Basic circuit analysis laws

7/29/2019 Basic circuit analysis laws

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Chapter 2

Basic circuit analysis laws

EEEB 1113

CIRCUIT ANALYSISI


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Introduction to circuit element Resistor

Ohms law

Basic concepts of network topology (node, branch & loop )

Kirchhoffs laws

Resistors in series and parallel

Voltage divider and cur rent divider

Wye-delta transformations

TOPIC & STRUCTURE OF THE LESSON


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Georg Ohm (1787-1854)

In 1825-26 Ohm gave the concept

of first circuit elementResistor.

He related the voltage and current

by a constant known as the

resistance (v=Ri).

Resistors

FIRST CIRCUIT ELEMENT

RESISTOR (1825-26)


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The circuit element used to model the current resisting

behavior of a material is the resistor.

Resistanceis the capacity of materials to impede the

flow of current.

The resistance R of an element denotes its ability to

resist the flow of electric current; it is measured in ohms

()

Symbol: R1

1k

Resistors

+ --

i

+l

The voltage, vacross a resistor is directly proportional to the

current, iflowing through the resistor:

V=IR Resistance R of an element denotes its ability to resist the flow of

electric current, measured in Ohms ();where 1 = 1V/A.

Conductance, G is the ability of an element to conduct electric

current, measured in Siemens

G= 1/R =I/V


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Resistors

Power absorbed by a resistor Conductance G (unit: siemens S)

There are two extreme possible values of R:

Short Circuit when R = 0,

Open Circuit when R = .

Power dissipated in or absorbed by the resistor is always positive.

Theoretically - resistor is a passive element, cannot generate energy.


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Resistors

A short circuit is a circuit that has circuit element with resistance approaching 0 butthe current is NOT 0.

0 0


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Resistors

A short circuit is a circuit that has circuit element with resistance approaching 0 butthe current is NOT 0.

0

0

A open circuit is a circuit that has circuit element with resistance approaching .

lim 0


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For the circuit shown below, calculate the voltage v, the conductance G, and the

powerp.

Solution(a) v = iR = 2 mA x 10 k = 20 V(b) G = 1/R = 1/10 k = 100 S(c)p = vi= 20 volts x 2 mA = 40 mW

For the circuit shown below, calculate the Resistance, R.

Resistors


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Basic concepts of network topology

Electric circuits consist of connected basic circuit elements.

A branch represents a single element such as a voltage source or a

resistor.

A node is a point where two or more circuit elements join.

A loop/mesh is a closed path, starting and ending at the same nodewithout passing through any intermediate node more than once.

A loop is said to be independent if it contains a branch which is not inany other loop.

Two or more elements are in series if they are cascaded or connected

sequentially and consequently carry the same current. Two or more elements are in parallel if they are connected to the same

two nodes and consequently have the same voltage across them.


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node, branch & loop

Question: How many branches, nodes and loops are there?five branches, namely, the 10-V

voltage source, the 2-A current

source, and the three resistors.

By redrawing the circuit =>three nodes

Three independent loops


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node, branch & loop

Question: How many branches, nodes and loops are there?Should weconsider it asonebranch or twobranches?

Consider as two branches, as there are 2 elements 7 branches, 4 nodes and 4 loops 7=4+4-1


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Kirchhoffs laws

Ohm's law coupled with Kirchhoff's two laws gives a sufficient, powerful

set of tools for analyzing a large variety of electric circuits.

Kirchhoffs Law

Kirchhoffs

Current Law

(KCL)

Kirchhoffs

Voltage Law

(KVL)


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KCL: The algebraic sum of all the currents at any node in a

circuit equals zero .

Kirchhoffs laws

=

N = no. of branches connected to the node;

in= nth current entering the node.

Note that KCL also applies to a

closed boundary. This may be

regarded as a generalizedcase, because a node may be

regarded as a closed surface

shrunk to a point. In two

dimensions, a closed

boundary is the same as a

closed path. The total currententering the closed surface is

equal to the total current

leaving the surface.


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Alternatively , the sum of currents entering a node equals thesum of currents leaving a node.

Kirchhoffs laws

+ + 0

+

+

+


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Alternatively , the sum of currents entering a node equals thesum of currents leaving a node.

Kirchhoffs laws

+ + 0

+

+

+ 1

i

5i

2i

3i

4i


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KCLcan be applied to obtain the combined current, whencurrent sources are connected in parallel .

Kirchhoffs laws

Original CircuitEquivalent Circuit


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Determine the current I for the circuit shown in the figure

below.

Kirchhoffs Current laws

We canconsider thewholeenclosed areaas one node.

+ 2 + 3 + 4 0 3

This indicates thatthe actual current for I is flowingin the opposite direction.


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Find voand io in the circuit given below.

Kirchhoffs Current laws


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KVL: the algebraic sum of all voltages around a closed path(or loop) is zero .

Kirchhoffs laws

1

0M

m

m

v

where M = the number of voltages in the loop

vm = the mth voltage in the loop.

Sum of voltage drop= sum of voltage rises

+ + + 0 + + +


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KVLcan be applied to obtain the combined voltage, whenvoltage sources are connected in series .

Kirchhoffs Voltage laws

Original Circuit Equivalent Circuit


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Sign convention: The sign of each voltage is the polarity of the terminal firstencountered in traveling around the loop.

The direction of travel is arbitrary.

Clockwise:

Counter-clockwise:

0 1 2 0V V V

2 1 00V V V

0 1 2

V V V

V0

I

R1

R2

V1

V2

A+

+

-

-


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Find v1and v2in the circuit given below.


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Find vxand v0in the circuit given below.


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Kirchhoffs laws

Find the currents and voltages in the circuit shown in the circuit given below.


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Resistors in series

Recall: Elements are inseriesif theyshare a single node and carry the same

current

The equivalent resistance of any resistance connected in series is sum of individual

resistance. Thus for series resistors,

Applying KVL,

+ + 0 +

Apply Ohms Law

+(+)

+ + + +


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Resistors in Parallel

Two or more elements are in parallel if they are connected to the same two nodes

and consequently have the same voltage across them.

The total current, iis shared by the resistors in inverse proportion to their resistance

Applying KCL at node a,

+

Apply Ohms Law

+

1

+ 1

+ //


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By combining the resistors in the circuit below, find Req

( ( ( ( 4 + 5 + 3 )//4 ) + 3 )//6 ) + 2 + 1 6


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By combining the resistors in the circuit below, find Rab

( ( 4 + 2 )// 6 ) + 811


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Voltage divider

If R1 and R2 are connected in series with a voltage source v volts, the voltage

drops across R1 and R2 are0 0

1 2s

V VI

R R R

0

2 2 2

1 2

VV IR R

R R

22 0

1 2

RV V

R R

1

1 0

1 2

AlsoR

V VR R

V0

I

R1

R2

V1

V2

A


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Voltage divider

Note that source voltage vis divided among the

resistors in direct proportion to their

resistances.principle of voltage divisionIn general, if a voltage supply has N resistors in

series with the source voltage v, the nth resistor

(Rn) can be expressed as:

+ + + We call this Voltage

Divider rule


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Current divider

i(t) R1

i

R2i1 i2 v(t)

+

-

1 2

1 2

( ) ( ) ( )pR R

v t R i t i t R R

21

1 1 2

( )( ) ( )

Rv ti t i t

R R R

12

2 1 2

( )( ) ( )

Rv ti t i t

R R R

Current divides in inverse proportion to the resistances

To determine the current through each resistor;Ohms Law

W

ecallthis

Voltage

D

ividerrule


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Current divider

N resistors in parallel

1 2

1 1 1 1

p nR R R R ( ) ( )pv t R i t

( )( ) ( )pjj j

Rv ti t i t R R

Current injth branch is


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Voltage divider& Current divider

Find v1and v2for the circuit shown below. Also calculate i1and i2and the power

dissipated in the 12 and 40 resistors.

12//6 = [6x12][6+12] = 4

10//40 = [10x40][10+40] = 8

Equivalent circuit:


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Equivalent circuit:

Use voltage division,v1= [4/(4+8)](15) = 5V

v2= [8/(4+8)](15) = 10V

i1 = v1/12 = 5/12 = 416.7 mA

i2= v2/40 = 10/40 = 250 mA

P1= v1i1= 5x(5/12) = 2.083 W

P2= v2i2= 10x(0.25) = 2.5 W


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2.4

Step 1: Simplifying the circuit with series-parallel reductions.

4

io = [16/(16+4)]10 = 8 A,

i6 = [4/(6+4)]8 = 3.2 A,p = (3.2)2 6 = 61.44 W.


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Find the power dissipated at the 6-resistor.

Hint: Find the current i6, then usep = i2R to calculate the power.


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Wye-delta transformations

Situations often arise in circuit analysis

when the resistors are neither in parallel nor

in series.

For example, consider the bridge circuit

shown.

How do we combine resistors R1 - R6 when

the resistors are neither in series nor in

parallel?

Many circuits of the type shown can be

simplified by using three-terminal networks.These are the wye (Y) or tee (T) network, and

the delta () or pi () network as will be

shown next.

These networks occur by themselves or as

part of a larger network.


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Wye-delta transformationsY, T, and Resistors Connections


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Wye-delta transformationsDelta to Wye Conversion

Each resistor in theYnetwork is a product of the resistors in adjacent branches,

divided by the sum of the three resistors.


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Wye-delta transformationsWye to Delta Conversion

Each resistor in the network is the sum of all possible products of Y resistors

taken two at a time, divided by the opposite Y resistor.


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Wye-delta transformationsExample

Transform the wye network, in the figure shown below, to a delta network.


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For the bridge network in the figure below, find Raband i.


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For the bridge network in the figure below, find Raband i.

We first find the equivalent resistor R

Then convert sub-network to wye connection

form as shown:


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Wye-delta transformationsExercise

For the circuit given, find the source current i.


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Obtain the equivalent resistance at the terminals a-b of the circuit shown.

Answer: 36.25.


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Answer: 36.25.


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End of Chapter 2

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Basic circuit analysis laws