 −12 + 6𝑖1 + 12 𝑖1 − 𝑖2 = 0 → (1)

For loop 2 𝑖2 = −2𝐴 → (2)

From 1,2 𝑖1 =−2

3𝐴


Spring 2016

Since

−12 + 6𝑖1 + 3𝑖2 + 𝑉𝑇ℎ = 0

−12 + 6−2

3+ 3(−2) + 𝑉𝑇ℎ = 0

∴ 𝑉𝑇ℎ = 22𝑉

Therefore the maximum power

𝑃𝑚𝑎𝑥 =𝑉𝑇ℎ2

4 𝑅𝑇ℎ=(22)2

4 × 9= 13.44𝑊

Ch 5: 1Capacitors & inductors

Fall 2018


Chapter 5

Capacitors & inductors


Fall 2018

Capacitors & inductors are two other important passiveelements, besides resistors.

Capacitors & inductors do not dissipate energy like resistors,but instead they store energy.

Capacitor stores energy in its electric field. Inductor stores energy in its magnetic field. Capacitors & inductors are also known as “energy storage

elements”.

Applications

Capacitors are used in electronics, communications, computersand power systems.

Inductors are used in power supplies, transformers, radios, TVsand electric motors.

Introduction


Fall 2018

Capacitors

A capacitor consists of two conducting plates separated byinsulator (or dielectric).In many practical applications, the plates may be aluminum foil,while the dielectric may be air, ceramic, paper, or mica

A typical capacitor


Fall 2018

Capacitor with applied voltage v

When a voltage source v is connected to the

capacitor, the source deposits a positive charge

q on one plate and a negative charge -q on the

other. The capacitor is said to store the electric

charge. The amount of charge stored,

represented by q, is directly proportional to the

applied voltage so that:

𝑞𝑞 = 𝐶𝐶 𝑣𝑣q: the charge in coulombs (C) C: the capacitance in farads (F)v: the applied voltage in volts (V)


Fall 2018

Capacitance can be calculated in terms of physical dimensions of the capacitor as:

𝐶𝐶 =𝜖𝜖 𝐴𝐴𝑑𝑑

Where,

A: is the surface area of each plate,

d: is the distance between the plates, and

ε: is the permittivity of the dielectric material between the plates.

Typical commercial values of C are in the Pico farad (pF) to

micro farad (µF) range.


Fall 2018

(a) fixed capacitor, (b) variable capacitor.

Circuit symbols for capacitors

On charging: Current flows into positive terminal (vi > 0).While,On discharging: Current flows into negative terminal (vi < 0)


Fall 2018

Actual capacitors


Fall 2018

Current-voltage relationship of the capacitor

𝑞𝑞 = 𝐶𝐶 𝑣𝑣𝑖𝑖 =𝑑𝑑𝑞𝑞𝑑𝑑𝑑𝑑

𝑖𝑖 =𝑑𝑑𝑑𝑑𝑑𝑑

𝐶𝐶𝑣𝑣 = 𝐶𝐶𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑

𝑣𝑣 =1𝐶𝐶� 𝑖𝑖 𝑑𝑑𝑑𝑑 + 𝑣𝑣(0)

Therefore

Where v(0) is the voltage across the capacitor at time=0

Power

Energy

𝑃𝑃 = 𝐶𝐶𝑣𝑣𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑

𝑊𝑊 = �𝑝𝑝𝑑𝑑𝑑𝑑 = �𝐶𝐶𝑣𝑣𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = �𝐶𝐶𝑣𝑣 𝑑𝑑𝑣𝑣 =

12𝐶𝐶𝑣𝑣2


Fall 2018

E#1 What is the voltage across a 3µF capacitor if the charge on one plate is0.12 mC? How much energy is stored?

E#2 If a 10µF capacitor is connected to a voltage source with v(t) = 50 sin2000t V, determine the current through the capacitor.


Fall 2018

E#3 Find the energy stored in each capacitor in the following circuit under dcconditions.

𝑖𝑖 =39 × 6 = 2 𝑚𝑚𝐴𝐴 𝑣𝑣1 = 2 × 10

−3 × 2 × 103 = 4 𝑉𝑉

𝑣𝑣2 = 2 × 10−3 × 4 × 103 = 8 𝑉𝑉

𝑊𝑊1 =12 × 2 × 10

−3 × (4)2= 0.016 𝐽𝐽 = 16 𝑚𝑚𝐽𝐽

𝑊𝑊2 =12 × 4 × 10

−3 × (8)2= 0.128 𝐽𝐽 = 128 𝑚𝑚𝐽𝐽


Fall 2018

Answer 810 µJ, 135 µJ

E#4 Under dc conditions, determine the energy stored in each capacitor inthe following circuit.


Fall 2018

Capacitors in parallel

𝑖𝑖 = 𝑖𝑖1 + 𝑖𝑖2 + 𝑖𝑖3 + ⋯+ 𝑖𝑖𝑁𝑁

𝐶𝐶𝑒𝑒𝑒𝑒 = 𝐶𝐶1 + 𝐶𝐶2 + 𝐶𝐶3 + ⋯+ 𝐶𝐶𝑁𝑁

𝑖𝑖 = 𝐶𝐶1𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑

+ 𝐶𝐶2𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑

+ 𝐶𝐶3𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑

+ ⋯+ 𝐶𝐶𝑁𝑁𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑

= 𝐶𝐶𝑒𝑒𝑒𝑒𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑


Fall 2018

Capacitors in series

𝑣𝑣 = 𝑣𝑣1 + 𝑣𝑣2 + 𝑣𝑣3 + ⋯+ 𝑣𝑣𝑁𝑁

1𝐶𝐶𝑒𝑒𝑒𝑒

=1𝐶𝐶1

+1𝐶𝐶2

+1𝐶𝐶3

+ ⋯+1𝐶𝐶𝑁𝑁

𝑣𝑣 =1𝐶𝐶1� 𝑖𝑖 𝑑𝑑𝑑𝑑 +

1𝐶𝐶2� 𝑖𝑖 𝑑𝑑𝑑𝑑 +

1𝐶𝐶3� 𝑖𝑖 𝑑𝑑𝑑𝑑 + ⋯+

1𝐶𝐶𝑁𝑁

� 𝑖𝑖 𝑑𝑑𝑑𝑑 =1𝐶𝐶𝑒𝑒𝑒𝑒

� 𝑖𝑖 𝑑𝑑𝑑𝑑


Fall 2018

E#5 Find the equivalent capacitance seen between terminals a and b of thefollowing circuit.

Answer 20 µF


Fall 2018

E#6 Find the equivalent capacitance of the following circuit.

Answer 40 µF


Fall 2018

Inductors

A typical inductor

An inductor is designed to store energy in its magnetic field (whilecapacitor stores energy in its electric field).

Any conductor has inductive properties.

An inductor consists of a coil of conducting wire – (this form enhancesinductive effect).

Typical inductor structure is solenoid.

Core material can be iron, steel, plastic, air or ceramic.


Fall 2018

Actual inductors


Fall 2018

(a) Air-core inductor(b) Iron-core inductor(c) Variable Iron-core inductor

Representation of inductor


Fall 2018

If a current is allowed to pass through an inductor: The voltage acrossthe inductor is directly proportional to the time rate of change of current.

WhereL: the constant of proportionality called inductance, in henry (H)v: the voltage across the inductor in volts (V)Also, the current can be calculated as:

𝑣𝑣 = 𝐿𝐿𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑

𝑣𝑣 ∝𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑

Current-voltage relationship of the inductor

𝑖𝑖 =1𝐿𝐿�𝑣𝑣 𝑑𝑑𝑑𝑑 + 𝑖𝑖(0)


Fall 2018

Power

Energy

𝑃𝑃 = 𝑣𝑣𝑖𝑖 = 𝐿𝐿𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑

𝑖𝑖

𝑊𝑊 = �𝑝𝑝𝑑𝑑𝑑𝑑 = �𝐿𝐿𝑖𝑖𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = �𝐿𝐿𝑖𝑖 𝑑𝑑𝑖𝑖 =

12𝐿𝐿𝑖𝑖2

Inductance (L): the property whereby an inductor exhibits opposition to the changeof current flowing through it, measured in henry (H).Recall: 1 henry = 1 Volt-second/Ampere (H = V.s/A or Wb/A).Inductance depends on physical dimension and construction.

Inductance

For solenoid inductor𝐿𝐿 =

𝑁𝑁2 𝜇𝜇 𝐴𝐴𝑙𝑙

Where N: number of turns, µ: permeability of the core material, A : cross-sectional area, l : length. Typical values of L are ranging from a few micro henrys (µH) to tens of henrys.


Fall 2018

E#8 If the current through a 1 mH inductor is i(t) = 20 cos 100t mA. Find the terminalvoltage and the energy stored.

E#9 The terminal voltage of a 2 H inductor is v = 10(1- t) V. Find the current flowingthrough it at t = 4 s and the energy stored in it at t = 4 s. Assume i(0) = 2 A.


Fall 2018

E#10 Determine vc, iL, and the energy stored in the capacitor and inductorunder dc conditions.


Fall 2018

Inductors in series

𝐿𝐿𝑒𝑒𝑒𝑒 = 𝐿𝐿1 + 𝐿𝐿2 + 𝐿𝐿3 + ⋯+ 𝐿𝐿𝑁𝑁

𝑣𝑣 = 𝐿𝐿1𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑

+ 𝐿𝐿2𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑

+ 𝐿𝐿3𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑

+ ⋯+ 𝐿𝐿𝑁𝑁𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑

= 𝐿𝐿𝑒𝑒𝑒𝑒𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑

𝑣𝑣 = 𝑣𝑣1 + 𝑣𝑣2 + 𝑣𝑣3 + ⋯+ 𝑣𝑣𝑁𝑁


Fall 2018

Inductors in parallel

𝑖𝑖 = 𝑖𝑖1 + 𝑖𝑖2 + 𝑖𝑖3 + ⋯+ 𝑖𝑖𝑁𝑁

𝑖𝑖 =1𝐿𝐿1�𝑣𝑣 𝑑𝑑𝑑𝑑 +

1𝐿𝐿2�𝑣𝑣 𝑑𝑑𝑑𝑑 +

1𝐿𝐿3�𝑣𝑣 𝑑𝑑𝑑𝑑 + ⋯+

1𝐿𝐿𝑁𝑁

�𝑣𝑣 𝑑𝑑𝑑𝑑 =1𝐿𝐿𝑒𝑒𝑒𝑒

�𝑣𝑣 𝑑𝑑𝑑𝑑

1𝐿𝐿𝑒𝑒𝑒𝑒

=1𝐿𝐿1

+1𝐿𝐿2

+1𝐿𝐿3

+ ⋯+1𝐿𝐿𝑁𝑁


Fall 2018

E#11 Find the equivalent inductance for the inductive ladder network.


Fall 2018

E#12 Find the equivalent inductance for the inductive ladder network.


Fall 2018

Ch 6: 1First Order Circuits

Spring 2018


Chapter 6

First Order Circuits


Spring 2018

First Order Circuits

Two types:1) Circuits with resistors and capacitors (RC circuits)2) Circuits with resistors and inductors (RL circuits)

Why are they called “first order circuits”?• Because when we analyze the circuits using KVL and KCL

first order differential equations are yielded

Circuits with both inductors and capacitors (RLCcircuits) are called “second order circuits”


Spring 2018

First Order Circuits Two ways to excite first order circuits:

1) Source-free circuit Energy is initially stored in capacitor or inductor Stored energy causes current-flow in the circuit

2) Independent sources DC sources AC sources

Charging Discharging

Ex:


Spring 2018

Source-Free RC CircuitSource-free RC circuit occurs when DC source is suddenly

disconnected energy stored in capacitor is released to the resistor

Consider the source-free RC circuit shown: Since capacitor is initially charged at t = 0, v(0) = V0 , and energy wC (0) = CV02/2

Applying KCL:

To solve this eq:

RCv

dtdv -=

AtRC

v ln+1

-=ln

0=+

0=+

Rv

dtdv

C

ii Rc

∫∫ =

=

dtRCv

dvRCdt

vdv

1-

-⇒

RCt

RCt

eVtv

VvA

eAv

tRCA

v

-

0

0

-

=)(

=)0(=

=

1-=lnfirst order

differential

equation

butIntegration

constant


Spring 2018

Source-Free RC Circuit

Voltage response of RC circuit is exponential decay of initial voltage• The response is called natural response because it isn’t due to

external sourceRapidity with which the voltage decreases is expressed by the

time constant, τ = RC

RCt

eVtv-

0=)(


Spring 2018


Circuit response as a function of τ:

• The less (smaller) time constant the more rapid voltage decrease faster response

See table (7.1) – page (256) in the text book


Spring 2018


Power dissipated in resistor:p = v2/R

Energy absorbed by resistor

Notice that at t →∞, wR(∞) = CV02/2 = wC(0)

This means: The energy stored initially in the capacitor is eventually dissipated in the resistor

τt

eR

V22

0 −=

[ ]

)-1()(

pdt )(

/2-20

0/2-

2

0

0

/220

0

21

2

τ

ττ τ

tR

ttt

tt

R

etw

edtetw

CV

RV

RV

=

−=== ∫∫ −

wRwC

wR(∞) = wC(0)


Spring 2018


Example: (pp (7.1) – p(257))

Let vc(0) = 45 V. Determine vc, vx and io for t ≥ 0.


Spring 2018


Example: (Ex (7.2) – p(258))

The switch in the circuit has been closed for a long time, and it is opened at t = 0. Find v(t) for t ≥ 0. Calculate the initial energy stored in the capacitor.


Spring 2018

Source-Free RL CircuitSource-free RL circuit occurs when DC source is suddenly

disconnected energy stored in inductor is released to the resistor

Consider the source-free RL circuit shown: Since inductor is initially charged at t = 0, i(0) = I0 , and energy wL (0) = LI02/2

Applying KVL:

To solve this eq:

0ln+-=ln ItLR

i

0=+

0=+

Ridtdi

L

vv RL

dtLR

iditi

I∫∫t

0

)(

-0

=

)0(==)(

=

-=ln

0

-0

-

0

0

iIeIti

eIi

tLR

Ii

LRt

LRt

first order

differential

equation

Integration

constant

iLR

dtdi

-=

integrate

where


Spring 2018

Source-Free RL Circuit

Current response of RL circuit is exponential decay of initial current• The response is called natural response because it isn’t due to

external source

Rapidity with which the current decreases is expressed by the time constant, τ = L/R

LRt

eIti-

0=)(


Spring 2018

Source-Free RL CircuitPower dissipated in resistor:

p = i2R

Energy absorbed by resistor

Notice that at t →∞, wR(∞) = LI02/2 = wL(0)

This means: The energy stored initially in the inductor is eventually dissipated in the resistor

[ ]

)-1()(

-)(

/2-20

0/2-

20

0

/2-20

0

21

2∫∫τ

ττ τ

tR

ttt

tt

R

eLItw

eRIdteIRpdttw R

=

===

wRwC

wR(∞) = wL(0)

τ-

0 .==t

R eRIiRv

τ2

-20 .=⇒

t

eRIp

Since


Spring 2018


Example:

The switch in the circuit of Figure shown has been closed for a long time, find i(t) for t ≥ 0. Calculate the initial energy stored in the inductor.


Spring 2018


Example:

In the circuit shown in Figure, find i0, v0 and i for all time, assuming that the switch was open for a long time..

Ch 7: 1AC Circuits Analysis

Spring 2018


Chapter 7

AC Circuits Analysis


Spring 2018

Sinusoids

Sinusoid: a signal that has a form of sine or cosine function.• Sinusoidal current is called alternating current (ac)• Any circuit driven by sinusoidal current or voltage is called ac

circuit

Main advantages of sinusoids:1) Nature is sinusoid2) Easy to generate and transmit3) Dominant form in electronics and communications industry4) Easy to handle mathematically


Spring 2018

Sinusoids

Consider a sinusoidal voltage:

v(t) = Vm sinωt ------------- (1)Vm: amplitude of the sinusoidω: angular frequency in (rad/s)ωt: argument of the sinusoid

Period (T): the time of one complete cycle.If ωt =2π T = 2π/ω

Frequency (f): the number of cycles per second in (s-1) or hertz (Hz).

f = 1/T

From relations above ω=2πf

Sinusoid repeats itself every T seconds


Spring 2018

Sinusoids

More general expression for sinusoidal voltage:v(t) = Vm sin(ωt+φ)

φ: the phaseωt+φ: the argument

Consider two sinusoids:v1(t) = Vm sinωt

v2(t) = Vm sin(ωt+φ)

v2 occurs first v2 leads v1 by φ v1 lags v2 by φ

If φ ≠ 0 v1 and v2 are out of phase If φ = 0 v1 and v2 are in phase

} Can be in radians or degrees


Spring 2018

Sinusoids It is possible to transform a sinusoid from sine to cosine and vise

versa using:

tttt

tttt

tttt

ωωωω

ωωωω

ωωωω

cos-)180cos(sin-)180sin(

sin)90-cos(sin-)90cos(

cos-)90-sin(cos)90sin(

=°±=°±

=°=°+

=°=°+

22 += BACAB

θ 1-tan=

)-cos(=sin+cos θtωCtωBtωA


Spring 2018

Sinusoids

Example: (pp (9.1) – p(375))

Given the sinusoid 5 sin(4π t - 60). Calculate its amplitude, phase, angular frequency, period, and frequency.


Spring 2018

Example: (Ex (9.2) – p(375))

Calculate the phase angle between v1 = -10 cos(ωt + 50°) and v2 = 12 sin(ωt - 10°). State which sinusoid is leading.

Solution:

use

V1 = -10 cos(ωt + 50°) = 10 sin(ωt + 50° – 90°) = 10 sin(ωt – 40°)

therefore V2 is leading V1 by -10°–(-40°) = 30°

Sinusoids

tωtω cos-=)°90-sin(


Spring 2018

Sinusoids

cont.

v1 and v1 can be plotted as shown.

v1 = -10 cos(ωt + 50°)

v2 = 12 sin(ωt - 10°)


Spring 2018

Phasor

Sinusoids are easily expressed in terms of phasors which is more convenient that sine/cosine functions

Phasor: is a complex number that represents amplitude and phase of sinusoid

Phasor representation:

)+cos(=)( θtωVtv m (in time domain)

(in phasor domain)θmV=V


Spring 2018

[ e.g. ]

Phasor

Review of complex:• Since phasor is complex, it can be written in three different forms:

1) Rectangular:

x: is the real part of z

y: is the imaginary part of z

2) Polar:

3) Exponential:r: the amplitude of V,

ϕ: the phase of V,

• To convert rectangular into polar: , • To convert polar into rectangular: x = r cosϕ , y = r sinϕ

Thus

1-=j

22 += yxr xy

φ 1-tan=

jyxV +=

ϕrV =φjreV =

jyxV += ϕr= )sin+(cos= φjφr 1=j 90o


Spring 2018

Phasor

Complex number Operations:Given: V = x + jy = r , V1 = x1 + jy1 = r1 , V2 = x2 + jy2 = r2

•Addition and subtraction: (can only be done in rectangular form) V1 + V2 = (x1 + x2) + j(y1 + y2) V1 – V2 = (x1 – x2) + j(y1 – y2)

•Multiplication: in rectangular form: V1V2 = x1x2 – y1y2 + j(x1y2 + y1x2)

in polar form: V1V2 = r1r2

•Division: in rectangular form: V1/V2 = ( x1x2 + y1y2 + j(y1x2 – x1y2) )/(x22 – y22)

in polar form: V1/V2 = r1/r2

ϕ ϕ1 ϕ2

ϕ1 + ϕ2

ϕ1 - ϕ2


Spring 2018

Phasor In summary:

• For addition (+) & subtraction (-) of complex numbers: use rectangular only

• For multiplication (×) & division (÷) of complex numbers: use rectangular or polar, but polar is easier

Other operations:

•Reciprocal:

• Square root:

•Conjugate:

-ϕrV1

=1

ϕ /2rV =

rjyxV =-=* -ϕ φjre-=

jj

-=1


Spring 2018

Phasor

Summary:

Example:Time domain Phasor domain

v(t) = Vmcos (ωt + φ)v(t) = Vmsin (ωt + φ)i(t) = Imcos (ωt + θ)i(t) = Imsin (ωt + θ)

)+cos(=)( θtωVtv m(Time domain representation) (Phasor domain representation)

θmV=V⇔

Im θ

Vm φVm φ - 90

o

Im θ - 90o

Note: ωt factor is suppressed in phasor domain because ω is assumed constant. However, response depends on ω phasor domain is called frequency domain.


Spring 2018

Phasor

Example:

Represent the following phasors graphically using phasordiagram:V = Vm , I = Imφ -θ


Spring 2018

Phasor

Differentiation:

Integration:

dtdv

(Time domain) (Phasor domain)

ωj V⇔

∫ vdt

(Time domain) (Phasor domain)

ωjV⇔

Note: phasor analysis applies only if frequency is constant.


Spring 2018

Phasor

Example: (pp (9.3)-(a) – p(382))

Evaluate the complex number:

[(5 + j2)(-1 + j4) - 5 ]*60o


Spring 2018

Phasor

Example: (pp (9.4) – p(382))

Express these sinusoids as phasors:

(a) v = 7 cos(2t + 40o) V(b) i = -4 sin(10t + 10o) A


Spring 2018

Phasor

Example: (Ex (9.5) – p(383))

Find the sinusoid represented by these phasors:

(a) I = -3 + j4 A(b) V = j8e-j20 V

o


Fall 2018

AC Circuit Elements – Resistor

Ohm’s law

if a current flows through a resistor R, is i(t) = Imcos (ωt + φ)

voltage across the resistor is v = iR = R Imcos (ωt + φ) – (in time-domain)

The phasor form is V

Since phasor form of i is I

Then V = RI – (in phasor-domain)

ϕmRI=

ϕmI=

Voltage and currentAre in-phase Ohm’s law applies in phasor

domain as in time domain


Fall 2018

AC Circuit Elements – Capacitor

Capacitor

If voltage across capacitor is v(t) = Vmcos(ωt + φ) , (or V = Vmejφ = Vm )

current through capacitor is i = Cdv/dt

= -CVmω sin(ωt + φ) = ωCVm cos(ωt + φ + 90°)

I = ωCVmej(φ + 90°) , (or I = ωCVm )

= CVmejφej90°

from ej90° = = j I = jωCVmejφ

I = jωCV

Thus V = I/jωC

• Note that current leads voltage by 90° (since V = Vm while I = ωCVm )

φ

90°This is voltage-current

relationship for

capacitor in phasor-

domain

φ + 90o

φ φ + 90o


Fall 2018


Capacitor – cont.

Phasor diagram for the capacitorIn time-domain In phasor-domain

I leads V by 90°V lags I by 90°


Fall 2018


Example: (pp (9.8) – p(386))

If voltage v(t) = 10 cos(100t + 30°) V is applied to a 50 µF capacitor. Calculate the current through the capacitor.


Fall 2018

AC Circuit Elements – Inductor

Inductor

If current through inductor is i(t) = Imcos(ωt + φ) , (or I = Imejφ = Im )

voltage across inductor is v = Ldi/dt

= -LImω sin(ωt + φ) = ωLIm cos(ωt + φ + 90°)

V = ωLImej(φ + 90°) , (or V = ωLIm )

= LImejφej90°

from ej90° = = j V = jωLImejφ

V = jωLI

Thus I = V/jωL

• Note that voltage leads current by 90° (since I = Im while V = ωLIm )

φ

90°This is voltage-current

relationship for

inductor in phasor-

domain

φ + 90o

φ φ + 90o


Fall 2018


Inductor – cont.

Phasor diagram for the inductorIn time-domain In phasor-domain

V leads I by 90°I lags V by 90°


Fall 2018


Example: (Ex (9.8) – p(386))

If voltage v(t) = 12 cos(60t + 45°) V is applied to a 0.1-H inductor. Find the steady-state current through the inductor.


Fall 2018

Impedance and Admittance

Impedance (Z) of a circuit: a ratio of phasor voltage V to phasor current I, measured in ohms (Ω).

Z = V/I or V = ZI

• Impedance represents the opposition the circuit exhibits to the flow of sinusoidal current (ac)

• Impedance is a complex and can be rewritten as:1) Z = R + jX - (in rectangular form)

R is resistance (the real part of Z, R = Re Z)

X is reactance (the imaginary part of Z, X = Im Z)

2) Z = Z - (in polar form)

Z =

θ

,+ 22 XRRX

θ 1-tan=

If X = 0 Z = R


Fall 2018


Admittance (Y) of a circuit: the reciprocal of impedance, measured in Siemens (S).

Y = 1/Z = I/V

•Admittance is a complex and can be rewritten in rectangular formas:

Y = G + jB =

G is conductance (the real part of Y, G = Re Y)B is susceptance (the imaginary part of Y, B = Im Y)By rationalization of the above equation (i.e. multiply by conjugate)

– If X = 0 G = 1/R, otherwise, X ≠ 0 but Y = 1/Z

jXR +1

,+

= 22 XRR

G ,+

-= 22 XRX

B


Fall 2018


Impedance of resistor

from V = RI and Z = V/I Z = R

Also, from Z = R + jX, since X = 0 for resistor Z = R

Impedance of capacitor

from V = I/jωC and Z = V/I Z = 1/jωC (or Z = -j/ωC = (1/ωC) )• If ω = 0 (i.e. dc source) Z ∞ capacitor acts like open circuit• If ω ∞ (i.e. ac with high frequency) Z = 0 capacitor acts like short circuit

Impedance of Inductor

from V = jωLI and Z = V/I Z = jωL (or Z = ωL )• If ω = 0 (i.e. dc source) Z 0 inductor acts like short circuit• If ω ∞ (i.e. ac with high frequency) Z = ∞ Inductor acts like open circuit

-90°

90°


Fall 2018


Summary:

Element Impedance Admittance

R Z = R Y = 1/R

C Z = 1/jωC Y = jωC

L Z = jωL Y = 1/jωL

V = ZIThis is Ohm’s law in phasorform for any type of element


Fall 2018

Kirchhoff’s Laws in Frequency DomainKCL

since I = Imejφ , and ejωt ≠ 0

I1 + I2 + I3 + ………+ IN = 0 KCL in frequency (phasor) domain

following similar procedure as above

KVL

V1 + V2 + V3 + ………+ VM = 0 KVL in frequency (phasor) domain

01

=∑=

M

mmv

)+cos(+....+)+cos(+)+cos(=+.....++=0= 2211211=∑ NmNmmN

N

nn φtωIφtωIφtωIiiii

0=)Re(+.....+)Re(+)Re(⇒ 21 21jwtφj

mNjwtφj

mjwtφj

m eeIeeIeeI N

0=)]+.....++(Re[⇒ 21 21 Nφj

mNφj

mφj

mjwt eIeIeIe


Fall 2018

Kirchhoff’s Laws in Frequency Domain

Summary:

KCL In = 0

KVL Vm = 0

Kirchhoff’s laws are applicable in the frequency domain it is easy to do things like impedance combination, nodal & mesh analysis, Thevenin & Norton theorems ...etc.

∑1=

N

n

∑1=

M

m


Fall 2018

Impedance Combination

Series impedances:

Zeq = Z1 + Z2 + … + ZN

• Voltage division:

V1 = V Z1/(Z1 + Z2)

V2 = V Z2/(Z1 + Z2)


Fall 2018

Impedance Combination

Parallel impedances:

1/Zeq = 1/Z1 + 1/Z2 + … + 1/ZN

Yeq = Y1 + Y2 + … + YN

• Current division:

I1 = I Z2/(Z1 + Z2)

I2 = I Z1/(Z1 + Z2)


Fall 2018

AC Circuit Elements – Examples

Example: (pp (9.9) – p(389))

Determine v(t) and i(t) in the circuit.


Fall 2018


Example: (pp (9.10) – p(394))

Determine the input impedance of the circuit at ω = 10 rad/s.


Fall 2018


Example: (pp (9.11) – p(394))

Calculate vo in the circuit.


Fall 2018


Exercise:

Use nodal analysis to find v1 and v2 in the circuit.

EEN 100 Electrical Circuits I 2018 Ch_4.pdf�EEN 100�Electrical Circuits I�Chapter 4 �Capacitors & inductorsIntroductionCapacitorsSlide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15InductorsSlide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Slide Number 25Slide Number 26Slide Number 27

EEN 100 Electrical Circuits I 2018 Ch_5.pdf�EEN 100�Electrical Circuits I�Chapter 5 �Capacitors & inductorsIntroductionCapacitorsSlide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15InductorsSlide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Slide Number 25Slide Number 26Slide Number 27

EEN 100 Electrical Circuits I 2018 Ch_6.pdf�EEN 100�Electrical Circuits I�Chapter 6 �First Order CircuitsFirst Order CircuitsFirst Order CircuitsSource-Free RC CircuitSource-Free RC CircuitSource-Free RC CircuitSource-Free RC CircuitSource-Free RC CircuitSource-Free RC CircuitSource-Free RL CircuitSource-Free RL CircuitSource-Free RL CircuitSource-Free RL CircuitSource-Free RL Circuit

EEN 100 Electrical Circuits I 2018 Ch_7_1.pdf�EEN 100�Electrical Circuits I�Chapter 7 �AC Circuits AnalysisSinusoidsSinusoidsSinusoidsSinusoidsSinusoidsSlide Number 7SinusoidsPhasorPhasorPhasorPhasorPhasorPhasorPhasorPhasorPhasorPhasor

EEN 100 Electrical Circuits I 2018 Ch_7_2.pdfAC Circuit Elements – Resistor AC Circuit Elements – Capacitor AC Circuit Elements – CapacitorAC Circuit Elements – CapacitorAC Circuit Elements – Inductor AC Circuit Elements – InductorAC Circuit Elements – InductorImpedance and AdmittanceImpedance and AdmittanceImpedance and AdmittanceImpedance and AdmittanceKirchhoff’s Laws in Frequency DomainKirchhoff’s Laws in Frequency DomainImpedance CombinationImpedance CombinationAC Circuit Elements – Examples AC Circuit Elements – ExamplesAC Circuit Elements – ExamplesAC Circuit Elements – Examples

een 100 electrical circuits i - kau. een 100... · 2019. 6. 30. · an electric circuit is an...

Documents