1 - ee2001 circuit analysis lectures 1-6.pdf

EE2001 Circuit AnalysisEE2001 Circuit AnalysisEE2001 Circuit AnalysisEE2001 Circuit Analysis

Professor Wang Youyi, PhD.Office: S2-B2b-47Tel: 67904537Tel: 67904537Email: [email protected]

AY2009/10 S t 2AY2009/10 Semester 2

Prescribed Text and References• TEXTBOOKS• TEXTBOOKS

– "Electric Circuits", by James W Nilsson & Susan A Riedel, 8th Edition, Prentice Hall, 2007Riedel, 8 Edition, Prentice Hall, 2007

• REFERENCES

– "Engineering Circuit Analysis", by William H Hayt, Jack E Kemmerly & Steven M Durbin, McGraw Hill, 7th Rev. Edition, 2007.

– "Fundamentals of Electric Circuits" by Charles K Alexander & Matthew Sadiku, 3rd

Edition, McGraw-Hill, 2006.

Course Syllabus• Circuit theorems and analysis techniques (WYY)• Circuit theorems and analysis techniques (WYY)

– 6 lectures, 3 tutorials– Chapters 1-5

• Energy storage elements and transient responses (WYY)• Energy storage elements and transient responses (WYY)– 4 lectures, 2 tutorials– Chapters 6-8

Si id l AC i i (WYY)• Sinusoidal AC circuits (WYY)– 4 lectures, 1 ½ tutorials– Chapters 9-10

• Laplace transform techniques (LKG)– 6 lectures, 3 tutorials

• Network functions and two-port networks (WYY)– 6 lectures, 2 tutorials

Course Assessments• Examination – 80%• Examination – 80%

– Closed book – 2 hours

4 compulsory questions– 4 compulsory questions– Conducted in exam hall at end of semester

• Q i 20%• Quiz – 20%– Closed book– 30 minutes

5 t 6 lti l h i ti– 5 to 6 multiple-choice questions– Conducted in your registered tutorial group– Week 8 - the week following semester break

M k i– Make up quiz • Only for those with valid reason/s and supporting document/s• To be completed within 1 week (i.e. by end of Week 9)

Basic Concepts• Electrical engineering is the profession concerned with systems that produce• Electrical engineering is the profession concerned with systems that produce,

transmit and measure electric signals.

• Electrical engineering combines the physicist’s models of natural phenomena with• Electrical engineering combines the physicist s models of natural phenomena with the mathematician’s tools for manipulating those models to produce the systems that meet practical needs.

• Electrical systems are part of our lives.

Ci it l i b t i t t i it ill h l i d t di• Circuit analysis becomes paramount important since it will help in understanding the behavior of circuit model and its ideal circuit component.

Th t f t ti t f i it l i t k h t t t th• The most frustrating parts of circuit analysis are to know how to start the problem and obtain a complete set of equations and organizing them in such a way as to appear manageable.

Basic Strategy• Figure on right is designed as a guide to

Read the problem statement slowly and carefully.

Identify the goal of the problem• Figure on right is designed as a guide to overcoming the common obstacles of starting a problem and managing the solution process

Identify the goal of the problem.

Collect the known information.

• Several of the steps may seem obvious, but it is the chronological order as well as the performance of each task that leads to

Devise a plan.

Construct an appropriate set of equationsperformance of each task that leads to success.

• The real key to success in circuit analysis is PRACTICE especially in a relaxed low-stress

Construct an appropriate set of equations.

Determine if additional information

i i d

Yes

PRACTICE, especially in a relaxed, low-stress environment.

• Experience is the best teacher and

is required.

Attempt a solution.

No

• Experience is the best teacher, and learning from mistakes will always be a part of the process of becoming a competent electrical engineer

Verify the solution. Is it reasonable or

expected?

No

competent electrical engineer. p

End.

Yes

EE2001 – LECTURE 1

Voltage and Current (1.4)Ideal Basic Circuit Element (1.5)Power and Energy (1.6)Power and Energy (1.6)Passive Sign Convention (1.5, 1.6)Circuit Elements (2.1, 2.2)

Voltage and Current• Electric charge• Electric charge

– Electric charge is the basis for describing all electrical phenomena– Charge is bipolar (positive and negative charges)

Like charges repel and unlike charges attract– Like charges repel and unlike charges attract– Charge exists in discrete quantities which are integral multiples of the electronic

charge 1.6022 x 10-19 Coulombs. – Electrical effects are attributed to both the separation of charge and charges in motionElectrical effects are attributed to both the separation of charge and charges in motion

• In circuit theory, the separation of charge creates an electric force (voltage), and the motion of charge creates an electric fluid (current)the motion of charge creates an electric fluid (current)– The concepts of voltage and current are useful because they can be expressed

quantitatively. – Energy is expended whenever positive and negative charges are separatedEnergy is expended whenever positive and negative charges are separated.

• Voltage is the energy per unit charge created by the separation of the positive and negative charges. It is expressed as follows:

dwvdq

=where v – voltage in volts

w – the energy in joulesq – the charge in coulombs

• The electrical effects caused by charges in motion depend on the rate of charge flow and this is known as electric current. It can be expressed as follows:p

dqidt

=where i – the current in amperes

q – the charge in coulombst – time in seconds

• These two equations are definitions for the magnitude of voltage and current respectivelyrespectively.

• The bipolar nature of electric charge requires that we assign polarity references to these variables.

Ideal basic circuit element• An ideal basic circuit element has three attributes:• An ideal basic circuit element has three attributes:

– it has only two terminals, which are points of connection to other circuit components;

– it is described mathematically in terms ofit is described mathematically in terms of current and/or voltage; and

– it cannot be subdivided into other elements.• v is voltage across terminals of the elementv is voltage across terminals of the element

– Polarity reference for voltage is indicated by plus and minus signs• i is current in the element

– Reference direction for current is shown by arrow– Reference direction for current is shown by arrowPositive Negative

vvoltage drop from terminals 1 to 2

orvoltage rise from terminals 1 to 2

orv orvoltage rise from terminals 2 to 1

orvoltage drop from terminals 2 to 1

ipositive charge flowing from terminals 1 to 2

ornegative charge flowing from terminals 2 to 1

positive charge flowing from terminals 2 to 1or

negative charge flowing from terminals 1 to 2negative charge flowing from terminals 2 to 1 negative charge flowing from terminals 1 to 2

• The flow of current is conventionally regarded as a flow of positive charges, although charge flow in metal conductors results from electrons with a negative chargecharge.– Direction of current is opposite to the direction of electron flow

• The assignment of the reference polarity of voltage and the reference direction for t ti l bitcurrent are entirely arbitrary.

• Example 1.1:– i1 is a current flowing from terminal ‘a’ to ‘b’ a b

i1

– i1 = - i2, i.e., the variable i2 represents the same current i1.

• Example 1.2:i2

3A 3A– Numeral values show the actual direction

• Example 1.3:

3A -3A

– For the element in the figure, v1 = 17V. Determine v2.

– (Ans: v2 = -v1 = - 17 V)

+

-

-

+v1

v2

Power and energy• Power is the time rate of expending or absorbing energy• Power is the time rate of expending or absorbing energy.

dwpdt

=where p - the power in watts,

w- the energy in joules, and t – the time in seconds.

• Substitute in voltage and current definitions, the power equation is obtained as follows:

dw dw dq⎛ ⎞⎛ ⎞dw dw dqpdt dq dt

p vi

⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=

where p – the power in watts, v – the voltage in volts, and i – the current in amperes.

– Power is energy per unit of time and is equal to the product of terminal voltage and current

• Power is a quantity associated with a pair of terminals and thus we have to be

p

• Power is a quantity associated with a pair of terminals, and thus we have to be able to tell whether power is being delivered to the pair of terminals or extractedfrom it (i.e. to/from element/s connected at the terminals).

Passive sign convention• Passive sign convention: Whenever the reference direction for the current in• Passive sign convention: Whenever the reference direction for the current in

an element is in the direction of the reference voltage drop across the element, use a positive sign in any expression that relates the voltage and the current. Otherwise, use a negative sign.O , g g

• Following passive sign convention:– If the power is positive (p > 0), power is being absorbed by the circuit inside the box.

f h i i ( 0) i b i i f h i i i id h– If the power is negative (p < 0), power is being delivered from the circuit inside the box.

• Example 1.4:) S h h l d h l i f h i (b) Gi i 4 Aa) Suppose that we have selected the polarity references as shown in (b). Given i = 4 A

and v = -10 V, then the power associated with the terminal pair 1,2 isp = - (-10)(4) = 40 W

Thus, the circuit inside the box is absorbing 40 W

b) Suppose that your colleagues have chosen the polarity references shown in (c) Givenb) Suppose that your colleagues have chosen the polarity references shown in (c). Given i = -4 A and v = 10 V, then the power is

p = - (10)(-4) = 40 W

Thus, the circuit inside the box is still absorbing 40 W

• Example 1.5:– A high-voltage direct-current (HVDC) transmission line between Celilo, Oregon and

Sylmar California is operating at 800 kV and carrying 1800 A Calculate the power atSylmar, California is operating at 800 kV and carrying 1800 A. Calculate the power at Oregon end of the line and state the direction of power flow.

– At Celilo, it is non-passive sign convention

( )( )800 1.8 1440 MWp vi= − = − = −

– Power is negative, and hence power is delivered from Celilo, Oregon– Therefore, direction of power flow is from Celilo to Sylmar

– Note: At Sylmar, it is passive sign convention

• Positive power indicates that power is absorbed by Sylmar

( )( )800 1.8 1440 MWp vi= = =

Positive power indicates that power is absorbed by Sylmar

• Try Chapter Problems 1.12 to 1.15

Circuit elements• All circuit elements are broadly classified as active or passive• All circuit elements are broadly classified as active or passive.

– An active element is one that is capable of generating energy. Examples are batteries, voltage sources and current sources.

– A passive element is one that cannot generate energy. Examples are resistors,A passive element is one that cannot generate energy. Examples are resistors, capacitors and inductors.

• Independent and dependent Sources– An independent source establishes a voltage or current in a circuit without relying on

voltages or currents elsewhere in the circuit.– A dependent source establishes a voltage or current whose value depends on the value

of a voltage or current elsewhere in the circuit.

• Ideal independent voltage and current sourcesA id l lt i i i l h i i ib d l– An ideal voltage source is a circuit element that maintains a prescribed voltage across its terminals regardless of the current flowing in those terminals.

– An ideal current source is a circuit element that maintains a prescribed current through its terminals regardless of the voltage across those terminalsthrough its terminals regardless of the voltage across those terminals.

• Examples of ideal independent sourcesa) A time varying independent voltage source b) A i d d lb) A constant independent voltage source c) A time varying independent current sourced) An independent voltage source

v(t) V i(t)

( ) (b) ( )e) An independent current source

• Common practices but not always adhere to

(a) (b) (c)

– Lower case for time varying (AC, square wave), e.g. v and i.

– Upper case for constant (DC), e.g. V.vs is

– Subscripts used to differentiate one sourcefrom another (d) (e)

• Ideal dependent (or controlled) voltage and current sources– Output is determined by a voltage or

current at a specific location in a circuitcurrent at a specific location in a circuit. – A dependent source is unilateral, i.e.

the output variables have no influence on the input variables. o e pu v b es.

a) Voltage-controlled voltage source• μ is dimensionless μ

b) Current-controlled voltage source• ρ is in ohms (Ω)

c) Voltage-controlled current source• α is in mhos (Ω-1)

d) Current-controlled current source• β is dimensionless

Series and Parallel Connections• Most circuits comprise of elements in a mixture of• Most circuits comprise of elements in a mixture of

series and parallel connections– Those connected in series share the same current– Those connected in parallel share the same voltage– Those connected in parallel share the same voltage

• Example 1.6:

– Resistors 50 Ω and 10 Ω are in series– Resistors 20 Ω and 30 Ω are in parallelp

• They are also in parallel to the series combination of 50 Ω and 10 Ω resistors– Resistors 40 Ω and 70 Ω are in series with the 60 V voltage source

• They are also in series to the parallel combination of the remaining resistors

• Note: A parallel connection of voltage sources or a series connection of current sources is forbidden unless the sources are pointing in the same direction and have exactly the same valuesexactly the same values.

Valid ValidInvalid ValidValid Valid

InvalidValid

T Ch t P bl 2 2 d 2 3

Invalid Invalid

• Try Chapter Problems 2.2 and 2.3

Valid

v1

vv1+v2 v1=v2v1 v2

v2

v1R

Voltage sources in series Voltage sources in parallel

v2

v1-v2 vv R

g g p

i1i1=i2 i1 i2 i1+i2

i2

R

Current sources in parallelCurrent sources in series

ii i1 i2 i1-i2

Current sources in parallelCurrent sources in series


Ohm’s Law (2.2)Kirchhoff’s Laws (2.4, 2.5)Operational amplifier (5.1, 5.2, 5.3)Operational amplifier (5.1, 5.2, 5.3)

Ohm’s Law• Resistance is the capacity of materials to impede the flow of current or more• Resistance is the capacity of materials to impede the flow of current or, more

specifically, the flow of electric charge. The circuit element used to model this behavior is the resistor.

• Ohm’s law is the algebraic relationship between voltage and current for a resistor• Ohm s law is the algebraic relationship between voltage and current for a resistor and it is expressed as follows:

v iR=where v - the voltage in volts,

i - the current in amperes

– Express voltage as function of current

v iR i the current in amperes, R - the resistance in ohms.

• For the purpose of circuit analysis, we must reference the current in the resistor to the terminal voltage. – We can do so in two ways, according to

passive sign convention or otherwise

• Ohm’s law equivalent - express current as function of voltage

vi Gvwhere v - the voltage in volts,

i h i

– Conductance is simply the inverse of resistance.

i GvR

= = i - the current in amperes, G - the conductance in mhos.

• It is represented by the symbol G. • The unit of conductance is Siemens (S).

Sometimes it is also known as mho (Ω-1).

Po er calc lation at the terminals of a resistor can be e pressed in se eral a s:

1 S = 1 A/V = 1 Ω-1

• Power calculation at the terminals of a resistor can be expressed in several ways:p = vi when v = iR and p = -vi when v = -iR

• or p = vi = (iR)i = i2R , and p = -vi = -(-iR)i = i2R

• orp = v2/R which is independent of the polarity referencep p p y

• Note: regardless of voltage polarity and current direction, the power at the terminals of a resistor is always positive. – Therefore, a resistor always absorbs power from the circuit.Therefore, a resistor always absorbs power from the circuit.

• Example 2.1: In each of the circuit, calculate the value of v or i and determine the power dissipated in each resistor.

) P i i tia) Passive sign conventionva = 1 × 8 = 8 Vp = 12 × 8 = 8 W

b) Passive sign conventionib = 50 × 0.2 = 10 Ap = 502 × 0.2 = 500 W

c) Non passive sign conventionvc = -1 × 20 = -20 V

( )2 /p = (-20)2 / 20 = 20 W

d) Non-passive sign conventioni 50 / 25 2 Aid = -50 / 25 = -2 Ap = 502 / 25 = 100 W

• Example 2.2:a. What value of vg is required in order for

the interconnection to be valid?the interconnection to be valid?ib = - 8 A vg = 0.25 (- 8) = - 2 V

b. For this value of vg, find the power delivered/absorbed associated withdelivered/absorbed associated with the 8A source.p = vg i = -2 × 8 = -16 W (delivered)

• Example 2.3:a) What value of α is required in order for

the interconnection to be valid?vx = -25 V and αvx = -15 A α = 0.6

b) For the value of α calculated in part (a), find the power delivered/absorbed associated with the 25 V source.p = vi = (vx)(αvx) or 25 × 15 = 375 W (absorbed)

T A P bl 2 3 d 2 4 d Ch P bl 2 6 d 2 8• Try Assessment Problems 2.3 and 2.4, and Chapter Problems 2.6 and 2.8

Kirchhoff’s Laws• Ohm’s law by itself is not sufficient to analyze circuits• Ohm s law by itself is not sufficient to analyze circuits.

– However, when it is coupled with Kirchhoff’s two laws, we have sufficient, powerful set of tools for analysing a large variety of electric circuits.

• These two laws are formally known as Kirchhoff’s Current Law (KCL) andThese two laws are formally known as Kirchhoff s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL).

• Constructing a circuit model:Label voltage and current associated– Label voltage and current associated with each resistor

– Label current of voltage source– Terminal dots are start and end pointsTerminal dots are start and end points

of an individual element– A node is where two or more elements meet: e.g. a, b, c, and d

• 7 unknowns requires 7 independent equations7 unknowns requires 7 independent equations– Ohm’s law The remaining 4

gives 3 of equations come fromthe equations applying Kirchhoff’s law

1 1 1

c c c

v i Rv i Rv i R

==

1

1

node a 0node b 0node c 0

s

c

c l

i ii i

i i

− =

+ =− − =

l l lv i R=node d 0l si i− =

Kirchhoff’s Current Law (KCL)• KCL states that the algebraic sum of currents entering a node in a circuit is zero• KCL states that the algebraic sum of currents entering a node in a circuit is zero.

10

N

nn

i=

=∑where N is the number of branches connected to the node and in is the nth current entering (or leaving the node)

– By this law, currents entering a node may be regarded as positive, while currents leaving the node may be taken as negative or vice versa.

1n= leaving the node).

• Example 2.4: Consider the node in the figure below.– Applying KCL gives

i1 + (-i2) + i3 + i4 + (-i5) = 0or

(-i1) + i2 + (-i3) + (-i4) + i5 = 0

i1 i2

( i1) i2 ( i3) ( i4) i5 0or

i1 + i3 + i4 = i2 + i5

Sum of entering currents = Sum of leaving currents

i3

i4 i5

Sum of entering currents Sum of leaving currents

Kirchhoff’s Voltage Law (KVL)• KVL states that the algebraic sum of all voltages around a closed path (or loop) in• KVL states that the algebraic sum of all voltages around a closed path (or loop) in

a circuit is zerowhere M is the number of voltages in the loop (or the number of branches in the loop) and vm is the0

M

mv =∑

• Example 2.5: Consider a single-loop circuit shown below.

the number of branches in the loop) and vm is the mth voltage.1

mm=∑

– Applying KVL yields

-v1 + v2 + v3 – v4 + v5 = 0

+ v2 - + v3 -

orv1 - v2 - v3 + v4 - v5 = 0

or

v4v1

- v5 +v2 + v3 + v5 = v1 + v4

– which may be interpreted as

5

Sum of voltage drops = Sum of voltage rises

• Example 2.6:– Applying KCL,

node a i1 + i4 – i2 – i5 = 0,node b i2 + i3 – i1 – ib – ia = 0,node c ib – i3 – i4 – ic = 0,node d i5 + ia + ic = 0.

• Example 2.7:– Applying KVL,

path a -v1 + v2 + v4 – vb – v3 = 0,path b -va + v3 + v5 = 0,path c vb – v4 – vc – v6 – v5 = 0,path d -va – v1 + v2 – vc + v7 – vd = 0.

• Example 2.8: The currents i1 and i2 in the circuit shown are 20A and 15A respectively.

) Fi d th li d b ha) Find the power supplied by each voltage source.

b) Show that the total power supplied equals to the total power dissipated

i0230V+ V1

2Ω

8Ω

i1i5

equals to the total power dissipated in the resistors.

(a) – V2 – V1 + V0 = 0 ⇒ V0 = V1 + V2

-4Ω

260V 16Ω

80Ω+ V2

+ V0-

i4

⇒ V0 = 20(8) + 16(15) = 400V⇒ i0 = 400/80 = 5A ⇒ i5 = 20 + 5 = 25A⇒ i3 = 15 + 5 = 20A ⇒ i4 = 20 – 15 = 5AP230V = -230(25) = -5750W (delivering)

2Ω -i2

i3

P230V 230(25) 5750W (delivering)P260V = -260(20) = -5200W (delivering)

(b) ΣPabs = (25)2(2) + (20)2(8) + (5)2(4) + (15)2(16) + (20)2(2) + (5)2(80) = 10950WΣPdel = 5750 + 5200 = 10950W∴ ΣPdel = ΣPabs = 10950W

• Try Chapter Problems 2 16 to 2 19• Try Chapter Problems 2.16 to 2.19

Circuit with dependent sources• Find voltage v• Find voltage v0

– 3 unknowns of iΔ , i0, and v0

– Knowing any 2 of the unknownswould find the 3rd onewould find the 3 one

• Solving for the currentsKVL loop a b c a ( )500 5 20 1i i= + L– KVL loop a-b-c-a,

– KCL at node b,

( )0500 5 20 1i iΔ= + L

( )0 5 6 2i i i iΔ Δ Δ= + = L

– Solving by substituting (2) into (1) for i0,

4A20 480V

iv iΔ = ⎫= =⎬ 0 0

0

20 480V24A

v ii

= =⎬= ⎭

• Example 2.9: Find the total power developed in the circuit shown if Vo = 100V.

A l i d A i ld Node 1Applying KVL around Loop A yields-60 + V1 = 100 ⇒ V1 = 160V

Likewise around Loop B igiΔ

2iΔ80V +

+V1-

Node 1

-V2 + 80 + 100 = 0 ⇒ V2 = 180Vand we know that iΔ= 4A.

60V 4A

V0

-+V2

LoopA

LoopB

Applying KCL at Node 1 yields-ig + iΔ + 2iΔ = 0 ⇒ ig = 3iΔ = 12A

-

Σ Pdel = 180(4) + 100(8) + 60(12) = 2240 W

– CHECK: Σ Pabs = 160(12) + 80(4) = 2240W

• Try Chapter Problems 2.28 to 2.30

Operational amplifier (op-amp)• An op-amp is an example of voltage-controlled dependent voltage source• An op-amp is an example of voltage-controlled dependent voltage source

– Output voltage is a function of the difference between the input voltages

– If a positive voltage is supplied to the non-inverting input, op-amp yields a positive output .

– If a positive voltage is supplied to the inverting input op-amp yields a negative outputIf a positive voltage is supplied to the inverting input, op amp yields a negative output. This op-amp circuit is known as inverting op-amp.

– Output voltage is constrained to a value between the negative and positive power supply voltage.

• Our interest is on terminal behavior of the op-amp– Black-box approach to its operation as affected by external circuit connections– Ignore the internal structure and voltage/current except that it imposes constraints on g g p p

the terminals

Op-amp terminals

( )( ) ( )0

CC p n CC

p n CC p n CC

V A v v V

v A v v V A v v V

⎧− − < −⎪⎪= − − ≤ − ≤ +⎨⎪

( ) ( )( )CC p n CCV A v v V

⎪+ − > +⎪⎩

• Actual op-amp– Input resistance between positive and negative input terminals is very large (105 to 1012

Ω)– Amplification factor (gain) is very high, ranges from 105 to 107 Ω– Output resistance is very low, ranges from 1 to 50 Ω

• Ideal op-ampp p– Input resistance is infinite– Amplification factor is infinite– Output resistance is zerop

Analyzing ideal op-amp• Two important equations for ideal op-amp• Two important equations for ideal op-amp

– Since input resistance is infinite,

If the output is to remain bounded within the power supply voltage then as the gain

0p ni i= =

– If the output is to remain bounded within the power supply voltage, then as the gain approaches infinity, the voltage across the input terminals must simultaneously become infinitesimally small, 0p n p nv v or v v− ≈ ≈

• If the gain is infinite, it is impossible to control the output– In actual application, voltage control is accomplished through feedback via a resistor

which feeds the output signal back to the inverting inputp g g p

• Example 2.10: Given va = 1 V and vb = 0 V, find vo

h i i l h l– Both input terminals have same voltage,

– Input currents are zero,( )0 V 1n p bv v v= = = L

( )– Taking note that,

( )25 100 0 A 2ni i i+ = = L

( ) 1 mAa nv vi

−= =

( )25

100

mA25k 25

mA100k 100o n o

i

v v vi

= =

−= =

– Substituting into (2),1 0 4 V25 100

oo

v v+ = ⇒ = −

100k 100

• Try Chapter Problems 5.3, 5.5 and 5.6

25 100


Resistors in Series and Parallel (3.1, 3.2)Voltage Division and Current Division (3.4)Delta-to-Wye Transformations (3.7)Delta to Wye Transformations (3.7)

Resistors in Series• Series-connected circuit elements carry the same current• Series-connected circuit elements carry the same current• Applying KVL around the whole loop

1 2 3 4 5 6 7 0s s s s s s s sv i R i R i R i R i R i R i R− + + + + + + + =

( )1 2 3 4 5 6 7s s s eq

orv i R R R R R R R i R

with R R R R R R R R

= + + + + + + =

= + + + + + +1 2 3 4 5 6 7eqwith R R R R R R R R= + + + + + +

• For k resistors connected in series, the equivalent single resistor has a resistance equal to the sum of the k resistances

k

R R R R R= = + + +∑ L1 21

eq i ki

R R R R R=

= = + + +∑ L

Resistors in Parallel• Parallel-connected circuit elements have the same voltage across their terminals• Parallel-connected circuit elements have the same voltage across their terminals• Applying KCL at node a,

1 2 3 41 1 1 1s s s s sv v v v vi i i i i v

⎛ ⎞= + + + = + + + = + + + =⎜ ⎟1 2 3 4

1 2 3 4 1 2 3 4s s

eq

i i i i i vR R R R R R R R R

+ + + + + + + + +⎜ ⎟⎝ ⎠

• For k resistors connected in parallel,1 1 1 1 1k

= = + + +∑ L

• It is more convenient to use conductance for parallel-connected resistors1 1 2ieq i kR R R R R=

+ + +∑

k

G G G G G+ + +∑ 1 21

eq i ki

G G G G G=

= = + + +∑ L

• For two resistors in parallel,

1 2 1 21 1 1 R R R RR+= + = ⇒ =

• Example 3.1: Find the equivalent resistance Rab

1 2 1 2 1 2eq

eq

RR R R R R R R

+ ⇒+

1(15)(30) 1015 30eqR = = Ω

+

7Ωa

2

15 301 1 1 1 1

24 30 20 8

8

q

eqR

R

+

= + + =

⇒ Ω

24Ω 30Ω 20Ω15Ω 30Ω

b

2 8

(10)(7 8) 610 (7 8)ab

eqR

R

⇒ = Ω

+∴ = = Ω

+ +

7Ω

R

a

R R +7Ω

a

RReq2

b

Req1 Req2+7Ω

b

Req1

• Example 3.2: Find the equivalent resistance between terminals a-b.

(48)(10 6)+10Ω

a a (48)(10 6) 1248 (10 6)

(15)(12 18) 1015 (12 18)

eq

ab

R

R

+= = Ω

+ ++

∴ = = Ω+ +

a

15Ω 6Ω48Ω

a

15Ω Req

• Example 3 3: For the circuit shown find

15 (12 18)+ +b

18Ωb

18Ω

• Example 3.3: For the circuit shown, find a) the voltage v,b) the power delivered to the

circuit by the current sourcecircuit by the current source, c) the power dissipated in the

10 Ω resistor

(a) Work out the equivalent resistance,

( ) ( )1(64)(6 10) (30)(7.2 12.8)12.8 1264 (6 10) 30 (7.2 12.8)eq eqR R+ +

= = Ω ⇒ = = Ω+ + + +( ) ( )( ) ( )

a) With Req, v = 5 × 12 = 60 V

b) For the 5A source, the 5A current and voltage v is of non-passive sign convention,

p5A = -5 × v = -5 × 60 = -300 W (delivered)p5A 5 × v 5 × 60 300 W (delivered)

c) The power dissipated can be found by finding the current through the 10 Ω resistor,30

7.2

10

305 3A30 20

643 2.4A64 16

i

i

Ω

Ω

= × =+

= × =10

210

64 162.4 10 57.6Wp

Ω

Ω

+= × =


Voltage Division• The principle of voltage division – the voltage v is divided among the resistors in• The principle of voltage division – the voltage v is divided among the resistors in

direct proportion to their resistances– the larger the resistance, the larger the voltage drop across it.

1 21 2

1 2 1 2

;R Rv v v vR R R R

= =+ +v

+ v1 -

R1 R2

+ v2 -

• In general, if a voltage divider has N resistors (R1, R2, …, RN) in series

1 2 1 2R R R R+ +

g g ( 1 2 N)with the source voltage v, the nth resistor (Rn) will have a voltage drop of

;N

n ni

R Rv v v R R= = =∑

• Condition to apply voltage division - the resistors must have the same current

11 2

;n eq iiN eq

v v v R RR R R R =+ + + ∑

L

• Example 3.4: Find v1, v2, and v3 in the circuit shown below.

i bi h Ω d Ω i ll l 14Ω 10Ω– First combine the 10Ω and 15Ω in parallel

Ω=+

= 61510

)15(10eqR

40V

+ v1 -

14Ω 10Ω

+ v3 -

15Ω+v2

– Then apply the voltage division,

+1510

14 (40) 28V

40V 5 v2-

1

2 3

(40) 2814 6

6 (40) 1214 6

v V

v v V

= =+

= = =+

• Common mistake

14 6+

• Example 3.5:a) Find no-load voltage v0

b) i d h i kΩb) Find v0 when RL is 150 kΩc) How much power is dissipated in

the 25 kΩ resistor if the load terminals are accidently shortedterminals are accidently shorted.

075) (200) 150 V

25 75a v = =

+( )( )( )

25 7575 150

) 50 k75 150eqb R

+

= = Ω+

0

2

50 (200) 133.3 V25 50

200

v = =+

25200) 1.6 W25kkc p Ω = =

Current Division• The principle of current division – the total current i is shared by the resistors in

inverse proportion to their resistances– the lower the resistance, the higher the current through it

R2

2 11 2

1 2 1 2

;R Ri i i iR R R R

= =+ +v

i2

R1

i i1

• In general, if a current divider has N resistors (R1, R2, …, RN) in parallelwith the source current i the nth resistor (R ) will have a current ofwith the source current i, the nth resistor (Rn) will have a current of

1 1;N

eqn

Rvi iR R R R

= = =∑• Condition to apply current division - the resistors must have the same voltage• or using conductance,

1in n eq iR R R R=

N

11 2

;N

n nn eq i

iN eq

G Gi i i G GG G G G =

= = =+ + + ∑

L

• Example 3.6:Find i1 through i4 in the circuit shown below.

We first combine resistors in parallel,

110(40) 8eqR = = Ω40Ω i

10Ω

30Ω

20Ωi4

i

i2

1

2

810 4020(30) 1220 30

eq

eqR

+

= = Ω+

20A

i1i3

Using the current division rule:

1 2 18 20(20) 8 (8) 3.2i i A i A+ = = ⇒ = =1 2 1

2

( ) ( )8 12 50

30 (8) 4.850

i A

+

= =

3 4 312 10(20) 12 (12) 2.4

8 12 5040 (12) 9 6

i i A i A

A

+ = = ⇒ = =+

440 (12) 9.650

i A= =

• Practice problems(a) Find the equivalent resistance Rab.

3Ω

1 5Ω

6Ω

(Ans: Rab = 10 Ω)ab

1.5Ω5Ω

12Ω

1Ω40Ω45Ω

(b) For the circuit shown,

5.2Ω15Ω

(b) For the circuit shown,use current division to find i0, and use voltage division to find vo.

(A i 2A 18V)(Ans: io = 2A; vo = 18V)

• Try Chapter Problems 3.21, 3.22 and 3.23

Delta-to-Wye Equivalent Circuits• In the analysis of electric networks engineers often encounter the Y (star) and Δ• In the analysis of electric networks, engineers often encounter the Y (star) and Δ

(delta) networks.– Delta interconnections

• R1 R2 and RR1 , R2 , and Rm

• R3 , Rs , and Rm

– Wye interconnections• R1 , R3 , and Rm1 3 m

• R2, Rs , and Rm

• Delta is also called pi (π)p ( )– Δ can be shaped into π without

disturbing electrical equivalence

• Star is also called tee (T)– Y can be shaped into T without

disturbing electrical equivalenceg q

Delta-Wye Transformation• For the two networks to be equivalent at each corresponding pair of terminals the• For the two networks to be equivalent at each corresponding pair of terminals, the

resistance at each respective pair of terminals must be equal while the third terminal is open-circuited.

( )1 2

c a bab

R R RR R R

R R R+

= = ++ +

( )2 3

a b c

a b cbc

a b c

R R RR R R

R R RR R R

+ +

+= = +

+ +

( )3 1

a b c

b c aca

a b c

R R RR R R

R R R+

= = ++ +

• Straightforward algebraic manipulation of previous equations gives values for Y-connected resistors in term of Δ-connected resistors and vice versa.

1 2 2 3 3 11

1

b ca

a b c

R R R R R R R RR RR R R R

R R R R R R R R

+ += =

+ +

+ +1 2 2 3 3 12

2

1 2 2 3 3 1

c ab

a b c

a b

R R R R R R R RR RR R R R

R R R R R R R RR R

+ += =

+ +

+ += =

• Generally, the equations used for star-delta transformation can be expressed in the f ll i f

33

ca b c

R RR R R R

= =+ +

following forms:

YProduct of the two nearest branches

Sum of resistances R Δ

=Δ

Sum of cyclic products of two Y branchesResistance of furthest Y branch

RΔ =

• For the balanced case,

1 2 3 andY a b cR R R R R R R RΔ= = = = = =1 and 33Y YR R R RΔ Δ= =

• Example 3.7: Consider the circuit shown , find the input resistance seen by the source.

– Four transformations are possible for the abcd network:

• Δabc, 1Ω2Ω 2Ω

v

a

b• Δbcd, • Y at b, and • Y at c.

2Ω 1Ω

v

1Ω

b c

dd

(a) Δabc transformation to Y

0.8Ω

a

2 2 0.82 2 1anR ×

= = Ω+ +

v0.4Ω 0.4Ω

b

n2 2 1

2 1 0.42 2 1

2 1

bnR

+ +×

= = Ω+ +

(b) Summing up resistances in paths nbd and ncd

2Ω 1Ω1Ω

b c

d

2 1 0.42 2 1cnR ×

= = Ω+ +

nbd and ncd(c) Combining resistances between

nodes n and d

d

a a

(a)

( ) 2.4 1.42 4 ||1 4 ×Ω Ω =

The input resistance,

0.8Ωn

v

0.8Ωn

v

( )2.4 ||1.42.4 1.40.884

Ω Ω =+

= Ω

• Try Chapter Problems 3 52 and 3 54 1Ω d

1.4Ω2.4Ω

1Ω d

0.884Ω0.8 0.884 1

2.684inputR = + +

= Ω• Try Chapter Problems 3.52 and 3.54 1Ω d 1Ω d

(b) (c)


Node Voltage Method or Nodal Analysis (4.2, 4.3, 4.4)

Planar Circuits• Planar circuits are those that can be drawn on a plane with no crossing branches• Planar circuits are those that can be drawn on a plane with no crossing branches

Planar circuit

Non-planar circuit

Descriptions of a circuitName Definition Examples

node (n) A point where 2 or more elements join

a, b, c, d, e, f, g

essential node

A node where 3 or more elements join

b, c, e, gnode (ne)

elements join

path A set of elements may be traversed in order

i h i h

many

• To solve be unknowns( 1) KCL ti

without passing the same node twice

branch (b)

A path that connects 2 nodes

v1, v2, R1, R2, R3, R4, R5, R6, – (ne-1) KCL equations

– Needs be - (ne-1) KVL equations

( ) 3, 4, 5, 6,R7, I

essential branch (b )

A path that connects 2 essential nodes without passing through an

v1-R1, R2-R3, v2-R4, R5, R6, R I

1 2 6 00

i i i Ii i i

− + + − =(be) passing through an

essential nodeR7, I

loop A closed path many

mesh A loop that does not v1-R1-R5-R3-R2,( )

1 3 5

3 4 2

1 1 1 2 5 3 2 3

00

i i ii i i

v i R i R i R R

− − =+ − =

= + + +mesh A loop that does not

enclose any other loopsv1 R1 R5 R3 R2, v2-R2-R3-R6-R4, R5-R7-R6, R7-I

( )2 3 2 3 4 6 5 4

2 5 6 7 4 60v i R R i R i R

i R i R i R= − + + +

= − + −

Examples of circuit analysis solutions• 3 essential nodes (n = 3) a b• 3 essential nodes (ne = 3)

– denoted by a, b, c• 5 essential branches (be = 5)

(10V 1Ω) 2Ω 5Ω 10Ω d 2A

i1 i2i3 i4

– (10V-1Ω), 2Ω, 5Ω, 10Ω and 2A• (ne-1) or 2 KCL equations at nodes a and b, c

1 2 3 0i i i− − =

• Since 1 branch current is already given (i.e. 2A), we need to have((be-1)-(ne-1)) or 2 KVL equations,

2 42 0i i+ − =

• 4 unknowns can be solved using the above set of 4 independent equations

1 3

3 2 4

10 5 05 2 10 0

i ii i i− − =

− − =4 unknowns can be solved using the above set of 4 independent equations– Solution gives currents through all of the branches

• i1 = 0.91A, i2 = -0.91A, i3 = 1.82A, i4 = 1.09A– Voltage across all branches can subsequently be computedVoltage across all branches can subsequently be computed

• vac = 9.09V, vbc = 10.91V, vab = -1.82V

Cramer’s method of solving set of equations• Value of unknown variable in set of equations expressed as ratio of 2 determinants• Value of unknown variable in set of equations expressed as ratio of 2 determinants• After substituting for i4 as i4 = i2+2,

3 equations are left for i1, i2 and i3

1 2 3

1 3

05 10

i i ii i− − =

+ =

• To find i12 312 5 20i i− + =

0 1 110 0 520 12 51

10 0 ( 50) ( 100) 120 0 70 0 91Ni

− −

− − − − + − + −= = = = =

• and i2, i3

1 1 1 11 0 50 12 5

1 0 1 1 1 0

0.910 ( 60) ( 5) 0 12 0 77

i − −

−

− −

Δ − − − − + + −

1 10 5 1 0 100 20 5 0 12 2032 50 100 20 120 20

2 377 771 1 1 1 1 11 0 5 1 0 50 12 5 0 12 5

0.91 ; 1.82NNi i −− − +− − − −= = = = − = = = =

Δ Δ

• Determinant:1 2 3

1 2 3 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1

a a ab b b a b c a b c a b c a b c a b c a b c= − − + + −

0 12 5 0 12 5− −

• Read Appendix A.2, A.3 and A.41 2 3c c c

Nodal Analysis or Node-Voltage Method• The circuit layout must be neat (no crossover) and mark clearly the essential nodes• The circuit layout must be neat (no crossover) and mark clearly the essential nodes

on the circuit diagram.• Select one node to be a reference node (e.g node connected to the greatest

number of branches or ground node)number of branches or ground node).• Define a voltage between each remaining node and the reference node ((N-1)

voltages defined in an N-node circuit).• Appl KCL at each of the nodes res lting in a set of (N 1) node eq ations• Apply KCL at each of the nodes, resulting in a set of (N-1) node equations.• Following the same method, formulate a set of independent equations. Solve the

equations simultaneously.v1 v2

• With node 3 as the reference, voltagesat nodes 1 and 2 are:

1 2v1 v2

+ +

1 13

2 23

v vv v== 3

v13 v23

- -

• Example 4.1: For the circuit shown,compute the voltages across each of the current sources R2=8Ωv1 v2Node 1 Node 2of the current sources.– 3 essential nodes– Setting one as the reference node

and left with 2 nodes

ig11A R1=8Ω R3=16Ω

ig2-2A

i2i1 i3

and left with 2 nodes– Apply KCL at 2 remaining nodes,

giving two equations involving v1 and v2. Node 3 (reference node)

– KCL @node 1:

( )1 1 21 1 2 1 1 2

1 2

0 0 2 8 1g gv v vi i i i v vR R

−− − = ⇒ − − = ⇒ − + = − L

– KCL @node 2:

( )1 2 22 3 2 2 1 2

2 3

0 0 2 3 32 2g gv v vi i i i v v

R R−

− − = ⇒ − − = ⇒ − = − L

– Solve for v1 and v2 by first adding (1) and (2),

– Substitute for v2 in (1),2 22 40 20Vv v− = − ⇒ =

1 1 12 20 8 2 28 14Vv v v− + = − ⇒ = ⇒ =

• Example 4.2: Use node voltage method onsimilar circuit in previous chapter

Ch b tt d f– Choose bottom node as reference

– KCL @node 1: KCL @node 2:10

( )

1 1 1 2

1 2

10 01 5 2

17 5 100 1

v v v v

v v

− −+ + =

− = L ( )

2 1 2

1 2

2 02 10

5 6 20 2

v v v

v v

−+ − =

− + = L

– From (2),

Sub into (1)

12

5 206

vv +=

– Sub. into (1),

( )

11 1 1

5 20 77 700 70017 5 100 9.09V6 6 6 77

5 9 09 20

vv v v+⎛ ⎞− = ⇒ = ⇒ = =⎜ ⎟⎝ ⎠

+

• Node voltage method is a simpler solution with 2 instead of 3 equationsT Ch P bl 4 9 4 10

( )2

5 9.09 2010.91V

6v

+= =

• Try Chapter Problems 4.9, 4.10

Node voltage method with dependent sources• For circuit with dependent sources the node-voltage equations must be• For circuit with dependent sources, the node-voltage equations must be

supplemented with constraint equations– Constraint equations are imposed by the dependent sources

• Example 4.3: Find power dissipated by the 5 Ω resistor in the circuit below.

– KCL equations:

( )1 1 1 220v v v v− − ( )

( )

1 1 1 21 2

22 1 21 2

20 0 0.75 0.2 10 12 20 5

80 0.2 0.8 4 2

5 10 2

v v v v v v

v iv v v v v iφφ

+ + = ⇒ − =

−−+ + = ⇒ − + =

L

L( )1 25 10 2 φ

– The constraint equation,

S b i i t (2) d l f d

1 2

5v viφ−

=

16V 10V– Sub. iφ into (2) and solve for v1 and v2,

– Subsequently,

1 216V ; 10Vv v= =

25

16 10 =1.2A 5 7.2W5

i p iφ φΩ

−= ⇒ = × =

• Homework: Make constraint equation as the 3rd equation and use Cramer’s method to solve for iφ

5


Node voltage method with voltage sources• For branches having voltage sources nodal analysis is difficult since currents• For branches having voltage sources, nodal analysis is difficult since currents

through voltage sources are not known. – Need to introduce a new variable i to

denote the current through the gunknown voltage source

– Note that v1=50V and thus left 1with 2 KCL equations at nodes 2 and 3:

32 250 0 ; 4 05 50 100

vv v i i−+ + = − + − =

– Constraint equations due to the 10iφ voltage source and it being dependent:

23 2

5010 and5

vv v i iφ φ−

= + =

– 4 equations to solve for the 4 unknowns• Introduction of a new variable requires an additional equation to solve• Instead of introducing a new variable, we can apply Supernode conceptg , pp y p p

Supernode concept• To eliminate the need to know the current through voltage sources while applying• To eliminate the need to know the current through voltage sources while applying

KCL, enclose the voltage sources and consider each enclosed surface as a generalized node or supernode.

• KCL holds good for ordinary node as well as for a supernode Use KCL Ohm’s• KCL holds good for ordinary node as well as for a supernode. Use KCL, Ohm s law and supernode to solve circuit analysis problem.

• Principle of supernode - a supernode is formed by enclosing a (dependent or independent) voltage source connected in between two reference nodes and anyindependent) voltage source connected in between two reference nodes and any elements connected in parallel with it.

• A supernode may be regarded as a closed surface enclosing the voltage source and its two nodesits two nodes.

• KCL @ supernode, ( )32 250 4 0 15 50 100

vv v−+ + − = L

( )10 2i• Constraint equation due to supernode,

• Constraint equation from dependent source control variable,

( )3 2 10 2v v iφ= + L

( )2 50 35

viφ−

= L

• Simplify to a set of 3 equations and solve for v2, v3 and iφ

( )5φ

2 322 1400v v+ = 1400 1 02 3

2 3

2

22 140010 0

5 50

v vv v i

v iφ

φ

+− + =

− =2

0 1 1050 0 5 7000 500 60V22 1 0 110 5 10

v

−− +

= = =2 22 1 0 110 5 101 1 101 0 5

+ +−

−

Note: 1 fe er eq ation to sol e hen sing S pernode concept

( )360 50 2A ; 60 10 2 80V

5i vφ

−= = = + =

• Note: 1 fewer equation to solve when using Supernode concept

• Example 4.4: Analyze the circuit with a supernode as shown.

KCL @ d4Ω

supernode– KCL @ supernode,

( )1 4 2 3

1 3 31 2 2 1

i i i iv v vv v v

+ = +−−

+ = + L

i1

v1v2

p

2Ω

5Vv3

i4

– Constraint equation,

( )12 4 8 6

+ = + L

( )2 3 5 2v v= + L

10V 8Ω 6Ωi2 i3

• Noting that v1 = 10V, (1) becomes ( )

( )32 55 15 38 12 2

vv+ = L

• Using (2) and (3),8 12 2

( )3 35 5 5 15 4 2Vv v v+

+ = ⇒ =3

2

4.2V8 12 24.2 5 9.2V

v

v

+ = ⇒ =

= + =

Properties of a Supernode• The voltage source inside the supernode provides a constraint equation needed to• The voltage source inside the supernode provides a constraint equation needed to

solve for the node voltage.• A supernode has no voltage of its own.

A d i th li ti f b th KCL d KVL• A supernode requires the application of both KCL and KVL.– KCL to relate to external circuit elements– KVL to relate the voltages of the two combined nodes

• Practice problem:Find v and i in the circuit shown. i+

4Ω 3V

(Ans: v = -0.2V, i = 1.4A)7V 3Ω 2Ω

+v-

6Ω


Summary – Node Voltage Method or Nodal Analysis

1. Select one node to be a reference node2. Define node voltages3 Apply KCL at each of the nodes3. Apply KCL at each of the nodes

Resulting in a set of (N-1) node equations.The equation variables are node voltages

Three cases:1. Normal circuits (without neither dependant sources

nor supernode)2. The circuit with dependant sources

Constraint equations (KVL) for control variablesConstraint equations (KVL) for control variables3. Supernode

Constraint equations (KVL)


Mesh Current Method or Mesh Analysis (4.5, 4.6, 4.7)Source Transformation (4.9)Thevenin and Norton equivalents (4.10)Thevenin and Norton equivalents (4.10)

Mesh Current Method or Mesh Analysis• Consider the closed path of elements in the circuit passing through no node or• Consider the closed path of elements in the circuit passing through no node or

element more than once as a mesh (or a loop).• Assume mesh current that flows around a mesh in a chosen direction (clockwise).

A i it l t b t i t o meshes• A circuit element may be present in two meshes. – The total current through the element is the algebraic sum of the mesh currents (taking

direction into account).• Apply KVL and Ohm’s law and obtain set of mesh equations• Apply KVL and Ohm’s law and obtain set of mesh equations. • Solve the equations simultaneously to determine the unknowns.

• Examples of meshes in a planar circuit– Each circuit element can be in

two meshes at the maximum– Branch currents

are defined in term of mesh currents e g

1

6

1

1 2

R

R

i i

i i i

i i i

=

= −currents, e.g.

8 2 3Ri i i= −

• To determine branch currents through all resistors.– be = 3, ne = 2 1 KCL and 2 KVL equations

1 2 3

1 1 1 3 3

2 2 2 3 3

i i iv i R i R

v i R i R

= +

= +− = −

– Need to solve for a set of 3 equations

A l i h h d

2 2 2 3 3v i R i R

• Applying mesh-current method– Identify the (be-(ne-1)) meshes and

define mesh currents (with direction)KVL f h i– KVL for mesh ia

– KVL for mesh ib

( )1 1 3a a bv i R i i R= + −

– Only need to solve a set of 2 equations for the mesh currents ia and ibTh b h t b d i t f th h t

( )2 3 2b a bv i i R i R− = − + 1

2

a

b

i ii i==

• The branch currents can be expressed in terms of the mesh currents,3 a bi i i= −

• Example 5.1: Use mesh-current method to find power delivered/absorbed by the 80 V source, and power dissipated by the resistorsby the resistors.

i i

ic

• Mesh-current equations:

h

ia ib

( ) ( )– Mesh ia,

Mesh i

( ) ( )( )

( ) ( )

80 5 26

31 26 5 80 1

26 90 8 0

a c a b

a b c

i i i i

i i i

i i i i i

= − + −

⇒ − − =

+ +

L

– Mesh ib,

– Mesh i

( ) ( )( )

( ) ( )

26 90 8 0

26 124 90 0 2

5 30 90 0

b a b c b

a b c

i i i i i

i i i

i i i i i

− + − + =

⇒ − + − =

− + + − =

L

Mesh ic, ( ) ( )( )

5 30 90 0

5 90 125 0 3c a c c b

a b c

i i i i i

i i i

+ + =

⇒ − − + = L

– Using Cramer’s rule,80 26 50 124 90

− −0 124 900 90 12531 26 526 124 905 90 125

592000 5A118400ai

−−− −

− −− −

= = = 80V

2

80400W (delivered)

ap i= − ×

= −5 90 125

31 80 526 0 905 0 125

31 26 5

296000 2.5A118400bi

−− −−

− −= = =

( )

( )

25

230

2

5 45W

30 120Wa c

c

p i i

p iΩ

Ω

= − × =

= × =31 26 526 124 905 90 125

31 26 8026 124 0

118400b− −− −

−

( )( )

226

290

2

26 162.5W

90 22.5Wa b

b c

p i i

p i iΩ

Ω

= − × =

= − × =26 124 05 90 0

31 26 526 124 905 90 125

236800 2A118400ci

−− −

− −− −− −

= = =2

8 8 50Wbp iΩ = × =


5 90 125

Mesh-current method with dependent sources• Is the strategy of mesh analysis still effective if the circuit contains current• Is the strategy of mesh analysis still effective if the circuit contains current

sources and/or dependent sources?– If the circuit contains dependent sources, the mesh current equations must be

supplemented with the constraint equations imposed by the presence of the pp q p y pdependent sources.

• Example 5.2: Use the mesh analysis to determine the power dissipated in the 4 Ω resistor in the circuit shown below.– Mesh-current equations,

( ) ( )1 2 1 350 5 20i i i i= − + −

( ) ( ) ( )( ) ( )2 1 2 2 3

3 1 3 2

0 5 1 4

0 20 4 15

i i i i i

i i i i iφ

= − + + −

= − + − +

– Constraint equation imposed by presence of 15iφ source,

1 3i i iφ = −

• Substitute iφ into the mesh-current equations,

( )1 2 325 5 20 50 1i i i− − = L( )( )( )

1 2 3

1 2 3

1 2 3

5 10 4 0 2

5 4 9 0 3

i i i

i i i

− + − =

− − + =

L

L

• To find power dissipated by 4Ω resistor, we compute mesh currents i2 and i3.25 50 20

5 0 43250

−− −

5 0 92 25 5 20

5 10 45 4 9

3250 26A125

i −− −

− −− −

= = =

( )24 3 2 4 16Wp i iΩ = − × =

25 5 505 10 05 4 0

3 25 5 20

3500 28A125

i

−−− −

− −= = =

( )

• Try Chapter Problems 4 37 and 4 38

5 10 45 4 9

125− −− −

Try Chapter Problems 4.37 and 4.38

• When a branch includes a current source the mesh analysis method requires

Mesh-current method with current sources• When a branch includes a current source, the mesh analysis method requires

some additional manipulations.• Example 5.3: Find the mesh currents ia, ib, and ic.

When we attempt to sum voltages around– When we attempt to sum voltages around either mesh ia or ic, we must introduce into the equations the unknown voltage across the 5 A current source. Thus,

– Mesh ia:

– Mesh ic:

( ) ( )100 3 6 1a b ai i v i= − + + L

( ) ( )2 50 4 2c b cv i i i= − + + L

– Combine (1) and (2) to eliminate v,

– KVL around mesh b

( )50 9 5 6 3a b ci i i= − + L

( ) ( ) ( )0 3 10 2 4i i i i i= − + + − L– KVL around mesh b,

– Taking note that:

( ) ( ) ( )0 3 10 2 4b a b b ci i i i i= + +

( )5 5a ci i− + = L

– Solve (3), (4) and (5) ia = 1.75 A, ib = 1.25 A, ic = 6.75A

Supermesh concept• We can create a kind of supermesh from two meshes where a current source is a• We can create a kind of supermesh from two meshes where a current source is a

common element; the current source is in the interior of the supermesh.• Apply KVL and Ohm’s law in meshes without the current source. Then solve the

equations obtained to find the unknownsequations obtained to find the unknowns.

• In Example 5.3, we can derive equation (3) itho t introd cing the nkno n oltagewithout introducing the unknown voltage

variable v by using supermesh concept.

Remo e the c rrent so rce from eq ation b– Remove the current source from equation bysimply avoiding it when writing the mesh-current equations

– KVL around the supermesh, ( ) ( )100 3 2 50 4 6b bi i i i i i= − + − + + +KVL around the supermesh,

– This equation simplifies to equation (3), which is solved with (4) and (5)• Try Chapter Problems 4 41 and 4 47

( ) ( )100 3 2 50 4 6a b c b c ai i i i i i+ + + +

Try Chapter Problems 4.41 and 4.47

• Taking time to look carefully at a circuit to identify shortcuts such as the above supermesh provides a big payoff in simplifying the analysis.I l d l h l i h d hi h i i l ( d• In general, we use nodal or mesh analysis method, whichever is simpler (and appropriate – what the question asks for: voltages or currents?)

• Practice problems(a) Use the nodal analysis method to

find the power delivered/absorbed b th 50 V ltby the 50 V voltage source.

Ans: P50V = - 150 W (delivered)

(b) Use the mesh analysis method to find the mesh current ia.

Ans: 15 A

Summary – Mesh Current Method

1. Select meshes 2. Define mesh currents3 Apply KVL at each of the meshes3. Apply KVL at each of the meshes

Resulting in a set of (be-(ne-1)) equations.The equation variables are mesh currents

Three cases:1. Normal circuits (without neither dependant sources

nor supermesh)2. The circuit with dependant sources

Constraint equations (KCL) for control variablesConstraint equations (KCL) for control variables3. Supermesh

Constraint equations

Useful circuit analysis techniques• There are many techniques available to simplify circuits before applying node-• There are many techniques available to simplify circuits before applying node-

voltage and mesh-current methods

F l if i th5Ω3Ω

• For example, if you are given the following circuit, are there any other alternative(s) to determine the voltage across 2Ω resistor?

20V8A2Ω+vo-

across 2Ω resistor?

• Some of the common techniques for circuit simplificationS i ll l d i– Series-parallel reductions

– Δ-to-Y transformations– Source transformations

Th i d N i l– Thevenin and Norton equivalents– Superposition

Source transformations• A voltage source in series with a resistor is equivalent to a current source in• A voltage source in series with a resistor is equivalent to a current source in

parallel with the same resistor provided vs = is × R

• Example 5.4: For the circuit shown, find the power associated with the 6 V source.

– There are several approaches to calculate the power associated with the 6V source. However, all the approaches will focus on finding just one branch current in the 6V source. As a result, we first simplify the circuit by using source transformations.

• Note we must reduce the circuit in a way that preserves the identity of the branch containing the 6V source. There is no need to preserve the identity of the branch containing the 40V source or any of the resistorscontaining the 40V source or any of the resistors. – Starting with this branch, we can transform the 40V source in series with 5Ω resistor to

8A source in parallel with 5Ω resistor, and so on.

6V 6V19.2 6 0.825A 6 0.825 4.95W (absorbed)4 12

i p−= = ⇒ = × =

+

• Example 5.5: Determine the equivalent circuit between the two terminals in figure below via source transformation.

4V1A 2Ω

2Ω1A

2Ω2A 2Ω 3A 1Ω

4V

1Ω3Ω2A

3Ω3A 1Ω 1A 0.75Ω

3V6V3Ω

3V

1Ω

1.75Ω1.75Ω2 1/7AOR

0.75V

0.75Ω3.75V

• Try Assessment Problem 4.15 and Chapter Problem 4.59

Ignoring independent sources• In circuit analysis we often need to ignore contribution from certain independent• In circuit analysis, we often need to ignore contribution from certain independent

sources. We can do that by turning them off (“killing” them).– Turning off independent voltage source, Vs = 0 replacing it with a short-circuit– Turning off independent current source I = 0 replacing it with a open-circuit– Turning off independent current source, Is 0 replacing it with a open-circuit

20Ω

1.6Ω

20Ω 60V

1.6Ω

T i ff ll lt

8Ω36A20Ω

5Ω6Ω

8Ω36A

120V 5Ω6Ω

C l t i it

Turning off all voltage sources

20Ω 60V

1.6Ω

Complete circuit

Turning off all current sources

8Ω120V 5Ω

6Ω

Turning off all current sources

Thevenin and Norton equivalents• At times in circuit analysis we want to concentrate on what happens at a specific• At times in circuit analysis, we want to concentrate on what happens at a specific

pair of terminals.

F l Wh l li i t l t i it tl t• For example: When we plug an appliance into an electricity outlet, we are interested primarily in the voltage and current at the terminals of the appliance. We have no interest in the effect that connecting the appliance has on voltages and currents elsewhere in the circuit supplying the outlet In other words we want tocurrents elsewhere in the circuit supplying the outlet. In other words, we want to focus on the behavior of the circuit supplying the outlet, but only at the outlet terminals.

• Thevenin and Norton Equivalents are circuit simplification techniques that focus on terminal behavior and thus are extremely valuable aids in circuit analysisanalysis.

Thevenin’s theorem• Any linear network can be replaced by an independent voltage source in series• Any linear network can be replaced by an independent voltage source in series

with a resistor such that the current-voltage relationship at the terminals is unchanged.

• In other words a linear two-terminal aI• In other words, a linear two-terminal

circuit (a) can be replaced by an equivalent circuit (b) consisting of a voltage source VTh in series with a

+V-

LoadLinear

two-terminal circuitg Th

resistor RTh,

– VTh is the open-circuit voltage at

b(a)

Th p gthe terminals

– RTh is the input or equivalent resistance at the terminals when the internal i d d d ff V

RTh

+V L d

aI

independent sources are turned off.

(b)

VTh V-

Load

b(b)

Norton’s theorem • Norton’s theorem is identical to Thevenin’s statement except that the equivalent• Norton s theorem is identical to Thevenin s statement except that the equivalent

circuit is an independent current source in parallel with a resistor.• It states that a linear two-terminal circuit

can be replaced by an equivalent circuitaI

can be replaced by an equivalent circuit of a current source IN in parallel with a resistor RN,

+V-

LoadLinear

two-terminal circuit

– IN is the short circuit current through the terminals

– RN is the input or equivalent resistance at

b(a)

aIRN is the input or equivalent resistance at the terminals when the internal independent sources are turned off.

IN RN

+V-

Load

a

• Note: Thevenin’s and Norton’s equivalent circuits are related by source transformation.

ThVR R I(b)

b

; ThN Th N

Th

R R IR

= =


Deriving equivalent circuits (4.10, 4.11)Maximum power transfer (4.12)Superposition (4.13)Superposition (4.13)

• If only independent sources are present calculate:

Method A for determining equivalent circuit• If only independent sources are present, calculate:

(a) Thevenin's equivalent resistance RTh with independent sources killed (i.e. short-circuit all voltage sources and open-circuit all current sources).

(b) Either the open-circuit terminal voltage (v = i RTh)(b) Either the open circuit terminal voltage (voc iscRTh) or the short-circuit current (isc = voc/RTh).

• Example 6 1: Using Thevenin’s theorem find the equivalent circuit to the left ofExample 6.1: Using Thevenin s theorem, find the equivalent circuit to the left of the terminals a-b in the circuit shown below.

6Ω6Ω a

i

4Ω2A12V 1Ω

i

– Ignoring the 1Ω resistor, determine an equivalent circuit to represent the circuit to its left (i.e. simplifying the circuit while retaining the 1Ω resistor).

b

• Solving for RTh (kill all the independent sources),6Ω6Ω

( )( )12 412 || 4 3

12 4ThR = = = Ω+4Ω RTh

• Solving for VTh, 6Ω+

6Ω+

6Ω 4Ω2A2A VTh

-

6Ω 4Ω4A VTh

-

6Ω6Ω

24V

+3Ω

a

6V4Ω24V VTh

-

( )4

6V

b

( )4 24 6V4 6 6ThV = =+ +

• Example 6.2: Consider the circuit below, find Io using Norton’s theorem.

3kΩ6kΩ

2kΩ 2mA6V 3kΩI

– Note: the 2kΩ resistor is considered as a load, which is ignored when deriving the equivalent circuit.

Io

q

• Solving for RN

( )( )( )6k || 3k 3k

6k 6kNR = Ω Ω+ Ω

Ω Ω

3kΩ6kΩ

3kΩ ( )( )6k 6k3k

6k 6kΩ Ω

= = ΩΩ+ Ω

RN

3kΩ

• Solving for IN

3kΩ6kΩ

( )6V 3k 2mA6k 3k 3k2mA

NI Ω= +

Ω Ω+ Ω=

2mA6V 3kΩIN

• Connect back the 2kΩ resistor to the Norton’s equivalent circuit and solve for Io

2mA=

– The Norton equivalent circuit is drawn to the left of the 2 kΩ resistor. Using current division, the current Io in the 2 kΩ resistor is

( )3k 2mA 1.2mA3k 2koI Ω

= =Ω+ Ω2mA 3kΩ 2kΩ

IIo

Method B for determining equivalent circuits• If dependent and independent sources are both present determine the Thevenin• If dependent and independent sources are both present, determine the Thevenin

equivalent by calculating:(a) Terminal open-circuit voltage voc and short-circuit current isc. (b) R = v /i(b) RTh voc/isc.

• Example 6.3: Find Vo for the circuit shown below.

2Vx

24V-

2Ω

4Ω 3A24V

+ Vo -

Vx+

– The 2Ω resistor is the load and not included in equivalent circuit.

• Find the open-circuit terminal voltage after the 2 Ω load is removed.

2V

3 4 12VXV = − × = −

2Vx

4Ω 3A24V-

Vx

24 2 0 12Voc X X ocV V V V+ + + = ⇒ =

+ Voc -

+

• Find the short-circuit current.

2Vx

24 2 0 8V

2A

X X X

X

V V VVI

+ + = ⇒ = −

= = −4Ω 3A24V-

Vx+ 2A

43 1ASC

I

I I= + =Isc

+ I

• Find RTh and form the Thevenin’s equivalent circuit

12 12 12VocVR V VΩ

– (Note: This is not equal to 4 Ω since the presence of the dependent source has altered th ff ti i t f th i t k )

12 ; 12V1

ocTh Th oc

sc

R V VI

= = = Ω = =

the effective resistance of the given network.)

• Connect the Thevenin equivalent circuit to the 2 Ω load resistor and solve for Vo

RTh = 12Ω

( )2 1212 V2 12 7oV = =+

Voc = 12V 2Ω+Vo-

Method C for determining equivalent circuit• If only dependent sources are present• If only dependent sources are present

(a) Apply a 1 A current source at the terminals and calculate the terminal voltage. Since v = 1 × RTh = RTh, take RTh = v ; orvo 1 × RTh RTh, take RTh vo ; or

(b) Apply a 1 V voltage source at the terminals and calculate the terminal current i. Take RTH = 1/i.TH

• Example 6.4: Find the Thevenin’s equivalent of the circuit shown.

– Since the rightmost terminals are already open-circuited, i = 0. Consequently, the dependent source is dead, so voc = 0.

3Ω

2Ω1 5i

i

p oc 2Ω1.5i

– We apply a 1 A source externally, then using the nodal analysis,

1.5 1 0 d 1v i v i−3Ω i 1 0 and 13 2

0.5 0.6V3 2

i

v v v

+ − = = −

∴ + = ⇒ =

3Ω

2Ω1.5i

i

1A+v

3 20.6ThR∴ = Ω

-

– We can also apply a current source of iin externally, then using the nodal analysis,

3Ω i3Ω

2Ω1.5i

i

iin+v-

• Summary: – Method to use depends on which is convenient and simpler– Method B can be generally applied to all circuits– Method C is only useful for determining the equivalent resistance, Methods A or C are

t d t i th Th i ’ lt N t ’ tnecessary to determine the Thevenin’s voltage or Norton’s current– Only independent sources can be “killed” when determining equivalent resistance

R d E l 4 10 d 4 11• Read Examples 4.10 and 4.11• Try Chapter Problems 4.63, 4.67 and 4.71

Maximum power transfer• Given a DC circuit as shown maximum power• Given a DC circuit as shown, maximum power

transfer takes place when the load resistance RL = RTh.

2⎛ ⎞

– Proof:

– PL is maximum when

2 ThL L L

Th L

VP i R RR R

⎛ ⎞= = ⎜ ⎟+⎝ ⎠

PL is maximum when

( ) ( )( )

222

4

20Th L L Th LL

ThL Th L

R R R R RdP VdR R R

⎛ ⎞+ − ⋅ += =⎜ ⎟

⎜ ⎟+⎝ ⎠

– This results in the following,

( ) ( )2 2Th L L Th LR R R R R+ = +

– The maximum power is:

L ThR R⇒ =2 2

,max 2 4Th Th

L LL L

V VP RR R

⎛ ⎞= =⎜ ⎟⎝ ⎠L L⎝ ⎠

• Example 7: For the circuit shown, find the value of RL that results in maximum power being transferred to RL and the value of this maximum power.

– Find Thevenin’s equivalent circuit,

( ) ( )30 15030 / /150 25R = = = Ω

( )

30 / /150 2530 150

150 360 300V150 30

Th

Th

R

V

= = = Ω+

= =+

– Power transfer is maximum when

150 30+

25L ThR R= = Ω

– The maximum power: 2300⎛ ⎞

25L ThR R Ω

,max300 25 900W50LP ⎛ ⎞= =⎜ ⎟

⎝ ⎠

– What percentage does this maximum power correspond to the total power delivered by the 360V source?the 360V source?

360V360 150150V 7A

2 30900

Thab

Vv i

P

−= = ⇒ = =

,max

360V

900 =35.71%360 7

LPP

∴ =×

• Try Assessment Problem 4.21 and Chapter Problem 4.79

Superposition• Linearity - A linear circuit is one whose output is linearly related (or directly• Linearity - A linear circuit is one whose output is linearly related (or directly

proportional) to its input.• A linear circuit is defined as any circuit that has only independent sources, linear

dependent sources and linear elementsdependent sources, and linear elements.– Homogeneity (scaling) property: v = i R k v = k i R– Additive property: v1 = i1 R ; v2 = i2 R v = (i1 + i2) R = v1 + v2

• Principle of Superposition In any linear circuit the current or voltage at any• Principle of Superposition - In any linear circuit, the current or voltage at any node may be calculated as the sum of the contributions of each independent source acting alone.

Superposition cannot be applied to power because power (p) is a nonlinear function– Superposition cannot be applied to power because power (p) is a nonlinear function (p = i v).

• Basic rules of applying superposition theorem– Superposition applies only to a linear circuit– Superposition applies only to a linear circuit– An ideal voltage source is “killed” or removed by replacing it with a short circuit.– An ideal current source is “killed” or removed by replacing it with an open circuit.

Dependent sources should remain in the circuit– Dependent sources should remain in the circuit.

Steps to apply superposition principle1 Turn off all independent sources except one source Find the output (voltage or1. Turn off all independent sources except one source. Find the output (voltage or

current) due to that active source using nodal or mesh analysis.– Independent voltage sources are replaced by 0 V (short circuit)– independent current sources are replaced by 0 A (open circuit)– independent current sources are replaced by 0 A (open circuit)

2. Repeat step 1 for each of the other independent sources.3. Find the total contribution by adding algebraically all the contributions due to

the independent sourcesthe independent sources.

• Note: Dependent sources are left intact because they are controlled by circuit i blvariables.

• Superposition reduces a complicated problem to several easier problems, each containing an independent source. – There is no advantage in using superposition to solve a network with dependent

sources because the dependent source is never killed.

• Example 6.6: Using superposition principle, find the currents through the resistors i1, i2, i3 and i4.

• First, find the currents resulting from the 120V voltage source– Kill 12A current source by replacing it with an open-circuitcu e t sou ce by ep ac g t w t a ope c cu t– Apply nodal analysis to solve for the currents,

1 1 1120 06 3 2 4

v v v−+ + =

+

1

'1

30V120 30 15A

6

v

i

⇒ =−

= =

'2

630 10A3

i = =

' '3 4

30 5A2 4

i i= = =+

• Secondly, find currents resulting from 12A current source– Kill 120V voltage source by replacing it with a short-circuit

A l d l l i l f h– Apply nodal analysis to solve for the currents,

( )3 3 3 4 0 16 3 2v v v v−+ + = L v3 v4

( )4 3 4 12 0 22 4

v v v−+ + = L

– Solving equations (1) and (2),

" "3 312 12v v− −3 412V ; 24Vv v= − = −

( )

" "3 31 2

" "3 4 43 4

12 122A ; 4A6 6 3 3

12 24 246A ; 6A

v vi i

v v vi i

= − = − = = = = −

− − −− −= = = = = = −3 4;

2 2 4 4

• Lastly, the branch currents in the original circuit are algebraic sums of the individual currents contributed by each of the independent sources.

' " ' "1 1 1 2 2 2

' " ' "3 3 3 4 4 4

15 2 17A ; 10 4 6A

5 6 11A ; 5 6 1A

i i i i i i

i i i i i i

= + = + = = + = − =

= + = + = = + = − = −

• Try Chapter Problems 4 91 and 4 93Try Chapter Problems 4.91 and 4.93

1 - ee2001 circuit analysis lectures 1-6.pdf

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