ch2 circuit elements

Ch2 Circuit Elements

Definition : An Electric Source is a device that is capable of converting nonelectric energy to electric energy and vice versa.

Example : Battery

Discharging battery convert chemical energy to electric energyCharging Battery convert electric energy to chemical energy

(AA )Four double-A rechargeable batteries

http://en.wikipedia.org/wiki/AA_battery

http://en.wikipedia.org/wiki/Rechargeable_battery

Electrical generator

An electrical generator is a device that converts mechanical energy to electrical energy

The source of mechanical energy may be a reciprocating or turbine steam engine, water falling through turbine or waterwheel, an internal combustion engine, a wind turbine or any other source of mechanical energy

Portable Electric GeneratorElectric Generator

http://en.wikipedia.org/wiki/Electrical_energy

Theses souses can either deliver power like the battery when operate devices like toys, radio, mobile phone, car electric instrument,….etc. or the electric generator that deliver power to houses , cities, countries. Or absorb power like charging the battery or operating electric drill.

We are going to use ideal sources to model practical sources as will be shown next.

There are two type of sources that we will disuse

Voltage Source Current Source

We are going to use ideal sources to model practical sources as will be shown next.

Those ideal sources do not exist as practical devices , they are idealized model of the actual devices

Ideal Voltage Source : is a circuit element that maintaining a prescribed voltage across its terminals regardless of the current flowing in those terminals

Example : if we consider the 1.5 volt dry battery you from the market as an ideal, then you will get a 1.5 volt across the battery regards of what is connected across it or in another way the battery will supply a steady 1.2 V regardless what current flowing through it

1.2 V

i

We should know that can not be possible since if the current i is large due to some load as we will see later when we discuss Ohms law, the power deliver by the small battery will be very large

We also will classified sources as Independent and Dependent sources

Independent source establishes a voltage or a current in a circuit without relying on a voltage or current elsewhere in the circuit

Dependent sources establishes a voltage or a current in a circuit whose value depends on the value of a voltage or a current elsewhere in the circuit

We will use circle to represent Independent source and diamond shape to represent Dependent sources

Independent sourceDependent sources

Independent and dependent voltage and current sources can be represented as

+

-

Independent voltage source Independent current source

+

-

5 V 3 A

Dedependent voltage sourceVoltage depend on current

Dedependent current sourceCurrent depend on voltage

4 ix V 4 vx A

were ix is some currentthrough an element

were vx is some voltageacross an element

+

-

4 vx V 7 ix A

were vx is some currentthrough an element

were ix is some voltageacross an element

The dependent sources can be also as

Dedependent voltage sourceVoltage depend on voltage Dedependent current source

Current depend on current

Example 2.1 for each of the following connections establish which interconnections are permissible and which violate the constrains by the ideal source

Connection is valid Connection is valid Connection is not permissible

Connection is not permissibleConnection is valid

Example 2.2 for each of the following connections establish which interconnections are permissible and which violate the constrains by the ideal source

Connection is not permissible Connection is valid

Connection is validConnection is not permissible

2.2 Electrical Resistance (Ohm’s)

Georg Simon Ohm, a German physicist who is famous for defining the fundamental relationship among voltage, current, and resistance through Ohm's law

Ohm's law states that, in an electrical circuit, the current passing through a conductor, from one terminal point on the conductor to another terminal point on the conductor, is directly proportional to the potential difference )i.e. voltage drop or voltage( across the two terminal points and inversely proportional to the resistance of the conductor between the two terminal points

Most materials exhibit measurable resistance to current. The amount of resistance depend on the material.

This is similar in some way when water flow in a pipe. If there is a dirt or material on the pipe it will impede ) يعيق( the flow of water.

were V = the voltage in volts )V( I = the current in amperes )A( R = the resistance in ohms )(

R

+

-

v

i

R

+

-

v

i

For purpose of circuit analysis, we must reference the current in the resistor to the terminal voltage. For the passive sign convention

v = Ri

Otherwise we introduce a minuses sign similar to what we did when we calculated power

v = - Ri

Example

3

+

-

v

2 A

v = )3()2( = 6 V 3

+

-

v

2 A

v = -)3()2( = -6 V

3

+

-

v

- 2 A

v = )3()- 2( = -6 V 3

+

-

v

- 2 A

v = -)3()- 2( = 6 V

The reciprocal of resistance )مقاومه( is conductance )موصل( and have the symbol G

1G =

R

and have the unit S for )siemens( or for )mho( is spelling backward for ohm

An 8 resistor is equivalent to 1/8= 0.125 S or

We may calculate the power at the terminal of a resistor in several way

First method : we use the defining equation as follows:

R

+

-

v

i

For the passive sign conventionp = vi

R

+

-

v

i

p = -vi

Otherwise we introduce a minuses sign

Second method : we express the power at the terminal of resistor in terms of the current and resistor as follows:

p vi 2) (iR i Ri

For the non passive sign convention we have

p vi- 2) (iR i Ri- -

R

+ -v

i

R

+ -v

i

Third method : we express the power at the terminal of resistor in terms of the voltage and resistor as follows:

p vi2

) (v v

vR R

R

+ -v

i

For the non passive sign convention we have

R

+ -v

i

p vi-2

) (v v

vR R

--

Fourth method : we express the power at the terminal of resistor in terms of the voltage or current and conductance G as follows:

p vi 2) (v Gv Gv R

+ -v

ip vi

2

) (i i

iG G

For the non passive sign convention we will have identical relation as the passive sign convention

R

+ -v

i

2p Gv2i

pG

Example 2.3 In each circuit find the followings:

1 A 8

+

-

va )1()8( 8 Vav 2 28 8)1( 8 Wp i

2 2)8(OR 8 W

8 8av

p

50 V 25 +

-

id

8, ?av p

25, ?d pi

502 A

25di-

-

2 225 25 25) 2( 100 Wdp i -

2

25

50OR 100 W

25p

2.3 Construction of a Circuit Model

This course will be focus on circuit analysis )i.e., finding voltages , currents andpowers of circuit elements)

However you would need to construct a model for the electric device as much asanalyzing it.

We are going to develop a circuit model based on the behavior of the circuit componentsand interconnections

2.4 Kirchhoff’s Law

The objective of this course is to find )or solve( for voltages and currents in every element in the circuit

Example consider the following circuit:

+-

2 3

6 5

1i

5 V

Suppose we want to find the current i1?

In this circuit we have 7 unknowns , namely and ) note i, , , s the same as ( 1 421 32 3 43i i i iv iv v v

+ -1v

+

-2v

2i

3i

+ -3v

+

-4v

+-

2 3

6 5

1i

5 V

+ -1v

+

-2v

2i

3i

+ -3v

+

-4v

To solve for the 7 unknown we need 7 equations

Ohms law can provide us with 4 equations, namely

= = =2 6 3 5 = 1 421 2 3 33i i iv v v iv

However Ohms equations can not be sufficient to solve for the 7 unknown ,we need still 3 equations , what are these equations?

Gustav Kirchhoff Russian scientist who first stated them in 1848 in a published paper and they are named after him as

Kirchhoff Current Law (KCL) Kirchhoff Voltage Law (KVL)

Kirchhoff's Current Law ( KCL):

The algebraic sum of all the currents at any node in a circuit equals zero.

First we have to define a node.

A node is a point where two or more circuit elements meet1i

2i

3i

Kirchhoff's Current Law ( KCL):

The algebraic sum of all the currents at any node in a circuit equals zero.

The algebraic signify a sign on the current that is positive or negative. Since the current is a reference quantity by direction. Then we can state the following

Current entering the node is positive and current leaving the node is negative

Current entering the node is negative and current leaving the node is positive

OR

1i2i

3i

Example

1i2i

3i

Current entering the node is positive and leaving the node is negative

) ( 0 - 1 2 3i i i 0 - 1 2 3i i i

Current entering the node is negative and leaving the node is positive

) 0( ) ( - - 1 2 3i i i 0 - 1 2 3i i i

Note the algebraic sign is regardless if the sign on the value of the current

Suppose

1 2 3 A A A - 1 2 3i i i- 1 A 2 A

3 ANow KCL

0- 1 2 3i i i ) ( 0-- 1 2 3

Since electric current is a rate flow of charges then Kirchhoff's Current Law is similar to the flow of water from different direction to a water valve ( (صمام

Water Valve

We have 4 nodes

Ohms law can provide us with 4 equations, namely = = =2 6 3 5 = 1 421 2 3 33i i iv v v iv

Now we go back to our circuit

+-

2 3

6 5

1i

5 V

+ -1v

+

-2v

2i

3i

+ -3v

+

-4v

+ -1v+

-2v

+ -3v

+

-4v

3i

2i

+-

Node 1Node 2

Node 3

shortshort

Node 4Same Node

2 3

5 6 5 V

1i

+-

Node 1

Node 2

Node 3

Node 4

2

3

6

5

5 V

Which can be redrawn as

+ -1v+

-2v

+ -3v

+

-4v

3i

2i

+-

Node 1Node 2

Node 3

shortshort

Node 4Same Node

2 3

5 6 5 V

1iNow if we apply KCL to each node we will have the followings

Node 1

1 1i i

0-1 1i i+-

Node 12

5 V

1i

1i

Nothing new!

Node 3

3 3i i0-3 3i i

Nothing new!

Node 33

5

3i

3i

Node 2

2 3

6

1i 3i

2i

Node 2

0- - 1 2 3i i i

Node 4

6

1i 3i

2i

0- 1 2 3i i i

Same as node 2

Node 4

Now we have 5 equations, namely

= = =2 6 3 5 = 1 421 2 3 33i i iv v v iv

+-

2 3

6 5

1i

5 V

+ -1v

+

-2v

2i

3i

+ -3v

+

-4v

Ohms law can provide us with 4 equations, namely

KCL provide us with 1 equations, namely

0- - 1 2 3i i i

We have now 5 equations , we still need two more equations

Kirchhoff Voltage Law (KVL) will provide us with the other two equations as will be shown nex

Kirchhoff Voltage Law (KVL)

The algebraic sum of all the voltages around any closed path in a circuit equals zero.

First we have to define a closed path

+-

a b c

def

A closed path or a loop is defined as starting at an arbitrary node, we trace closed path in a circuit through selected basic circuit elements including open circuit and return to the original node without passing through any intermediate node more than once

abea




+-

a b c

def


abea bceb




+-

a b c

def


abea bceb cdec




+-

a b c

def


abea bceb cdec aefa




+-

a b c

def


abea bceb cdec aefa abcdefa



The "algebraic" correspond to the reference direction to each voltage in the loop.

Assigning a positive sign to a voltage rise ) - to + (

Assigning a negative sign to a voltage drop ) to - (

5 V +-

2 3

6 5

+ -1v + -

3v+

-2v

+

-4v

Assigning a positive sign to a voltage drop ) to - (

Assigning a negative sign to a voltage rise ) - to (

OR

Example

5 V +-

2 3

6 5

+ -1v + -

3v+

-2v

+

-4v

Loop 1 0 - 1 2 5v v

Loop 2 0 - 43 2v v v

We apply KVL as follows:

Example

5 V +-

2 3

6 5

+ -1v + -

3v+

-2v

+

-4v

Loop 1 0 - 1 2 5v v

Loop 2 0 - 43 2v v v

Loop 3 0 - 1 43 5v v v

However we notice that KVL on Loop 3 is only the summation of Loop 1 and Loop 2

We apply KVL as follows:

Therefore only two KVL equations are valid

Loop 1 0 - 1 2 5v v

Loop 2 0 - 43 2v v v

Now we go back to our circuit

+-

2 3

6 5

1i

5 V

+ -1v

+

-2v

2i

3i

+ -3v

+

-4v

Now we have 7 equations, namely

= = =2 6 3 5 = 1 421 2 3 33i i iv v v iv

Ohms law provide us with 4 equations, namely

KCL provide us with 1 equations, namely

0- - 1 2 3i i i

KVL provide us with 2 equations, namely

0 - 1 2 5v v0 - 43 2v v v

Now we can solve the 7 equations to obtain the 7 variables

, , , and 1 42 31 2 3i iv v v vi

We can think of KVL as a gas that distribute pressure gas to different houses. When it leaves the station it is full of gas. Then it start delivering the gas to each house depend on the size of the house

Gas Station

House 1 House 2

House 3

House 4

House 5 House 6

Example 2.6 for the circuit shown apply KCL to each node a, b, c, and d.

Note there is no connection dot at the center of the diagram )i.e, no node( .

The branch crosses the branch containing the ideal current ia

Applying KCL to nodes a, b, c, and d , we have

0Node a - -1 4 52 i i i i 0 Node b - - - a12 3 b i i i i i

0 Node c - - c4b 3 i i i i 0Node d a c5 i i i

Example 2.7 for the circuit shown apply KVL to each designated path in the circuit


4 31 2 0bpath v v va vv - - -


3 5 0a v vpath b v-


4 6 5 0cbpath c v v vv v- - - -


2 71 0a c dvpath d v v vv v- - - -


4 31 2 0bpath v v va vv - - -

3 5 0a v vpath b v-

4 6 5 0cbpath c v v vv v- - - -

2 71 0a c dvpath d v v vv v- - - -

Example 2.8 for the circuit shown use Kirchcoff’s laws and Ohm’s law to find io ?

Solution

We will redraw the circuit and assign currents and voltages as follows

Since io is the current in the 120 V source , therefore there is only two unknown currents

in the circuit namely:

oi

io and i1

Therefore we need two equations relating io and i1

Applying KCL to the circuit nodes namely a,b and c will give us the following

Node a o oi i0- o oi i Nothing new!

oi

Node b 6 0 - - o 1i i

Node c 6 0 - o1i i The same as node b

Therefore KCL provide us with only one equation relating io and i1 namely

6 0 - - o 1i i

We need another equation to be able to solve for io

That equation will be provided by KVL as shown next

oi

We have three closed loops

However only one loop that you can apply KVL to it namely abca

Since the other two loops contain a current source namely 6 A and since we can not relatethe voltage across it to the current through it , therefore we can not apply KVL to that loop

1120 10 50 0oi i-

Applying KVL around loop abca clockwise direction assigning a positive sign to voltage drops ) + to - ( and negative sign to voltage rise , we have

Combinning this with the KCL equation 6 0 - - o 1i i

we have two equations and unknowns which can be solved simultaneously to get io= -3 A i1= 3 A

For the circuit shown we want apply Kirchhof’s and Ohm law to find vo ?

Solution

Since vo=20 io

We are going to device a strategy for solving the circuit

Let io be the current flowing on the 20 resistor

→ therefore we seek io

2.5 Analysis of a circuit containing dependent sources

KCL will provide us with one equation relating io with iΔ namely

5 0oiii -

We need two equations to be able to solve for io

KVL and Ohm’s law will provide us with the additional equation relating io with iΔ as will be shown next

6 0oii - Node b

The circuit has three closed loops

However only the loop abca is the one that you can apply KVL to it

Since the other two loops contain a current source that, you will not be able to write the voltage across it in terms of the current )i.e, you will not be able to write KVL around the loop the one that you can apply KVL to it (

Therefore

500 5 20 0oi i- 5 20 500oii

Therefore we have two equations relating io with iΔ namely

6 0oi i - KCL at node b

5 20 500oi i KVL around loop abca

Two equations and two unknowns io and iΔ , we can solve simultaneously and get

io = 24 A and iΔ = 4 A

Therefore vo=20 io = )20()24( = 480 V

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