KVL KCL NV

7/28/2019 KVL KCL NV

1/103

Lecture 1

Circuit Elements, Ohms Law,Kirchhoffs Laws


2/103

Overview

What a circuit element is?

Electrical Quantities Resistors and Ohms Law

Independent and dependent voltage

sources and current sources KVL & KCL


3/103

Circuit Elements In circuits, we think about basic circuit

elements that are the building blocksof our circuits. This is similar to whatwe do in Chemistry with chemical

elements like oxygen or nitrogen. A circuit element cannot be broken down

or subdivided into other circuitelements.

A circuit element can be defined in

terms of the behavior of the voltage andcurrent at its terminals.

We define an electric circuit as aconnection of electrical devices thatform one or more closed paths.


4/103

Basic Electrical Quantities

Basic Quantities: current, voltage andpower Current: Time rate of change of electric charge

I = dq/dt

1 Amp = 1 Coulomb/sec

Voltage: electromotive force or potential, V 1Volt = 1 Joule/Coulomb = 1 Nm/coulomb

Power: P = I V

1 Watt = 1 VoltAmp = 1 Joule/sec


5/103

Current I

Normally we talk about the movement ofpositive charges although we know that, ingeneral, in metallic conductors current resultsfrom electron motion (conventionally positiveflow)

The sign of the current indicates thedirection of flow

Types of current:

Direct Current (DC): batteries and some specialgenerators Alternating Current (AC): household current which

varies with time

I(t)


6/103

Voltage V

Voltage is the difference in energy levelof a unit charge located at each of two

points in a circuit, and therefore,represents the energy required to movethe unit charge from one point to theother

Circuit Element(s)

+ V(t)


7/103

Default Sign Convention

Passive sign convention : current shouldenter the positive voltage terminal

Consequence for P = I V Positive (+) Power: element absorbs power

Negative (-) Power: element supplies power

Circuit Element+

I


8/103

Electrical Analogies (Physical)

Electric Hydraulic

Base

quantity Charge (q) Mass (m)

Flow

variableCurrent (I) Fluid flow (G)

Potentialvariable

Voltage (V) Pressure (p)


9/103

Active vs. Passive Elements

Active elements can generate energy Voltage and current sources

Batteries Passive elements cannot generate

energy Resistors

Capacitors and Inductors (but CAN storeenergy)


10/103

The Basic Circuit Elements

There are 5 basic circuit elements:

1. Voltage sources

2. Current sources

3. Resistors

4. Inductors

5. Capacitors


11/103

Voltage Sources

A voltage source is a two-terminalcircuit element that maintains avoltage across its terminals.

The value of the voltage is thedefining characteristic of a voltagesource.

Any value of the current can go

through the voltage source, in anydirection. The current can also bezero. The voltage source does notcare about current. It caresonly about voltage.


12/103

Voltage Sources Ideal and

Practical A voltage source maintains a voltage across its

terminals no matter what you connect to thoseterminals.

We often think of a battery as being a voltagesource. For many situations, this is fine. Othertimes it is not a good model. A real battery willhave different voltages across its terminals insome cases, such as when it is supplying a largeamount of current. As we have said, a voltagesource should not change its voltage as the

current changes. We sometimes use the term ideal voltage source

for our circuit elements, and the term practicalvoltage source for things like batteries. We willfind that a more accurate model for a battery isan ideal voltage source in series with a resistor.


13/103

Voltage Sources

There are 2 types of voltage sources:

1. Independent voltage sources

2. Dependent voltage sources, of whichthere are 2 forms:i. Voltage-dependent voltage sources

ii. Current-dependent voltage sources


14/103

Voltage Sources Schematic Symbol for

Independent Sources

The schematicsymbol that we usefor independentvoltage sources isshown here.

Independent

voltage

source

+

-

vS=

#[V]

This is intended to indicate that the schematic symbol can belabeled either with a variable, like vS, or a value, with somenumber, and units. An example might be 1.5[V]. It could alsobe labeled with both.


15/103

Voltage Sources SchematicSymbols for Dependent Voltage Sources

The schematic symbolsthat we use fordependent voltage

sources are shownhere, of which thereare 2 forms:

i. Voltage-dependent

voltage sourcesii. Current-dependent

voltage sources

Voltage-

dependent

voltage

source

+vS=

vX

-

Current-dependent

voltage

source

+vS=

iX

-


16/103

Voltage-

dependent

voltage

source

+vS=

vX

-

Notes on Schematic

Symbols for Dependent Voltage Sources

The schematic symbols that we use fordependent voltage sources areshown here, of which there are 2forms:

i. Voltage-dependent voltage

sourcesii. Current-dependent voltagesources

The symbol m is the coefficient of thevoltage vX. It is dimensionless. Forexample, it might be 4.3 vX. The vX is avoltage somewhere in the circuit.

Current-dependent

voltage

source

+vS=

iX

-The symbol r is the coefficient of the current iX.It has dimensions of [voltage/current]. Forexample, it might be 4.3[V/A] iX. The iX is a

current somewhere in the circuit.


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Current Sources

A current source is a two-terminal circuitelement that maintains a current through itsterminals.

The value of the current is the definingcharacteristic of the current source.

Any voltage can be across the currentsource, in either polarity. It can also be

zero. The current source does not careabout voltage. It cares only aboutcurrent.


18/103

Current Sources - Ideal A current source maintains a current through its

terminals no matter what you connect to thoseterminals.

While there will be devices that reasonably model

current sources, these devices are not as familiaras batteries. We sometimes use the term ideal current source

for our circuit elements, and the term practicalcurrent source for actual devices. We will findthat a good model for these devices is an idealcurrent source in parallel with a resistor.


19/103

Current Sources

There are 2 types of current sources:

1. Independent current sources

2. Dependent current sources, of which thereare 2 forms:

i. Voltage-dependent current sources

ii. Current-dependent current sources


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Independent

current

source

iS=#[A]

Current Sources Schematic Symbol for

Independent Sources

The schematic symbolsthat we use for

current sources areshown here.

This is intended to indicate that the schematic symbol can be labeled eitherwith a variable, like iS, or a value, with some number, and units. An examplemight be 0.2[A]. It could also be labeled with both.


21/103

Current Sources Schematic

Symbols for Dependent Current SourcesThe schematic symbols

that we use fordependent currentsources are shownhere, of which thereare 2 forms:

i. Voltage-dependentcurrent sources

ii. Current-dependentcurrent sources

Voltage-

dependent

current

source

iS=

gvX

Current-

dependent

current

source

iS=

iX


22/103

Current-

dependent

current

source

iS=

iX

Voltage-

dependent

current

source

iS=

gvX

Notes on Schematic

Symbols for Dependent Current Sources

The schematic symbols that we usefor dependent current sourcesare shown here, of which thereare 2 forms:

i. Voltage-dependent current

sourcesii. Current-dependent currentsources

The symbol g is the coefficient ofthe voltage vX. It has dimensionsof [current/voltage]. For example,it might be 16[A/V] vX. The vX is avoltage somewhere in the circuit.

The symbolb is the coefficient of thecurrent iX. It is dimensionless. Forexample, it might be 53.7 iX. The iX is a

current somewhere in the circuit.


23/103

Voltage and Current Polarities

Previously, we have emphasized theimportant of reference polarities ofcurrents and voltages.

Notice that the schematic symbols forthe voltage sources and current sourcesindicate these polarities.

The voltage sources have a + and a toshow the voltage reference polarity. The

current sources have an arrow to show thecurrent reference polarity.


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Why do we have

these dependent sources? Students who are new to circuits often questionwhy dependent sources are included. Somestudents find these to be confusing, and they doadd to the complexity of our solution techniques.

However, there is no way around them. We needdependent sources to be able to model amplifiers,and amplifier-like devices. Amplifiers are crucialin electronics. Therefore, we simply need tounderstand and be able to work with dependent

sources.


25/103

Resistors

A resistor is a two terminalcircuit element that has aconstant ratio of the voltageacross its terminals to thecurrent through its terminals.

The value of the ratio of voltageto current is the definingcharacteristic of the resistor.

Real-world devices that aremodeled by resistors:incandescent light bulbs, heatingelements (stoves, heaters, etc.),long wires

In many cases a light bulb

can be modeled with a

resistor.


26/103

Resistors Definition and Units

A resistor obeys the expression

where R is the resistance. If something obeys this

expression, we can think of it,and model it, as a resistor.

This expression is called OhmsLaw. The unit ([Ohm] or [W]) is

named for Ohm, and is equal to a[Volt/Ampere]. IMPORTANT: use Ohms Law

only on resistors. It does nothold for sources.

To a first-order approximation,

the body can modeled as aresistor. Our goal will be toavoid applying large voltagesacross our bodies, because itresults in large currentsthrough our body. This is not

good.

R

R

vR

i

+

R

viR -


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RX=

#[]

vX

iX

-+

Schematic Symbol for Resistors

The schematic symbols that we use forresistors are shown here.

This is intended to indicate that the schematic symbolcan be labeled either with a variable, like RX, or a

value, with some number, and units. An example

might be 390[]. It could also be labeled with both.

XX

X

vR

i


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Ohms Law

The Rest of

the Circuit

R

i(t)

+

v(t)


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Open Circuit

What if R = ?

i(t) = v(t)/R = 0

v(t)

The

Rest oftheCircuit

i(t)=0

+

i(t)=0


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Short Circuit

What if R = 0 ?

v(t) = R i(t) = 0

The

Rest oftheCircuit

v(t)=0

i(t)

+


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Series

Two elements are in series if the currentthat flows through one must also flowthrough the other.

R1 R2

Series

R1 R2

Not Series


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Parallel

Two elements are in parallel if they areconnected between (share) the same two(distinct) end nodes.

Parallel Not Parallel

R1

R2

R1

R2


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Resistor Polarities

Previously, we have emphasized the important ofreference polarities of current sources andvoltages sources. There is no correspondingpolarity to a resistor. You can flip it end-

for-end, and it will behave the same way.However, even in a resistor, direction matters inone sense; we need to have defined thevoltage and current in the passive signconvention to use the Ohms Law equationthe way we have it listed here.


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Getting the Sign Right with Ohms Law

If the reference current is in the

direction of the reference voltage

drop (Passive Sign Convention),

then

RX=

#[]

vX

iX

-+

XX

X

vR

i

If the reference current is in the

direction of the reference voltagerise (Active Sign Convention),then

RX=

#[]

vX

iX

-+

XX

X

vR

i


35/103

Why do we have to worry

about the sign in Ohms Law? It is reasonable to ask why the sign in Ohms Law

matters. We may be used to thinking that resistanceis always positive.

Unfortunately, this is not true. The resistors we use,particularly the electronic components we callresistors, will always have positive resistances.However, we will have cases where a device will have aconstant ratio of voltage to current, but the value ofthe ratio is negative when the passive sign convention

is used. These devices have negative resistance.They provide positive power.This can be done usingdependent sources.


36/103

Why do we have to worryabout the sign in Everything?

This is one of the central themes in circuit analysis.The polarity, and the sign that goes with thatpolarity, matters. The key is to find a way to getthe sign correct every time.

This is why we need to define reference polaritiesfor every voltage and current.

This is why we need to take care about whatrelationship we have used to assign reference

polarities (passive sign convention and active signconvention).


37/103

In this part of the module, we willcover the following topics:

Some Basic Assumptions

Kirchhoffs Current Law (KCL)

Kirchhoffs Voltage Law (KVL)


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Some Fundamental Assumptions

Wires Although you may not have stated it,or thought about it, when you havedrawn circuit schematics, you haveconnected components or devices

with wires, and shown this with lines. Wires can be modeled pretty well as

resistors. However, their resistanceis usually negligibly small.

We will think of wires as connectionswith zero resistance. Note that thisis equivalent to having a zero-valuedvoltage source.

This picture shows wires

used to connect electrical

components. This particular

way of connectingcomponents is called

wirewrapping, since the

ends of the wires are

wrapped around posts.


39/103

Some Fundamental Assumptions Nodes

A node is defined as a placewhere two or morecomponents are connected.

The key thing to rememberis that we connectcomponents with wires. Itdoesnt matter how many

wires are being used; it onlymatters how manycomponents are connectedtogether.

+

-

vA

C

RD

iB

RF

RE


40/103

Basic Laws of Electric CircuitsNodes, Branches, and Loops:

A node: A node can be defined as a connection point betweentwo or more branches.

A branch: A branch is a single electrical element or device.

A circuit with 5 branches.

A circuit with 3 nodes.

2


41/103


If we start at any point in a circuit (node), proceed throughconnected electric devices back to the point (node) fromwhich we started, without crossing a node more than one time,

we form a closed-path.

A loop is a closed-path.

An independent loop is one that contains at least

one element not contained in another loop.


42/103


The relationship between nodes, branches and loopscan be expressed as follows:

# branches = # loops + # nodes - 1

or

B = L + N - 1

In using the above equation, the number of loops arerestricted to be those that are independent.


43/103

How Many Nodes?

To test ourunderstanding ofnodes, lets look at

the example circuitschematic givenhere.

How many nodes are

there in thiscircuit?

+

-vA

C

RD

iB

RF

RE


44/103

How Many Nodes Correct Answer

In this schematic,there are threenodes. These nodesare shown in dark blue

here. Some students countmore than threenodes in a circuit likethis. When they do, itis usually because

they have consideredtwo points connectedby a wire to be twonodes.

+

-

vA

RC

RD

iB

RF

RE


45/103

How Many Nodes Wrong Answer

In the example circuitschematic given here,the two red nodes arereally the same node.There are not fournodes.

Remember, two nodesconnected by a wirewere really only onenode in the first place.

+

-

vA

RC

RD

iB

RF

RE

Wire connecting two nodesmeans that these are really asingle node.


46/103

Some Fundamental Assumptions

Closed Loops A closed loop can be

defined in this way: Startat any node and go in anydirection and end up where

you start. This is a closedloop.

Note that this loop doesnot have to followcomponents. It can jumpacross open space. Most

of the time we will followcomponents, but we willalso have situations wherewe need to jump betweennodes that have noconnections.

+

-

vA

RC

RD

iB

RF

RE

vX

+

-


47/103

How Many Closed Loops To test our

understanding of

closed loops, letslook at theexample circuitschematic given

here. How many closedloops are therein this circuit?

+

-

vA

RC

RD

iB

RF

RE

vX

+

-


48/103

How Many Closed Loops

An Answer There are several closed

loops that are possiblehere. We will show a few

of them, and allow you tofind the others.

The total number ofsimple closed loops in thiscircuit is 13.

Finding the number will not

turn out to be important.What is important is torecognize closed loopswhen you see them.

+

-

vA

RC

RD

iB

RF

RE

vX

+

-


49/103

+

-

vA

RC

RD

iB

RF

RE

vX

+

-

Closed Loops Loop #1 Here is a loop we

will call Loop #1.

The path isshown in red.


50/103

+

-

vA

RC

RD

iB

RF

RE

vX

+

-

Closed Loops Loop #2 Here is Loop #2.

The path is shown inred.


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+

-

vA

RC

RD

iB

RF

RE

vX

+

-

Closed Loops Loop #3 Here is Loop #3.

The path is shown inred.

Note that this pathis a closed loop that

jumps across the

voltage labeled vX.This is still a closedloop.


52/103

+

-

vA

RC

RD

iB

RF

RE

vX

+

-

Closed Loops Loop #4 Here is Loop #4. The

path is shown in red.

Note that this path isa closed loop that

jumps across thevoltage labeled vX.This is still a closed

loop. The loop alsocrossed the currentsource. Rememberthat a current sourcecan have a voltage

across it.


53/103

+

-

vA

RC

RD

iB

RF

RE

vX

+

-

A Not-Closed Loop The path is shown in

red here is notclosed.

Note that this pathdoes not end whereit started.


54/103

Kirchhoffs Current Law (KCL)

With these definitions, we are

prepared to state KirchhoffsCurrent Law:

The algebraic (orsigned) summation of

currents through aclosed surface mustequal zero.


55/103

Kirchhoffs Current Law(KCL) Some notes.

The algebraic (or signed)summation of currentsthrough any closed surface

must equal zero.

This definition is often stated as applying to nodes. It applies to any closedsurface. For any closed surface, the charge that enters must leavesomewhere else. A node is just a small closed surface. A node is theclosed surface that we use most often. But, we can use any closedsurface, and sometimes it is really necessary to use closed surfaces that

are not nodes.

This definition essentially means that charge does not build up at aconnection point, and that charge is conserved.


56/103

KCL (Kirchhoffs Current Law)

The sum of currents entering the node is zero:

Analogy: mass flow at pipe junction

i1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

j

j ti1

0)(


57/103

Current Polarities

Again, the issue of thesign, or polarity, or direction,of the current arises. Whenwe write a Kirchhoff CurrentLaw equation, we attach asign to each referencecurrent polarity, dependingon whether the referencecurrent is entering or leavingthe closed surface. This canbe done in different ways.


58/103

Kirchhoffs Current Law (KCL) a Systematic Approach

The algebraic (or signed) summation ofcurrents through any closed surface mustequal zero.

For this set of material, we will always assign a positive sign to aterm that refers to a reference current that leaves a closedsurface, and a negative sign to a term that refers to a referencecurrent that enters a closed surface.

For most students, it is a good idea to choose one way to writeKCL equations, and just do it that way every time. The idea isthis: If you always do it the same way, you are less likely to getconfused about which way you were doing it in a certain equation.


59/103

Kirchhoffs Current Law (KCL) For this set of material, we

will always assign a positive

sign to a term that refers toa current that leaves aclosed surface, and anegative sign to a term thatrefers to a current thatenters a closed surface.

In this example, we havealready assigned referencepolarities for all of thecurrents for the nodesindicated in darker blue.

For this circuit, and using my

rule, we have the followingequation:

+

-

vA

RC

RD

iB

RF

RE

iA

iB

iC

iE

iD

0A C D E Bi i i i i


60/103

Kirchhoffs Current Law (KCL) Example Done Another Way

Some prefer to write thissame equation in a differentway; they say that thecurrent entering the closedsurface must equal the

current leaving the closedsurface. Thus, they write :

+

-

vA

RC

RD

iB

RF

RE

iA

iB

iC

iE

iD

0A C D E Bi i i i i

A D B C Ei i i i i

Compare this to theequation that we wrote inthe last slide:

These are the same

equation. Use either


61/103

Kirchhoffs Voltage Law (KVL)

Now, we are prepared to state KirchhoffsVoltage Law:

The algebraic (or signed)summation of voltages arounda closed loop must equal zero.


62/103

Kirchhoffs Voltage Law

(KVL) Some notes.

The algebraic (or signed)summation of voltages around aclosed loop must equal zero.

This applies to all closed loops. While we usually write equations forclosed loops that follow components, we do not need to. The onlything that we need to do is end up where we started.

This definition essentially means that energy is conserved. If wemove around, wherever we move, if we end up in the place westarted, we cannot have changed the potential at that point.


63/103

Voltage Polarities

Again, the issue of thesign, or polarity, or direction, of

the voltage arises. When wewrite a Kirchhoff Voltage Lawequation, we attach a sign toeach reference voltage polarity,depending on whether the

reference voltage is a rise or adrop. This can be done indifferent ways.


64/103


(KVL) a Systematic Approach

The algebraic (or signed) summation of voltagesaround a closed loop must equal zero.

For this set of material, we will always go around loops clockwise. We willassign a positive sign to a term that refers to a reference voltage drop,and a negative sign to a term that refers to a reference voltage rise.

For most students, it is a good idea to choose one way to write KVLequations, and just do it that way every time. The idea is this: If youalways do it the same way, you are less likely to get confused aboutwhich way you were doing it in a certain equation.

(At least we will do this for planar circuits. For nonplanar circuits,clockwise does not mean anything. If this is confusing, ignore it for now.)


65/103

+

-

vA

RC

RD

iB

RF

RE

vX

+

-

vF

+

-

vE

- +


(KVL) an Example For this set of material, we will

always go around loops clockwise.We will assign a positive sign to aterm that refers to a voltage drop,

and a negative sign to a term thatrefers to a voltage rise.

In this example, we have alreadyassigned reference polarities forall of the voltages for the loopindicated in red.

For this circuit, and using our rule,starting at the bottom, we havethe following equation:

0A X E Fv v v v


66/103

+

-

vA

RC

RD

iB

RF

RE

vX

+

-

vF

+

-

vE

- +


(KVL) Notes For this set of material, we will

always go around loopsclockwise. We will assign apositive sign to a term that refers

to a voltage drop, and a negativesign to a term that refers to avoltage rise.

Some students like to use thefollowing handy mnemonicdevice: Use the sign of the

voltage that is on the side of thevoltage that you enter. Thisamounts to the same thing.

0A X E Fv v v v

As we go up through thevoltage source, we enter thenegative sign first. Thus, vAhas a negative sign in the

equation.


67/103

+

-

vA

RC

RD

iB

RF

RE

vX

+

-

vF

+

-

vE

- +


(KVL) Example Done Another Way

Some textbooks, and somestudents, prefer to write this same

equation in a different way; they

say that the voltage drops mustequal the voltage rises. Thus, they

write the following equation:

0A X E Fv v v v

X F A Ev v v v

Compare this to the equation thatwe wrote in the last slide:

These are the same equation.Use either method.


68/103

How many of these equations

do I need to write?

This is a very important question. In general, it boils down to theold rule that you need the same number of equations as you haveunknowns.

Speaking more carefully, we would say that to have a singlesolution, we need to have the same number of independentequations as we have variables.

At this point, we are not going to introduce you to the way toknow how many equations you will need,or which ones to write. It is assumed thatyou will be able to judge whether you havewhat you need because the circuits will befairly simple. Later we will developmethods to answer this question specifically and efficiently.


69/103

How many more laws

are we going to learn?

This is another very important question. Until, we get toinductors and capacitors, the answer is, none.

Speaking more carefully, we would say that most of the

rules that follow until we introduce the other basicelements, can be derived from these laws.

At this point, you have the tools to solve many, manycircuits problems. Specifically, you have Ohms Law, andKirchhoffs Laws. However, we need to be able to use these

laws efficiently and accurately.


70/103

Example #1

Lets do an exampleto test out our new

found skills.

In the circuit shownhere, find the

voltage vXand thecurrent iX.

R4=

20[]

R3=

100[]

vS1

=

3[V]

+

-

vX

+

-

iX


71/103

Example #1 Step 1

The first step insolving is to define

variables we need.

In the circuit shownhere, we will define

v4 and i3.

R4=

20[]

R3

=

100[]

vS1

=

3[V]

+

-

vX

+

-

iX

v4+ -

i3


72/103

Example #1 Step 2

The second step insolving is to write some

equations. Lets startwith KVL.

1 4

4

0, or

3[V] 0.

S X

X

v v v

v v

R4=

20[]

R3

=

100[]

vS1

=

3[V]

+

-

vX

+

-

iX

v4+ -

i3


73/103

Example #1 Step 3

Now lets write OhmsLaw for the resistors.

4 4

3 3

, and

.

X

X

v i R

v i R

Notice that there is a sign in OhmsLaw.

R4=

20[]

R3

=

100[]

vS1

=

3[V]

+

-

vX

+

-

iX

v4+ -

i3


74/103

Example #1 Step 4

Next, lets write KCL forthe node marked in

violet.

3

3

0, or

.

X

X

i i

i i

Notice that we can write KCL for a node,or any other closed surface.

R4=

20[]

R3

=

100[]

vS1

=

3[V]

+

-

vX

+

-

iX

v4+ -

i3


75/103

Example #1 Step 5

We are ready to solve.

We have substituted into our KVLequation from other equations.

3[V] 20[ ] 100[ ] 0, or

3[V]25[mA].

120[ ]

X X

X

i i

i

R4=

20[

]

R3=

100[]

vS1

=

3[V]

+

-

vX

+

-

iX

v4

+ -

i3


76/103

Example #1 Step 6

Next, for the otherrequested solution.

We have substituted into Ohms Law,using our solution for iX.

3 3 3, or

25[mA] 100[ ] 2.5[V].

X X

X

v i R i R

v

R4=

20[

]

R3=

100[]

vS1

=

3[V]

+

-

vX

+

-

iX

v4

+ -

i3


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Example #2 Lets do

anotherexample. Findthe voltage vX,the currents iXand iQ, and thepowerabsorbed byeach of the

dependentsources.

Problem 2.28 is on page 61 of the text. The


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Example #3 Problem 2.28

Problem 2.28 is on page 61 of the text. Thedependent source coefficient has units of [A/V].

Problem 2.20 is on page 59 of the text.


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Example #4 Problem 2.20

p g


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Kirchhoffs Voltage Law:

Basic Laws of Circuits

Consideration 1: Sum of the voltage drops around a circuitequal zero. We first define a drop.

We assume a circuit of the following configuration. Notice that

no current has been assumed for this case, at this point.

+

+

+

+

_

_

_

_

v1

v2

v4

v3

Figure 18/11/2012


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Kirchhoffs Voltage Law: Consideration 1.We define a voltage drop as positive if we enter the positive terminaland leave the negative terminal.

+ _v1

The drop moving from left to right above is + v1.

+_ v1

The drop moving from left to right above is v1.

Figure 2

Figure 3

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Kirchhoffs Voltage Law:

+

+

+

+

_

_

_

_

v1

v2

v4

v3Figure 4

Consider the circuit of Figure 4 onceagain. If we sum the voltage drops in the clockwise direction around thcircuit starting at point a we write:

- v1 v2 + v4 + v3 = 0

- v3 v4 + v2 + v1 = 0

a

drops in CW direction starting at a

drops in CCW direction starting at a

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Kirchhoffs Voltage Law:Consideration 2: Sum of the voltage rises around a circuit

equal zero. We first define a drop.

We define a voltage rise in the following diagrams:

+_ v1 Figure 5

+ _v1 Figure 6

The voltage rise in moving from left to right above is + v1.

The voltage rise in moving from left to right above is - v1.8/11/2012

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Kirchhoffs Voltage Law: Consider the circuit of Figure 7 onceagain. If we sum the voltage rises in the clockwise direction around thecircuit starting at point a we write:

+

+

+

+

_ _

_

v1

v2

v4

v3Figure 7

a

+ v1 + v2 - v4 v3 = 0

+ v3 + v4 v2 v1 = 0

rises in the CW direction starting at a

rises in the CCW direction starting at a

_

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Kirchhoffs Voltage Law:Consideration 3: Sum of the voltage rises around a circuit

equal the sum of the voltage drops.

Again consider the circuit of Figure 1 in which we start atpoint a and move in the CW direction. As we cross elements1 & 2 we use voltage rise: as we cross elements 4 & 3 we usevoltage drops. This gives the equation,

+

+

+

+

_

_

_

_

v1

v2

v4

v3

v1 + v2 = v4 + v3

1

2

3

4

p

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Kirchhoffs Voltage Law: Comments.

We note that a positive voltage drop = a negative voltage rise.

We note that a positive voltage rise = a negative voltage drop.

We do not need to dwell on the above tongue twisting statements.

There are similarities in the way we state Kirchhoffs voltageand Kirchhoffs current laws: algebraic sums

However, one would never say that the sum of the voltagesentering a junction point in a circuit equal to zero.

Likewise, one would never say that the sum of the currentsaround a closed path in an electric circuit equal zero.

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Kirchhoffs Voltage Law: Further details.For the circuit of Figure 8 there are a number of closed paths.Three have been selected for discussion.

+

+

+

+ +

+

+

+

+

+

+

-

- -

-

-

-

--

-

-

-v1

v2

v4

v3

v12

v11 v9

v8

v6

v5

v7

v10

+

-

Figure 8Multi-pathCircuit.

Path 1

Path 2

Path 3

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Kirchhoffs Voltage Law: Further details.

For any given circuit, there are a fixed number of closed pathsthat can be taken in writing Kirchhoffs voltage law and stillhave linearly independent equations. We discuss this more, later.

Both the starting point and the direction in which we go around aclosed path in a circuit to write Kirchhoffs voltage law arearbitrary. However,one must end the path at the same point fromwhich one started.

Conventionally, in most text, the sum of the voltage dropsequal to zero is normally used in applying Kirchhoffsvoltage law.

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Kirchhoffs Voltage Law: Illustration from Figure 8.

+

+

+

+ +

+

+

+

+

+

+

-

- --

-

-

--

-

-

-v1

v2

v4

v3

v12

v11 v9

v8

v6

v5

v7

v10

+

-

a

Blue path, starting at a

- v7 + v10 v9 + v8 = 0

b

Red path, starting at b

+v2 v5 v6 v8 + v9 v11

v12 + v1 = 0

Yellow path, starting at b

+ v2 v5 v6 v7 + v10 v11- v12 + v1 = 0

Using sum of the drops = 0

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Kirchhoffs Voltage Law: Double subscript notation.Voltages in circuits are often described using double subscript notation.

a b

Consider the following:

Figure 9: Illustrating double subscript notation.

Vab means the potential of point a with respect to point b withpoint a assumed to be at the highest (+) potential and point bat the lower (-) potential.

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Kirchhoffs Voltage Law: Double subscript notation.Task: Write Kirchhoffs voltage law going in the clockwisedirection for the diagram in Figure 10.

b a

xy

Going in the clockwise direction, starting at b, using rises;

vab + vxa + vyx + vby = 0

Figure 10: Circuit for illustrating double subscript notation.

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Kirchhoffs Voltage Law: Equivalences in voltage notations

The following are equivalent in denoting polarity.

+

-

v1 v1

a

b

+ -v2

= =

v2 = - 9 volts means the right hand side

of the element is actually positive.

vab = v1

Assumes the upper terminal is positive in all 3 cases

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Kirchhoffs Voltage Law: Application.Given the circuit of Figure 11. Find Vad and

Vfc.

5 V

8 V

15 V

12 V

20 V 10 V

30 V

a b c

d

ef

+ _

+

+

_

_

++

+

+

_

_

_

_

Using drops = 0; Vad + 30 15 5 = 0 Vab = - 10 V

Vfc 12 + 30 15 = 0 Vfc = - 3 V

Figure 11: Circuit for illustrating KVL.

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Kirchhoffs Voltage Law: Single-loop circuits.We are now in a position to combine Kirchhoffs voltage and current

Laws to the solution of single loop circuits. We start by developing the

Voltage Divider Rule. Consider the circuit of Figure 12.

R2

R1

v2

v1

+ +

+

_

_

_

v i1

v = v1 + v2

v1 = i1R1, v2 = i1R2

Figure 12: Circuit for developing

voltage divider rule.

then,

v = i1(R1 + R2) , and i1 =

v

(R1 + R2)so,

v1 =vR1

(R1 + R2)

* You will be surprised by how much you use this in circuits.

*

16

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Kirchhoffs Voltage Law: Single-loop circuits.Find V1 in the circuit shown in Figure 13.

V

R3 R2

R1V1

I

+_

1 2 3

1 1

( )

, ,

VI

R R R

V IR so wehave

11

1 2 3( )

VRVR R R

Figure 3.13

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V

R3 R2

R1I

+_


Kirchhoffs Voltage Law: Single-loop circuits.Example 1: For the circuit of Figure 14, the following is known:

R1 = 4 ohms, R2 = 11 ohms, V = 50 volts, P1 = 16 watts

Find R3.

Figure 14: Circuit for example 1.

Solution:

P1 = 16 watts = I2R1

I = 2 amps

V = I(R1

+ R2

+ R3

), giving,

R1 + R2 + R3 = 25, then solve for R3,

R3 = 25 15 = 10 ohms

, thus,

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Kirchhoffs Voltage Law: Single-loop circuits.Example 2: For the circuit in Figure 15 find I, V1, V2, V3, V4 and the

power supplied by the 10 volt source.

+ +

+

+

+

+

+_ _

_

_

_

_

_V1

V4

V3 V2

30 V 10 V

15 40

5

20

20 V

I

"a"

Figure 15: Circuit for example 2.

For convenience, we start at point a and sum voltage drops =0 in thedirection of the current I.

+10 V1 30 V3 + V4 20 + V2 = 0 Eq. 3.1

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Kirchhoffs Voltage Law: Single-loop circuits. Ex.2 cont.We note that: V1 = - 20I, V2 = 40I, V3 = - 15I, V4 = 5I Eq. 3.2

We substitute the above into Eq. 3.1 to obtain Eq. 3.3 below.

10 + 20I 30 + 15I + 5I 20 + 40I = 0 Eq. 3.3

Solving this equation gives, I = 0.5 A.

Using this value of I in Eq. 3.2 gives;

V1 = - 10 V

V2 = 20 V

V3 = - 7.5 V

V4 = 2.5 V

P10(supplied) = -10I = - 5 W

(We use the minus sign in 10I because the current is entering the + terminal)

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Kirchhoffs Voltage Law:Single-loop circuits, Equivalent Resistance.

+ +

+

+

_

_

+_ _

_

+

+

_

_V1

V4

V3 V2

VS1 VS3

R2 R4

R3

R1

VS2

I

"a"

Given the circuit of Figure 3.16. We desire to develop an equivalent circuit

as shown in Figure 17. Find Vs and Req.

Figure 16: Initial circuit fordevelopment.

VSReq

+

_I Figure 17: Equivalent circuit

for Figure 16

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+ +

+

+

_

_

+_ _

_

+

+

_

_V1

V4

V3 V2

VS1 VS3

R2 R4

R3

R1

VS2

I

"a"

Figure 16: Initial circuit.

Starting at point a, apply KVL going clockwise, using drops = 0, we have

VS1 + V1 VS3 + V2 + VS2 + V4 + V3 = 0

or

- VS1 - VS2 + VS3 = I(R1 + R2 + R3 + R4) Eq. 3.4

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Consider again, the circuit of Figure 17.

VS

Req+

_ I

Figure 17: Equivalent circuit

of Figure 16.

Writing KVL for this circuit gives;

VS = IReq compared to - VS1 - VS2 + VS3 = I(R1 + R2 + R3 + R4)

Therefore;

VS = - VS1 - VS2 + VS3 ; Req = R1 + R2 + R3 + R4 Eq. 3.5

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Kirchhoffs Voltage Law: Single-loop circuits, Equivalent Resistance.

We make the following important observations from Eq. 3.5:

The equivalent source of a single loop circuit can beobtained by summing the rises around the loop of

the individual sources.

The equivalent resistance of resistors in series is equalto the sum of the individual resistors.

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Kirchhoffs Voltage Law: Single-loop circuits.

Example 3: Find the current I in the circuit of Figure 18.

+ +

_

_ _

+

10 V 40 V

15 10

5

20

20 V

I

Figure 18: Circuit for

example 3.

From the previous discussion we have the following circuit.

50 V 50 +_ I

Therefore, I = 1 A

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