Chapter%201_Introduction%20n%20Basic%20Laws

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1/18/2011

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Chapter 1 (Basic Laws)Chapter 1 (Basic Laws)

Contents – Kirchhoff's Law

– Series, parallel and series-parallel circuits

– Voltage and current division

– Introduction to Wye-Delta transformation

2Prepared by: Miss Rafidah_FKE_UiTM

Learning Outcome Week2

– Able to define Ohm’s Law and Kirchhoff's Laws

– Able to simplify circuit using series-parallel

simplification technique

– Able to apply Voltage and Current Division Rules – Able to use Y-∆ and ∆ -Y transformation


Nodes, Branches and Loops

• Branch represents a single element such as a

voltage source or a resistor.




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• Node is the point of connection between

two or more branches.

• The above circuit have 4 nodes.


• Loop is any closed path in a circuit.

• The above circuit have 3 loops. 6Prepared by: Miss Rafidah_FKE_UiTM

• Relationship between branches, nodes and

loops.

b = l + n – 1

b – branches

n - nodes

l – independent loops


Kirchhoff’s Current Law

8

The algebraic sum of currents entering a

node is equals to zero

Prepared by: Miss Rafidah_FKE_UiTM



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Kirchhoff current law (KCL) states that the algebraicsum of currents entering a node is zero. Or,

The sum of current entering a node = the sum of current leaving the node.

Based on KCL we can write the current eq. for allnode in a circuit.

Ex.:

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Ia Ib

Ic

IdIe

Try to write down the current equation

base on KCL…


• Remember!!!

10

Sum current in = sum current out….(KCL).

Therefore:

Ia + Ic = Ib+ Id + Ie +

or

Ia + Ic – Ib – Id – Ie = 0



Kirchhoff’s Voltage Law

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The algebraic sum of all the voltages around

any closed path, in a circuit equals zero




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Applying KVL around the loop

20V

2

3

+ V1 -

+

V 2

-i

Applying KVL around the loop gives

-20 +V1 -V2 = 0

13Prepared by: Miss Rafidah_FKE_UiTM 14Prepared by: Miss Rafidah_FKE_UiTM


Find the currents and voltages in the

circuit

Applying Ohm’s Law and

Kirchhoff’s Law

V1=8i1, V2=3i2, V3=6i3 ----(1)

At node ‘a’, KCL gives,

i1 = i2+i3 ----(2)

Applying KVL gives,

-30 + V1 + V2 = 0

-30 + 8i1 + 3i2= 0

i1= (30-3i2)/8 ----(3)

-V2 + V3 = 0 ---(4)

V2 = V3

+

V 3

-

+

V

2 -




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We express V1 and V2 in terms of i1 and i2 asequ (1), therefore equ (4) becomes

6i3 = 3i2, i3 = i2/2 ----(5)

Substitute equ 3 and 5 into 2 gives

(30-3i2)/8 – i2 – i2/2 = 0

i2 = 2 ASo i1=3A, i3=1A, V1=24V, V2=6V, V3=6V


Resistors in Series

A series circuit provides only one path for

current between two points so that the

current is the same through each series

resistor.


Current in a Series Circuit

The current is the same through all points in a

series circuit. The current through each

resistor in a series circuit is the same as the

current through all the other resistors that arein series with it.


Total Series Resistance

The total resistance of a series circuit is equalto the sum of the resistances of eachindividual series resistor.




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Series Resistance FormulaFor any number of individual resistors

connected in series, the total resistance is the

sum of each of the individual values.

RT = R1 + R2 + R3 + . . . + Rn


Voltage Sources in SeriesWhen two or more voltage sources are in series, the

total voltage is equal to the algebraic sum (including

polarities of the sources) of the individual source

voltages.


Power in a Series Circuit

The total amount of power in a series resistive

circuit is equal to the sum of the powers in

each resistor in series.

PT = P1 + P2 + P3 + . . . + Pn


Resistors in Parallel

Each current path is called a branch.

A parallel circuit is one that has more than onebranch.






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Two Resistors in Parallel• The total resistance for two resistors in

parallel is equal to the product of the two

resistors divided by the sum of the two

resistors.

RT = R1R2/(R1 + R2)


Current Sources in Parallel

The total current produced by all current sourcesis equal to the algebraic sum of the individualcurrent sources.


Power in Parallel Circuits

Total power in a parallel circuit is found by

adding up the powers of all the individual

resistors.

PT = P1 + P2 + P3 + . . . + Pn


6Ω||3 Ω = (6 x 3)/(6+3)= 2Ω

1Ω+5Ω=6Ω

2Ω + 2Ω = 4 Ω

4Ω || 6Ω = 2.4Ω

Req = 4Ω + 2.4Ω + 8Ω

= 14.4Ω

Req

4

2

36

8

1

5




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Determine Req

33

Ω4 Ω4 Ω4Ω4Ω4


Find the equivalent resistance at terminal

a-b for each networks

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Ω4

Ω4

Ω8

Ω4

Ω8


Find the equivalent resistance at

terminal a-b for each networks


Find the equivalent resistance at terminal

a-b for each networks




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Voltage-divider circuitVs = V1 + V2

= iR1 + iR2

i = Vs / (R1 + R2)

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V1 = iR1

= Vs [R1/ (R1+ R2)]

V2 = iR2

= Vs [R2/ (R1+ R2)]


Find Va


Current-divider circuit

V = i1R1 =i2R2

= [(R1*R2)/ (R1+R2)] is

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i1 = V/R1 = [R2/(R1 + R2)] is

i2 = V/R2 = [R1/(R1 + R2)] is


Find i




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Find i


a) Find the no-load value of V0 incircuit shown

b) Find V0 when RL is 450KΩ

c) How much power is dissipatedin the 30KΩ resistor if the loadterminals are accidentallyshort-circuited?

d) What is the maximum power

dissipated in the 50KΩ resistor.


Find the power

dissipated in the 6Ω

resistor


Determine V 0

and i in the circuit shown




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Introduction to Y and ∆

45Prepared by: Miss Rafidah_FKE_UiTM 46

∆ - Y Conversion

Rc Rb Ra

RbRc R

++=1

Rc Rb Ra

RcRa R

++=2

Rc Rb Ra

RaRb R

++=3


47

1

133221

R

R R R R R R Ra

++=

Y - ∆ Conversion

2

133221

R

R R R R R R Rb ++=

3

133221

R

R R R R R R Rc

++=


Obtain the equivalent resistance

Rab for the circuit shown and find i

48

+

-


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