Lecture Notes: QE 108: Electricity -...

Lecture Notes: QE 108: Electricity

Electric Circuits

Department of Electrical and Electronic Engineering University of Peradeniya

Peradeniya

December 2005

Prepared by: Kithsiri M. Liyanage, Department of Electrical and Electronic Engineering, University of Peradeniya © KML - DEEE, December 2005

QE 108: Electricity – Section 2 – Circuits

Department of Electrical and Electronic Engineering – University of Peradeniya – 2005 Page 0

Contents

Course Structure...................................................................................................................................................... 1

Lectures and Tutorials ........................................................................................................................................ 1

Laboratory Classes ............................................................................................................................................. 1

Assignments ......................................................................................................................................................... 1

1. Analysis of DC Circuits ...................................................................................................................................... 2

1.1. Elements and Laws ..................................................................................................................................... 2

1.1.1. Current...................................................................................................................................................... 2

1.1.2. Voltage, Electromotive Force and Potential Difference ......................................................................... 3

1.1.3. Power in d.c. Circuits............................................................................................................................... 4

1.1.4. Circuit Elements....................................................................................................................................... 5

1.1.5. Resistance Conductance and Ohm’s law ................................................................................................ 5

1.1.6. Sources ..................................................................................................................................................... 7

1.1.7. Kirchchoff’s Laws ................................................................................................................................... 8

1.1.8. Practical (non-ideal) Sources................................................................................................................... 9

1.2. Circuit Analysis .......................................................................................................................................... 11

1.2.1. Definitions and terminology.................................................................................................................. 11

1.2.2. Mesh analysis......................................................................................................................................... 12

1.2.3. Nodal analysis........................................................................................................................................ 16

1.2.4. Loop Analysis ........................................................................................................................................ 18

TUTORIAL 1........................................................................................................................................................... 21

2. Network Theorems ....................................................................................................................................... 23

2.1. Thevenin’s Theorem................................................................................................................................. 23

2.2. Norton’s Theorem..................................................................................................................................... 24

2.3. Maximum Power Transfer Theorem ........................................................................................................ 24

2.4. Y-∆ and ∆- Y Transformation ................................................................................................................. 25

2.5. Examples ................................................................................................................................................... 26

TUTORIAL 2........................................................................................................................................................... 27



Course Structure Lectures and Tutorials

1. Analysis of DC Circuits ( ~5 Hrs.)

Circuit Elements and Laws

Mesh and Nodal Analysis of DC Circuits

Tutorial

2. Circuit Theorems ( ~4 Hrs.)

Thevenin’s Theorem and Applications

Norton’s Theorem and Applications

Star Delta Conversion

Tutorials

Laboratory Classes

1. Circuits (4 Hrs.)

Assignments

1. End Semester Examination

2. Quizzes (Tentative)

3. Laboratory class



Fig 1. Ammeters Commonly Used in the Laboratory

1. Analysis of DC Circuits 1.1. Elements and Laws In this section we discuss about Current, Voltage, Resistance, Ohm’s law, Electric Power, Kirchhoff’s Laws and their applications 1.1.1. Current We know about electrostatic charge. Electric charge in motion results in an electric current.

e.g. Lightning discharge Current is a measure of the rate at which electric charge passes through the circuit

dtdqi = (instantaneous value)

Charge motion is from a low potential to a higher potential. However the more usual convention is that current flow from a point of higher potential to a point of lower potential.

Fig 1. shows ammeters commonly used to measure current in our laboratory. Instantaneous values are represented using lower-case characters (i). Steady state quantities are represented using upper-case characters (I)

E.g. Find the current waveform at the point of a circuit given that the charge passes through the point is as shown in Fig. 2. Current Direction Representation

Fig. 3 shows the conventions used to represent current directions in circuits.

q(t) (coulombs)

t (s)

Fig. 2. Charge Flow Waveform

Supply Source

Circuit

5A -5A

5A

i(t)

Supply Source

Circuit

5A -5A

5A

i(t)

(a) (b)

Fig. 3. Current Direction Representation



1.1.2. Voltage, Electromotive Force and Potential Difference The Voltage across (or between) a pair of terminals is a measure of the energy required to move charge through the element or circuit connected between the terminals. The voltage across an element is the energy required to move a charge of 1 Coulomb from one terminal to the other

We usually represent instantaneous voltage by v and steady state voltage V. Units of voltage is volt (Joules/ Coulombs!)

Fig. 4. shows voltmeters commonly used in our laboratory. The energy converted per unit charge in an electrical source is known an the electromotive force (e.m.f.) of the source, and the electrical potential difference (p.d.) between two points in a circuit is a measure of the energy required to move charge through the element General name given to e.m.f and p.d. is “voltage”. Representation of Voltage Voltage may be represented on a circular diagram either by a ‘+’and ‘-‘ pairs of symbols or by an arrow pointing from one terminal to another as shown in Fig. 5.

Circuit

+

-

8 V

Circuit

-

+

- 8 V

Circuit

8 V

Circuit

- 8 V

Fig. 5. Voltage Direction Representation (a) (b) (c) (d)

Fig. 4. Voltmeters Commonly Used in the Laboratory



Source

Load

5A

- 5A

i(t) -

+ +

Fig. 8. Current Flow Directions in a Source and a Load

1.1.3. Power in d.c. Circuits

Power is the rate of transfer of energy. Energy requited to move 1 coulomb across an element is p.d. between two terminals of the element. Therefore power (p) is equal to (energy required (v) ÷ time taken (t)) to move 1 coulomb across the element. For d.c. circuits current (i) = (charge ÷ time). In the case where charge is 1 coulomb i = 1/t. Therefore p = v/t = v.i

Since we deal with instantaneous quantities here we get instantaneous power. Average power is given by P = V.I. Power could be measured using a wattmeter. Such wattmeter is shown in Fig. 6.

There are mainly two types of elements in electric circuits.

Power source - Power-supplying elements

Loads - Power absorbing elements

Fig. 7 shows several power supply units used in our laboratory for experiment.

Current flows out of the positive terminal of a power source and current flows into the positive terminal of a load as shown in Fig. 8.

Fig. 7. Power Supplies Commonly Used in the Laboratory

Fig.6. A Wattmeter Used in the Laboratory



1.1.4. Circuit Elements Types of circuit elements

Linear - Nonlinear

Passive - Active

Linear elements - Voltage across the element very linearly with the current

Passive elements - Circuit Components that can only dissipate or store energy

Energy dissipating - Resistors

Energy Storing - Capacitors and inductors

However most practical elements are nonlinear 1.1.5. Resistance Conductance and Ohm’s law There is a resistance to flow of current in a material. This results is a loss of energy which is usually converted to heat

Conductors have low resistance

e.g. Copper, Brass, Aluminum

Insulators have a high resistance e.g. Wood, Plastic. Glass

Semi conductors have a resistance between that of conductors and insulators e.g. Silicon, Germanium

Super conductors resistance is zero (at very low temperatures) e.g. Tin, Lead, Thallium

Ohm’s Law Ohm’s law as describes relationship between potential difference (v) between and current (i) through a linear element. v= iR or V=IR R is the constant of proportionality called resistance the unit of resistance is Ohm (Ω) Another from of expressing the relationships between v (or V) and I (or I) is

Fig. 10. Representation of Resistors

(a) (b)

R R

(a) (b)

(c) (d)

Fig. 9. Circuit Elements (a) Resistors – passive, energy dissipating,

linear! (b) Capacitors – passive, energy storing,

linear! (c) Diodes – passive, nonlinear (d) Transistors – active, nonlinear



i= G v or I=G V where G is conductance. The unit of conductance is Siemens (S) Resistance or conductance is represented by means of a rectangular box in circuit diagram as shown in Fig. 10. Power in a resistive circuit

We know that p = v.i and for a resistive circuit v = i .R = i/G

Therefore for a resistive circuit: p = v.i = R.i2 = v2/R = v2.G = i2/G Watts(W)

Power is also measured in µW, mW, kW and MW depending upon the values involved. E.g.

1. Given that v = 10 V and i = -15 mA, what is the power consumed

(a) 150 W (b) – 150 W (c) 0.15 W (d) –0.15 W

Is this circuit a source or a load?

2. Given that P = 200 mW and I = 10 mA

(i) R = 20 Ω (ii) R = 20 kΩ (iii) G = 0.05 S (iv) G = 50 µS

(a) Only (i) is true (b) Only (i) and (iii) are true

(c) Only (ii) and (iv) are true (d) Only (ii) and (iii) are true



1.1.6. Sources Sources could be Voltage or Current and they could be Independent or Dependent Ideal Sources: Can provide infinite amount of power Independent Sources

Independent sources are represented using the symbols shown in Fig. 11.

Dependent Sources (Controlled Sources)

Dependent sources, also referred to as controlled sources as well, are represented using the symbols shown in Fig. 12.

Out put of a dependent Source is dependent on a voltage or current at some other point in the circuit

+

-

6 V

1.5 mA

12 V +

-

3 A

4 A

Voltage Source Battery Current Source

Fig. 11. Representation of Independent Sources

+

-

10 V1

-

+

10 I2

5 V3

20 I4

Voltage Sources Current Sources

Fig. 12. Representation of Dependent Sources



1.1.7. Kirchchoff’s Laws Kirchchoff’s Current law (KCL) The algebraic sum of currents entering any node is zero

i1 – i2 – i3 + i1 = 0

Σ in = 0 Another form:

The algebraic sum of currents entering any node is equal to the algebraic sum of the currents leaving the node.

i1 + i1 = i2 + i3

Σ iin = Σ iout Kirchchoff’s Voltage Law (KVL) The algebraic sum of e.m.f.s and p.d.s around any close circuit is zero

v1 - v2 - v3 = 0

Σ vn = 0 In any closed path the algebraic sum of the e.m.f.s is equal to the algebraic sum of p.d.s.

v1 = v2 + v3

Σ vemf = Σ vpd

E.g. Consider the circuit shown in Fig. 14. (i) VA = ? (a) 120 V (b) –120 V (c) 40 V (d) –40 V (ii) IB = ? (a) 12 A (b) 0.5 A (c) –0.5 A (d) 4.5 A (iii) R1 = ? (a) 22.33 Ω (b) 33.33 Ω (c) 43.33 Ω (d) 15 Ω (iii) R2 = ? (a) 20 Ω (b) 41.22 Ω (c) 31.33 Ω (d) 21 Ω

i1 I2

I3 i4

Fig. 13. A Circuit Node

10 Ω 12 A

R2

10 Ω

120 V

R1

+

-

8 A

0.5 A

+ - VA IB

Fig. 14. Circuit for Example



The Double–Suffix voltage Notation for specifying the voltage of one mode with respect to another mode The notation shown in Fig. 15 called double-suffix voltage notation is also used to describe the voltage of a node with respect to another node in the circuit. E.g. Consider the circuit shown in Fig. 16. VAB = ? (i) 28 V (ii) 26 V (iii) –28 V (iv) –26 V I = ? (i) 2 A (ii) 1 A (iii) 0 A (iv) –1 A 1.1.8. Practical (non-ideal) Sources

The characteristics of practical sources deviate from that of ideal sources as shown in Fig.17. This deviation is accounted for in the model of:

practical voltages source by introducing a resistance called internal (or source or output) resistance connected in series with an ideal voltage source

practical current source by introducing a conductance called internal (or source or output) conductance connected in parallel with an ideal current source as shown in Fig. 18 (compare them with the models shown in Fig11).

Fig. 15. Double-Suffix Notation

VAN

N S

A

B

C

VNA

VSA

VBS

VCB

VSN

3 Ω

4 Ω

I

10 V +

-

5 Ω


40 V -

+2 Ω

6 Ω

9 Ω

5 Ω A

B

Ideal Source

Practical Source

i

v

Es Ideal Source

Practical Source

v

i Voltage Source

Fig. 17. Source Characteristics Current Source

Fig. 18. Models of Practical Source

Es

Rs

vo Is Gs vo

io

Voltage Source Current Source



Transformation of Practical Sources vL = Es – iL Rs (for voltage source) vL = (Is – iL )/ Gs (for current source) i.e. Gs = 1/ Rs and Is = Es / Rs E.g. Consider the circuit shown in Fig. 20. Current I = ?

(a) 1/2 A (b) 1/3 A (c) 1/4 A (d) 1/5 A

Es Is Gs vL

iL Rs

vL

iL

≡Es

Fig. 19. Equivalent Sources

Es

10 Ω I

10 V 0.2 S


1 A



Fig. 22. A Circuit

R1

R2

R3

R4 Is

2 3

0

1

1.2. Circuit Analysis Here we learn about mesh and nodal analysis.

Anyone of these methods can be applied to almost any circuit and, in many cases there is no simple way of saying which is the 'best' method of solution; we cannot lay down simple rules to determine the 'best' approach. Knowledge of each method can only be gained by acquiring a sound understanding of the features of each type of solution. 1.2.1. Definitions and terminology

An electrical network is a system of interconnected circuit elements containing, for example, resistors, inductors, capacitors, voltage and current sources, transformers, amplifiers, etc.

If the network contains at least one closed path or mesh it is an electrical circuit

Every circuit is a network, but not all networks are circuits! (Fig. 21)

An ideal circuit element is one, which does not, strictly speaking, represent a practical element.

For example, a practical resistor has an ideal resistance element as part of its make-up, but it also has some inductance because the current passing through it produces a magnetic field, and it has some capacitance because it is insulated from, say, earth (the insulation acting as a dielectric).

In many situations the major feature of an element (such as the resistance of a resistor, or the capacitance of a capacitor) can be thought of as being at one point within the element. If this is the case, then we say we are dealing with a lumped-constant (parameter) element. In other cases we say that we are dealing with a distributed-constant (parameter) element.

A node is a point in a circuit, which is common to two or more circuit elements. A junction or principal node is a point in the circuit where three or more elements are connected together; nodes 0 and 2 in Fig. 22 are principal nodes.

In the majority of cases of circuit analysis, we choose one node (usually a principal node) to be a reference node or datum node, so that the voltage of other nodes can be defined with respect to it.

A branch in a circuit is a path containing one circuit element, and which connects one node to another.

A path in a network is a set of connected elements that may be traversed without passing through the same node twice. E.g. R1 and R2 are in the path that connects nodes A and C.

A loop in a circuit is a closed path within the network. E.g. Is – R1 – R3 – R4

Fig. 21. (a) Network (b) Circuit

(a)

(b)

Open Circuit

Short Circuit V

V



An element that can be connected in a circuit in either direction without changing the electrical performance of the circuit is known as a bilateral element.

E.g. resistors, inductors and capacitors.

The majority of networks are bilateral networks, that is, they contain only bilateral elements. Certain network theorems, such as the reciprocity theorem (we’ll see soon!), are only applicable to bilateral networks.

A planar network is one that may be drawn on a flat surface, so that none of the branches passes over or under any other branch. When this cannot be done, the network is non-planar. The circuit drawn in full line in Fig. 23 is a planar network but, if the branch containing R6 is introduced (shown broken), it becomes a non-planar circuit.

A mesh is a loop that does not contain any other loops within it. However in some cases it is possible to re-draw the circuit so that a loop which is not a mesh in one version of the circuit can become a mesh in another version. 1.2.2. Mesh analysis

Mesh analysis involves the concept of mesh current. Fig. 24 shows a circuit with mesh currents, I1 and I2, circulating around the periphery of each mesh in a clockwise direction. The relationship between the mesh current and the branch currents (IA, IB and IC) are

IA = I1, IB = I2, IC = I1 -I2

Applying KVL to each mesh we get

20 = 40I1 – 30I2

10 = -30I1 + 50I2

Solving the two simultaneous equations we get the answers

I1 = 0.636 A and I2 = 0.182 A.

Es

10 Ω

20 V

Fig. 24. A Two Mesh Network

10 V

20 Ω

30 Ω

IA IB

IC

I1 I2

Es R1

Vs

Fig. 23. Planar and Non-planar Network

R2 R3 R4

R5

R6



General rules for writing mesh equations

To understand how to write down mesh equations by just looking at a circuit, let’s consider a more complex circuit shown in Fig. 5.

Applying KVL to mesh 1, mesh 2 and mesh 3

V1 -I1R1 + I2R1 + V2 -I1R4 + I3R4 = 0

-V 2 + I1R1 -I2R1 -I2R2 -I2R3 + I3R3 = 0

I1R4 -I3R4 + I2R3 -I3R3 -I3R5 = 0

These equations can be rearranged to give the following form

El = R11I1 + R12I2 + Rl3I3

E2 = R21I1 + R22I2 + R23I3

E3 = R31I1 + R32I2 + R33I3

Where

E1 = V1 + V2,

E2 = -V2,

E3 = 0,

R11 = R1 + R4,

R12 = R21 = - R1,

R13 = R31 = -R4,

R22 = R1 + R2,

R23 = R32 = -R3,

R33 = R3 + R4 + R5

In general

Rij = (-1) x The resistance in the branch mutual to meshes in which Ii and Ij flow.

Rii = The sum of resistance in the mesh in which Ii flows.

Ei = The sum of source voltages driving Ii

The resistance matrix is a square matrix that is it has as many rows as it has columns. For a bilateral network it is symmetrical about the major diagonal, that is Rij = Rji. In the case of a bilateral network, all elements on the major diagonal of the resistance matrix are positive; elements not on the major diagonal are either zero or negative.

These comments do not always apply to a non-bilateral network or to networks containing sources other than independent voltage sources, for example, current sources.

R1

R2

R4R5

R3 V1

V2

I1

I2

I3

Fig. 25. A Three- Mesh Network

Mesh 1

Mesh 2

Mesh 3

=

3

2

1

333231

232221

131211

3

2

1

III

RRRRRRRRR

EEE

Resistance Matrix



Depending on the circuit, there are either three or four general rules, which need to be followed:

1. Draw a carefully labeled circuit diagram.

2. Assign mesh currents I1, I2, ..., Im to each mesh flowing in clockwise direction in the circuit.

3. If the circuit contains only voltage sources, apply KVL to each mesh and solve the resulting simultaneous equations for the unknown mesh currents (if there are m meshes, there are m equations). If the circuit contains dependent voltage sources, relate the dependent source voltages to the unknown mesh currents.

4. If the circuit contains one or more ideal current sources, replace each such source by an open-circuit (note: the mesh currents assigned in step 2 must not be changed). Each source current should then be related to the mesh currents assigned in step 2. The resulting simultaneous equations should then be solved

Where a circuit contains practical current sources each can be converted into its equivalent practical voltage source, and the problem solved as outlined in step 3 above. Solution of simultaneous equations Solution of simultaneous equations could be found using methods such as Inverse Matrix, Cramer’s Rule, Gauss-Jordan Elimination (Reducing the augmented matrix to reduced row echelon form) and Gauss Elimination (Reducing the augmented matrix to row echelon form) methods. Example 1: Consider the circuit shown in Fig. 26:

(i) I1 is

(a) 0.75 A (b) 0.77 A (c) 0.79 A (d) 0.81 A

(ii) I2 is

(a) -0.12 A (b) -0.13 A (c) -0.14 A (d) -0.15 A

(iii) I3 is

(a) 0.18 A (b) 0.20 A (c) 0.22 A (d) 0.24 A

6 Ω2 Ω

5 Ω4 Ω

7 Ω10 V

8 V

I1

I2

I3

Fig. 26. Circuit for Example 1

3 Ω



Example 2: Consider the circuit shown in Fig. 27: (i) I1 is

(a) 0.45 A (b) 0.55 A (c) 0.65 A (d) 0.75 A

(ii) I2 is

(a) 0.05 A (b) 0.06 A (c) 0.07 A (d) 0.08 A

(iii) I3 is

(a) 0.18 A (b) 0.19 A (c) 0.20 A (d) 0.22 A

Solution to Example 2

Applying KVL to the super-mesh in Fig. 28, we get

10 - 8 = 2I1 + 3I2 + 7(I2 - I3) + 5(I1 -I3)

0 = -5I1 –7I2 + 16I3

0.5 = I1 – I2

Example 3: Consider the circuit shown in Fig. 29: (i) I1 is

(a) 4.00 A (b) 4.25 A (c) 4.58 A (d) 4.78 A

(ii) I2 is

(a) -2.12 A (b) -2.53 A (c) -2.85 A (d) -2.97 A

(iii) I3 is

(a) 0.25 A (b) 0.27 A (c) 0.29 A (d) 0.31 A

Solution to Example 3

Vx = 5(I1 - I3)

10 + 3Vx = 13I1- 6I2- 5I3

-3Vx - 8 = -6I1 + 16I2- 7I3

0 = -5I1 –7I2 + 16I3

6 Ω2 Ω

5 Ω4 Ω

7 Ω 10 V

8 V

I1

I2

I3


3 Ω

0.5 A

6 Ω2 Ω

5 Ω4 Ω

7 Ω 10 V

8 V

I1

I2

I3

Fig. 28. Circuit for Solving Example 2

3 Ω

Open Circuit

6 Ω2 Ω

5 Ω4 Ω

7 Ω 10 V

8 V

I1

I2

I3


3 Ω

3 VX

VX



1.2.3. Nodal analysis

Nodal analysis uses KCL to evaluate the voltage at each principal node in the circuit, and is valid for all circuits, both planar and non-planar. In this case we write down and solve a set of simultaneous equations in terms of the unknown voltage at each node. If the circuit has n principal nodes, we need (n -1) simultaneous equations to solve the circuit. We can obtain (n-1) simultaneous equations by applying KCL to each non-reference node in turn.

2 = 2Vl + 3Vl2 = 2Vl + 3(Vl –V2) = 5Vl -3V2

-3 = 3V21 + 4V2 = 3(V2 –V1) + 4V2 = -3V1 + 7V2

By solving two equations above we get V1 = 0.192 V and V2 = -0.346 V.

General rules for writing nodal equations

Consider the circuit shown in Fig. 31.

IA = (GA + GE)V1 -GAV2

IB - IA – IC = -GAV1 + (GA + GB + GC)V2 -GCV3

IC = -GCV2 + (GC + GD)V3

These equations can be rewritten as

I1 = G11 V1 + G12V2 + G13V3

I2 = G21 V1 + G22V2 + G23V3

I3 = G31V1 + G32V2 + G33V3

In general

Gij = (-1) x The conductance linking node i to node j.

Gii = Total conductance terminating on node i.

Ii = The sum of currents entering node i.

The conductance matrix is a square matrix, and is symmetrical about the major diagonal, that is for all ij (i ≠ j), Gij = Gji that is, G12 = G21, G23 = G32 etc.

In the case of a bilateral network, all the elements on the major diagonal of the conductance matrix are positive. The elements not on the major diagonal are negative or zero.

Es

3 S

2 A

Fig. 30. A Three-Node Circuit

3 A 4 S 2 S

GA

GB GC

GD

GE

IA

IB

IC

Fig. 31. Four Node Network



Depending on the circuit, there are either three or four steps to be carried out:

1. Draw a carefully labeled circuit diagram.

2. Mark the principal nodes on the circuit, and select a reference node. If there are n principal nodes, (n -1) simultaneous equations are needed to solve the circuit.

3. If the circuit contains only independent current sources, apply KCL to each non-reference node. If the circuit contains dependent current sources, relate the source current to the unknown node voltages.

4. If the circuit contains ideal voltage sources, we cannot deal with it in the normal way because its internal resistance is zero. In that case, replace each voltage source by a short-circuit (the voltages assigned in step 2 should not be changed). Each source voltage should then be related to the unknown node voltages.

If the circuit contains a practical voltage source, it can be converted to its equivalent practical current source and dealt with as a normal current source

Example 4: Consider the circuit shown in Fig. 32: (i) V1 is

(a) 0.089 V (b) 0.096 V (c) 0.150 V (d) 0.174 V

(ii) V2 is

(a) 0.008 V (b) 0.009 V (c) 0.010 V (d) 0.011 V

(iii) V3 is

(a) 0.189 V (b) 0.284 V (c) 0.376 V (d) 0.432 V

Solution may be obtained by solving the following simultaneous equations.

0 = 12Vl - 2V2 - 4V3

-1 = -2Vl + 5V2 - 3V3

3 = -4Vl - 3V2 + 12V3

Example 5: Consider the circuit shown in Fig. 33:

(i) V1 is

(a) 1.4 V (b) 2.0 V (c) 2.5 V (d) 3.2 V

(ii) V2 is

(a) -1.4 V (b) -1.0 V (c) –0.8 V (d) –0.6.2 V

(iii) V3 is

(a) -1.4 V (b) -2.0 V (c) -2.5 V (d) -3.2 V

3 S

Fig. 33. Network for Example 5

4 S

5 S

6 S

2 S

2 A

3 A

4 V

Super Node

3 S


4 S

5 S

6 S

2 S

2 A

3 A



Solution may be found by solving the following simultaneous equations.

Total current leaving super node = 2(V1 -V2) + 6(V1 -Vo) + 3(V3 -V2) + 5(V3 -Vo)

Total current entering super node = 3 A

3 = 2(V1 -V2) + 6V1 + 3(V3 -V2) + 5V3 = 8V1 – 5 V2 + 8V3

-1 = -2Vl + 5V2 -3V3

4 = V1 –V2

Example 6: Consider the circuit shown in Fig. 34: (i) V1 is

(a) 0.071 V (b) 0.081 V (c) 0.091 V (d) 0.101 V

(ii) V2 is

(a) 0.015 V (b) 0.016 V (c) 0.017 V (d) 0.018 V

(iii) V3 is

(a) 0.275 V (b) 0.315 V (c) 0.357 V (d) 0.387 V

Solution may be found by solving the following simultaneous equations.

0 = 12V1 - 0.5V2 - 2.5V3 -1 = -2V1 + 5V2 - 3V3

3 = -4V1 - 3V2 + 12V3 – 1.5 (V2 - V3) = -4V1 – 4.5V2 + 13.5V3

1.2.4. Loop Analysis

Network topology It was stated earlier that mesh current analysis is applicable only to planar networks. However there is a similar approach -known as loop analysis, which allows us to solve non-planar networks.

Network topology is concerned with a mathematical discipline known as graph theory with special reference to electrical circuits. The 'graph' referred to here is not a conventional graph, but is a collection of points (nodes) and connecting lines (branches). When drawing the 'graph' of a network, the nature of the element in the branch between a pair of nodes is suppressed, and is replaced by a line or 'edge'.

Fig. 35 shows a four nodes and six branches circuit and its graph.

Given a graph, we define a tree ( or spanning tree)

3 S


4 S

5 S

6 S

2 S

2 A

3 A

1.5 V23

6 Ω2 Ω

5 Ω4 Ω

7 Ω 10 V

8 V

Fig. 35. A Circuit and its Graph

3 Ω



as any set of branches in the graph which connect every node to all other nodes in the graph, but not necessarily directly. Moreover, the tree does not contain a loop of any kind.

If the graph has N nodes, each tree has (N-l) branches in it. The four-node graph in Fig. 15 contains sixteen trees, one of which is shown in full line in Fig. 16. In loop current analysis, we select a normal tree, that is one containing all the voltage sources in the network, together with the maximum number of voltage-controlled dependent sources.

A cotree is a set of branches which do not belong to a tree; the cotree corresponding to the tree in Fig. 16 is shown in broken lines. A branch in a cotree is known as a link. A cotree is the complement of a tree, and a tree and its cotree form the complete graph of a network. An N-node network contains a number of trees, each with (N-l) branches; if B is the number of branches in the network, and L is the number of links in the cotree, the relationship between them is

B = L + (N -1)

If we re-position anyone of the links from the cotree in the tree, we will form a loop in the tree.

Since adding a link to the tree forms a loop, the number of links is equal to the number of independent voltage equations we need to form the loop equations of the network. Loop analysis

The following steps allow us to write a set of loop current equations for a circuit:

1. Draw a graph of the network and identify a normal tree.

2. Ensure that all voltage sources and, if possible, all control-voltage branches for voltage-controlled dependent sources are in the tree.

3. Ensure that all-current sources and, if possible, all control-current branches for current-controlled dependent sources are in the cotree.

4. Reposition in the tree, one at a time, each link in the cotree. Using KVL, write down for each loop the associated loop current equation; solve the equations.



IA = 3 A

0 = 7IB + 5(IA + IB - ID) + 4(IA + IB+ Ic) + 3(IB + IC)

0 = 5 + 5(IA + IB + IC) + 3(IB + IC) + 2IC

0 = 5 - 5(IA + IB – ID) - 6(IA - ID) + 8ID

Solving the above equations gives

IB = -0.524 A, IC = -1.482 A, ID = 1.336 A

IA

IB

IC

ID

4 Ω

7 Ω

8 Ω

3 Ω 5 Ω

6 Ω 2 Ω

5 V

3 A

IA + IB + IC

IA + IB - ID

IA – ID

ID

IC

IB + IC

IA

ID (a)

(b)

(c) (d) (a)

(b)

IB

1

1 1

1

1

2

22

2

4

44

3

3 3

3

4

4

2 2

5

5

6

6

5

5

6 6



TUTORIAL 1 1. The repetitive waveform of current entering a circuit is

shown in Figure 1.

(a) What is the value of the current at t = 0.01

s?

(b) What charge enters the circuit between t = 80

ms and t =150 ms?

(c) What total charge has entered the circuit at t =

210 ms?

2. The electrical charge entering a terminal in a circuit is 10 sin

100πt mC

(a) What charge has entered between –3 ms and 3 ms ?

(b) Calculate the current at t =2ms

3. In Figure 2 calculate VBC , VCA,VBD,VEB and VDC

4. Calculate the power consumed by each resistor in Figure 2.

5. Calculate the power absorbed by each of the circuit elements in

Figure 3.

6. The resistance, R, of a conductor in ohms is given by the equation

R=ρL/A where ρ is the resistivity (or volume

resistivity) of the conductor in Ω m, L is the

length of the conductor in m,and A is the cross–

sectional area of the conductor in m2 .if a

conductor of resistivity 0.027 µΩ m,which is 100

km long and of diameter 1 cm carries a current of

20 A, calculate the p.d.across the length of the

wire ,the power consumed in the wire and the

energy consumed in 20 mins.

7. For the circuit in Figure 4. Calculate

(a) I,

(b) V and

(c) the power absorbed by each

element. The calculation should

verify the principle of

conservation of energy.

10V

0.2 A

12 V

9e-5t

20 V

-10 V

2 A+

-5A

6V

Figure 3 5 Ω -4.25 A

20 V

R

V

5 A

10 Ω

30 V

Figure 4

Figure 1

0 100 200

0.5 A

t (ms)

B

Figure 2

40Ω

20Ω

25Ω

-30V

-100V

10Ω

A

B

D

E



8. In Figure 6, calculate V and the power supplied by the 50 mA source.

9. Calculate the resistance between A and B in Figure 7

10. If 10 V is applied between A and B in Figure 7, calculate the

voltage across and the current in each element.

10 VX 10 V 3Ω

VX

2Ω I

Figure 5

2 kΩ V

1 kΩ 50 mA 20 mA

Figure 6

7 Ω

8 Ω

2 Ω

3 Ω 4 Ω

5 Ω

A B

6 Ω

Figure 7



2. Network Theorems 2.1. Thevenin’s Theorem

A linear bilateral two-terminal network can be replaced by a Thevenin equivalent circuit

consisting of a voltage source and a resistor connected in series.

VTH is the open circuit voltage between terminals “a” and “b” with part B disconnected. RTH

is the resistance of part A between terminals “a” and “b” with all idea voltage sources short

circuited ideal current sources open circuited and partial sources represented by their internal

resistance (deactivation of independent sources).

A

B

a

b b

RTh

B

a

VTh

A

a

b Vs RTh = Vs/ Is

RTh = VTh/ Isc A

a

b

Isc

Is

With independent Source deactivated



2.2. Norton’s Theorem

A linear bilateral two-terminal network can be replaced by a Norton Equivalent Circuit

consisting of a current source and a conductance connected in parallel.

The Thevenin Equivalent Circuit can be replaced by an equivalent current source in parallel

with a conductance.

IN =VTh / RTh = ISC

GN = 1 / RTh 2.3. Maximum Power Transfer Theorem

The maximum power is received by a load from a linear bilateral dc circuit when the load

resistance is equal to the RTH of the circuit.

PRL = VTH

2 R (RTH + R)2 dPRL = VTH

2 d R dR dR (RTH + R)2 = VTH (RTH + R)2 – 2 R (RTH + R) (RTH + R)4 = (RTH + R) (RTH + R-2R) VTH (RTH + R)4 i.e R = RTH

RL VTh

GN

a

b

IN

RTh



2.4. Y-∆ and ∆- Y Transformation ∆- Y Transformation RA = RAB RAC RB = RBA RBC RC = RCA RCB RAB + RAC + RBC RAB + RAC + RBC RAB + RAC + RBC Y-∆ Transformation RAB = RARB + RARC + RBRC RC RAC = RARB + RARC + RBRC RB RBC = RARB + RARC + RBRC RA

RBC

RAB

RCA RB

RA

RC

C

B A C

B A



2.5. Examples 2.5.1. Consider the circuit shown below. Find the value of the resistor which when connected across terminal a and b draws a current of 5A. Answer 2.5.2. Find the Thevenin Equivalent Circuit of the following circuit Answer 2.5.2. Find the NOrton Equivalent Circuit of the following circuit Answer:

VC 40

25IB

0.0004VC 1 V

2 kΩ IB

5 Ω

100V

6 Ω

20 Ω

a

b

VTH = 80V RTH = 10Ω

∴Rtotal = 80/10 = 16

∴RL = 6Ω

30 IB 500 ΩIB

1 kΩ

10V

VTH = 10/1000 x 30 x 500 = -150V ISC = 30 x IB = 30 x 10/1000 A = 3/10 A ∴RTH = 150/3 * 10 = 500 Ω

25 IB x 40 = -VC ((1-0.0004VC)/2000) x 25 x 40 = -VC VC = -625 ISC = 25 x IB = 25x (1/2000) = 1/80 ∴ IN = 12.5 mA ∴GN = ISC / VC = 0.02 mS



TUTORIAL 2

1. If, in Figure 1 ,V1 = 10 V, V2 = 20 V, R1 = 1 Ω , R2 = 2 Ω, R3= 3

Ω, R4= 4 Ω and R5 = 5 Ω, calculate the mesh currents.

2. Calculate voltages across 2 S and 4 S elements in Figure 2 and

compute the total power consumed.

3. Using mesh analysis, calculate I1 and I2 in Figure 3.

4. Using nodal analysis, calculate V1 and V2 in Figure 3.

5. The mesh equations of a network are as follows:

Draw the corresponding circuit diagram.

7. Using mesh analysis, calculate v in Figure 4.

8. Use nodal analysis to calculate v in Figure 5.

9. In Figure 4,the 3 Ω, 5 Ω and 6 Ω resistors from a tree. Use loop

analysis with respect to this tree to calculate v.

10. Use mesh analysis to calculate v in Figure 5.

11. For the circuit in Figure 6, use mesh analysis to calculate

(a) the voltage gain V2 / V1

(b) the input resistance (V1/ I1) of the circuit.

12. For the simplified emitter follower equivalent circuit in Figure 7,

use mesh analysis to calculate

(a) the voltage gain of the circuit (V2 / V1)and

(b) the input resistance (V1/ IB).

13. Use nodal analysis to solve problem 11.

14. Solve problem 12 using nodal analysis.

15. Use nodal analysis to calculate I1, I2 and I3 in the circuit in Figure 8.

16. Use nodal analysis to determine the node voltages V1and V2 in Figure 9.

R1

R2

R4 R5

R3

V2

Figure 1

V1

Es

3 S

2 A

Figure 2

3 A4 S 2 S

−−−−−−

−−

=

−−

−

4

3

2

1

178308170230144

02411

1.13.59.197.10

IIII

6 Ω

4 Ω

5 Ω

6 V

4 V v 3 Ω

Figure 4

v

5 Ω3 Ω

6 Ω 2 A 2 Ω

3 A

Figure 5

I1 I2

6 Ω 4 Ω

7 Ω

5 Ω 2 A 10 V

V1 V2

Figure 3



V1 = 0.1 V

IB

100 kΩ

100×10-6 V2

1500 Ω

10 kΩ50 µS

V2 70IB

Figure 7

3 Ω

5 Ω

4 Ω

2 Ω

2 A I2 3V1 V1

I1

I3

Figure 8

10 V 4 Ω

3 Ω

15 A

2 Ω

5 Ω

-10 A

V2 V1

Figure 9

V1 VX 2 Ω V2

6 Ω10 VX

4 Ω

3 Ω

Figure 6

I1 5 Ω



Annex-1

Silver 1.59 x10^-8 0.0061 6.29Copper 1.68 x10^-8 0.0068 5.95Aluminum 2.65 x10^-8 0.00429 3.77Tungsten 5.6 x10^-8 0.0045 1.79Iron 9.71 x10^-8 0.00651 1.03Platinum 10.6 x10^-8 0.003927 0.943Manganin 48.2 x10^-8 0.000002 0.207Lead 22 x10^-8 ... 0.45Mercury 98 x10^-8 0.0009 0.1Nichrome(Ni,Fe,Cr alloy)Constantan 49 x10^-8 ... 0.2Carbon*(graphite)Germanium* 1-500 x10^-3 -0.05 ...Silicon* 0.1-60 ... -0.07 ...Glass 1-10000 x10^9 ... ...Quartz(fused)Hard rubber 1-100 x10^13 ... ...

Material Resistivity ρ (ohm m)

100 x10^-8 0.0004 0.1

Temperature coefficient per degree

C

Conductivity σ x 107

/Ωm

7.5 x10^17 ... ...

Mar-60 x10^-5 -0.0005 ...

Lecture Notes: QE 108: Electricity -...

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