Chap2 Beta1313 Part i

Chapter 2 1

CHAPTER 2:

DIRECT CURRENT (DC)

CIRCUITS

(BASIC LAWS)

Chapter 2 2

DC SOURCE

Dc a constant flow of electric charge with time.

For ideal voltage source and ideal current source, theysupply fixed voltage and fixed current respectively.

Ideal source do not exist. It use to simplify circuitanalysis.

Chapter 2 3

There are two types of voltage sources independentand dependent voltage source.

Dependent voltage source - the voltage produced maydepend on some other circuit variable such as current orvoltage.

The sign of voltage can be negative.

DC SOURCE contd.

Chapter 2 4

Voltage source can be connected in series. In

this connection the voltage value is added.

But can not connect in parallel. Could easily

cause component failure.

DC SOURCE contd.

Chapter 2 5

There are two types of current sources -

independent and dependent current source.

Dependent current source - the current

produced may depend on some other circuit

variable such as current or voltage.

DC SOURCE contd.

Chapter 2 6

The sign of current can be negative.

Ideal current sources cannot be connected in series.

Could easily cause component failure (smoke).

DC SOURCE contd.

Chapter 2 7

However it is allowed if the sources is connected

in parallel.

DC SOURCE contd.

Chapter 2 8

OHMS LAW

Free electrons that flow through a material will collidewith the materials atoms.

This collision will cause the electrons to loose someenergy and thus restrict its movement.

The more collisions, the more the flow of electrons isrestricted.

This restriction varies and is determined by the type ofmaterial.

The property of material that restricts the flow ofelectrons known as resistance.

Chapter 2 9

Resistance is represented by symbol R.

Conductor (e.g. wires) have very low resistance (< 0.1 ) that can usually be ignored.

It is assumed that wires have zero resistance.

Insulators (e.g. air) have very large resistance

(> 50M ).

OHMS LAW contd.

Chapter 2 10

The circuit element used to resist the flow of

current is known as resistor.

Resistors have medium range of resistance and

must be accounted for in the circuit analysis.

OHMS LAW contd.

Chapter 2 11

OHMS LAW contd.

Resistor is the simplest passive element.

Chapter 2 12

OHMS LAW contd.

Georg Simon Ohm (1787 1854) found the relationshipbetween current and voltage for a resistor. Thisrelationship known as Ohms law.

Ohms law the voltage v across a resistor is directlyproportional to the current i flowing through the resistor.

v i

v mi

=

Chapter 2 13

The slope m, is equal to the resistance, Rof an element.

Only linear resistor obey the Ohms law. Nonlinear does not.

Normally, resistor is assumed to be linear.

OHMS LAW contd.

Chapter 2 14

OHMS LAW contd.

Thus the equation becomes:

which is the mathematical form of Ohms law.

R is measured in the unit of ohms, designated

.

The resistance, R of an element denotes its

ability to resist the flow of electric current.

v iR=

Chapter 2 15

OHMS LAW contd.

Can also be written as:

R can be ranged from zero to infinity.

vR

i=

Chapter 2 16

OHMS LAW contd.

An element with R = 0 short circuit.

In short circuit, the voltage is always zero but the current is not.

In practice, a short circuit is always a connecting wire assumed to be a perfect conductor.

0v iR= =

Chapter 2 17

OHMS LAW contd.

An element with R = open circuit.

In open circuit, the current is always zero but the voltage is not.

0i =

Chapter 2 18

OHMS LAW contd.

Another quantity in circuit analysis conductance, denoted by G.

Conductance is a measure of how well an element will conduct electric current.

Inverse of resistance.

The unit of conductance mho, or siemens, S

1 S = 1 = 1 A/V

1 iG

R v= =

Chapter 2 19

OHMS LAW contd.

The same resistance can be expressed in ohm

or siemens. Example: 10 = 0.1 S

The power dissipate by the resistor can

expressed in term of R:

2

2 vp vi i R

R= = =

Chapter 2 20

OHMS LAW contd.

Or can also be expresses in term of G:

Note: The power dissipated in a resistor is alwayspositive.

Thus resistors always absorbed power from the circuit.

The direction of current and the polarity of voltage mustconform with the passive sign convention. Current enter at positive sign - +v, +p Current enter at negative sign - -v, -p

2

2 ip vi v G

G= = =

Chapter 2 21

RESISTOR

The resistor is far and away the simplest circuit element.

In a resistor, the voltage v is proportional to the current i,with the constant of proportionality R known as theresistance.

Resistor is an element denotes its ability to resist theflow of electric current, it is measured in ohms ().

=

=or

v i

v iR

vR

i

Chapter 2 22

RESISTOR contd.

Chapter 2 23

The above resistance is 1,000,000 or 1M.

The 10% means the actual resistance is between 900k and 1.1M .

RESISTOR contd.

Chapter 2 24

The above resistance is 150,000 or 150k.

The 5% means the actual resistance is between 142.5k and 157.5k.

RESISTOR contd.

Chapter 2 25

The above resistance is 3,300 or 3.3k.

The 5% means the actual resistance is between

3,135 and 3,465.

RESISTOR contd.

Chapter 2 26

EXAMPLE

1) An electric iron draws 2A at 120V. Find its resistance.

2) The essential of a toaster is an electrical element (a resistor) that converts electrical energy to heat energy. How much current is drawn by a toaster with resistance 12 at 240V?

12060

2

vR

i= = =

24020

12

vi A

R= = =

Chapter 2 27

3) In the given circuit, calculate the current i, the conductance G and the power p.

Chapter 2 28

Solution:

The voltage across the resistor is the same as the source voltage (30V) because the resistor and the voltage source are connected to the same pair of terminals. Hence the current is

The conductance is

The power can be calculated in various ways

306

5

vi mA

R k= = =

1 10.2

5G mS

R k= = =

30(6 ) 180p vi m mW= = =

Chapter 2 29

or

or

or

2 2(6 ) (5 ) 180p i R m k mW= = =

2 230 (0.2 ) 180p v G m mW= = =

2 230180

5

vp mW

R k= = =

Chapter 2 30

4) For the given circuit, calculate the voltage v, the conductance G and the power p.

Answer:

20V, 100S, 40mW

Chapter 2 31

KIRCHHOFFS LAWS

The foundation of circuit analysis is: The defining equations for circuit elements (e.g.

ohms law) Kirchhoffs current law (KCL) Kirchhoffs voltage law (KVL)

The defining equations tell how the voltage and current within a circuit element are related.

Kirchhoffs laws tell us how the voltages and currents in different branches are related.

Chapter 2 32

KIRCHHOFFS LAWS contd.

Kirchhoffs current law (KCL) that the algebraic sum of currents entering a node (or a closed boundary) is zero.

One can assume that currents entering a node as positive while leaving the node as negative or vice versa.

1

0N

nn

i=

= N = number of branches connected to a node.

Chapter 2 33


The sum of currents entering a node is equal to the sum of the currents leaving a node.

Common sense:

All of the electrons have to go somewhere.

The current that goes in, has to come out some place.

Example:

Applying KCL:

4 + i = 5 + 11

thus, i = 12A

Chapter 2 34


Consider the following figure where all the current source can be combined as in figure (b).

Chapter 2 35


Kirchhoffs voltage law (KVL) the algebraic sum of allvoltages around a closed path (or loop) is zero.

1. Start at any branch and go around theloop either clockwise / counterclockwise.

2. Check which terminal the loop (the redarrow) encounter first,

a) If positive terminal then +v.

b) If negative terminal then v.

1

0M

mm

v=

= M = number of voltages on the loop.

Chapter 2 36


Thus, KVL yields

rearranging terms gives

When voltage sources are connected in series, KVL canbe applied to obtained the total voltage.

The combined voltage is the algebraic sum of thevoltages of the individual sources.

1 2 3 4 5 0v v v v v + + + =

2 3 5 1 4v v v v v+ + = +

Chapter 2 37


Chapter 2 38

Much of the circuit analysis will be based on these three laws.

Ohms Law : KCL :KVL :

These laws alone are sufficient to analyze many circuits.

Important notes: If current enter at positive terminal

v = +(iR) p = +(vi)

If current enter at negative terminal v = -(iR) p = -(vi)


v iR=

0ni =0mv =

Chapter 2 39


Steps for circuit analysis:

1. Determine the currents direction that enter and leaving each node.

2. Assign the +ve and ve sign for each elements.

3. Obtain the voltage across each elements by applying Ohms law. If the current flow through:

Positive terminal/sign, thus v = +iR

Negative terminal/sign, thus v = -iR

4. Apply the KVL:

Choose the direction of loops either clockwise or counterclockwise. Each loops must not overlapping.

Obtain the KVL equation based on the chosen loop (clockwise or counterclockwise). If the loop enter:

Positive terminal/sign, +v

Negative terminal/sign, -v

Chapter 2 40

5. For KCL, obtain the equation by assuming if

a current:

Enter a node, +i

Leaving a node, -i


Chapter 2 41

EXAMPLE

1)Find voltages v1 and v2 for the given circuit.

Chapter 2 42

From Ohms law,

v1 = 2i, v2 = -3i (a)

Applying KVL around the loop gives

-20 + v1 v2 = 0 (b)

Chapter 2 43

Substituting equation a into b

-20 + 2i (-3i) = 0

Thus, i = 4A

Substituting i into equations a gives

v1 = 8V & v2 = -12V

Chapter 2 44

EXAMPLE

2) Find voltages v1 and v2 for the given circuit.

Ans: 12V & -6V

Chapter 2 45

EXAMPLE

3) Find voltages vx and v0 for the given circuit.

Ans: 10V & -5V

Chapter 2 46

EXAMPLE

4) Find current and voltages for the given circuit.

Ans: i1 = 3A, i2 = 2A, i3 = 1A, v1 = 24V, v2 = 6V, v3 = 6V

Chapter 2 47

SERIES RESISTORS AND VOLTAGE DIVISION

Recall: The current that pass through the serieselements has the same value.

Thus,

i1 = i2 = i3

Chapter 2 48

SERIES RESISTORS AND VOLTAGE DIVISION contd.

Any number of resistors connected in series is the sum of the individual resistances.

=

= + + + =e 1 21

...N

q N nn

R R R R R

N = number of resistors in series

Chapter 2 49

Consider the following figure:

where Req = R1 + R2


Chapter 2 50


Applying Ohms law to each of the resistors,

v1 = iR1 , v2 = iR2 (2.1)


Chapter 2 51

KVL (clockwise):

v1 + v2 v = 0 (2.2)

Combining both the above equation,

v = v1+ v2 = i(R1 + R2)

or (2.3)


1 2

vi

R R=

+

Chapter 2 52

To determine the voltage across each resistor, substitute eq 2.3 into 2.1

The above equation is called the principle of voltage division.

The source voltage v is divided among the resistors in direct proportion to their resistances; the larger the resistance, the larger the voltage drop.

11

1 2

Rv v

R R=

+

22

1 2

Rv v

R R=

+


Chapter 2 53

PARALLEL RESISTORS AND CURRENT DIVISION

Note that elements in parallel have the same voltage drops across them.

Chapter 2 54


where

PARALLEL RESISTORS AND CURRENT DIVISION contd.

1 2

1 2

1 2

1 1 1

eq

eq

R R R

R RR

R R

= +

=+

Chapter 2 55


Applying Ohms law to each of the resistors,

v = i1R1 , v = i2R2


Chapter 2 56

or

(2.4)

Applying KCL at node a gives the total current i as

i = i1 + i2 (2.5)

Substituting eq 2.4 into 2.5, yields

(2.6)


1

1

vi

R= 2

2

vi

R=

+= + = + =

1 2

1 2 1 2 1 2

1 1 R Rv vi v v

R R R R R R

Chapter 2 57

From eq 2.6

(2.7)

Substituting eq 2.7 into 2.4 gives,

(2.8)

Equation 2.8 is known as the principle of current division.


=

+

1 2

1 2

R Rv i

R R

2 11 2

1 2 1 2

iR iRi i

R R R R= =

+ +

Chapter 2 58

Impotant notes:

Notice that larger current flows through smallerresistance.

In electrical circuit, current will always flow through apath with least resistance.

If there is a short circuit, the entire current will flowthrough the short circuit.


Chapter 2 59

1) Find Req for the above figure.

Answer: Req = 14.4

EXAMPLE

Chapter 2 60

2) Find Req for the above figure.

Answer: Req = 11.2

Chapter 2 61

2) Find Rab for the above figure.

Answer: Rab = 11

Chapter 2 62

2) Find Rab and current i for the above figure.

Answer: Rab = 9.632 & i = 12.458A

Chapter 2 63

WYE-DELTA TRANSFORMATIONS

Some resistors are combined neither in series

nor parallel. For example,

Chapter 2 64

These type of connection can be simplified by using three-terminal equivalent network.

WYE-DELTA TRANSFORMATIONS contd.

Chapter 2 65

The wye (Y) / tee (T) network and the delta () /

pi ().

The wye network can be converted into the delta

network and vice versa.

This conversion will simplify the circuit analysis.

Note: This conversion did not take anything out

of the circuit or put in anything new.


Chapter 2 66

Delta-Wye conversion:


1b c

a b c

R RR

R R R=

+ +

3a b

a b c

R RR

R R R=

+ +

2a c

a b c

R RR

R R R=

+ +

Chapter 2 67

Wye-Delta conversion:


1 2 2 3 3 1

1

a

R R R R R RR

R

+ +=

1 2 2 3 3 1

2

b

R R R R R RR

R

+ +=

1 2 2 3 3 1

3

c

R R R R R RR

R

+ +=

Chapter 2 68

EXAMPLE

1) Convert the delta network to wye network

Chapter 2 69

1

10(25) 2505

15 10 25 50b c

a b c

R RR

R R R= = = =

+ + + +

3

15(10) 1503

15 10 25 50a b

a b c

R RR

R R R= = = =

+ + + +

2

25(15) 3757.5

15 10 25 50a c

a b c

R RR

R R R= = = =

+ + + +

Chapter 2 70

Converted delta to wye network:

Chapter 2 71

2) Convert the wye network to delta network

Answer: Ra = 140 ; Rb = 70 ; Rc = 35

Chapter 2 72

3) Obtain the equivalent resistance Rab for the circuit below.

Chapter 2 73

Answer: Rab = 9.632 & i = 12.458A

Chapter 2 74

CAPACITOR

Unlike resistor which dissipate energy, capacitor store energy, which can be retrieved at later time. It is a passive elements.

Also called storage elements. The energy is stored in its electric field.

The unit to measure the capacitance of a capacitor is farad (F).

Chapter 2 75

1. There is a capacitor in parallel with the resistorand light bulb. The way the capacitor functionsis by acting as a very low resistance load whenthe circuit is initially turned on. Note: In electricalcircuit, current will always flow through a pathwith least resistance

CAPACITOR contd.

2. Initially, the capacitor has a very low resistance,almost 0. Since electricity takes the path ofleast resistance, almost all the electricity flowsthrough the capacitor, not the resistor, as theresistor has considerably higher resistance.

Capacitor acts as a storage element:

Chapter 2 76

3. As a capacitor charges, its resistance increasesas it gains more and more charge. As theresistance of the capacitor climbs, electricitybegins to flow not only to the capacitor, butthrough the resistor as well.

CAPACITOR contd.

4. Once the capacitor's voltage equals that of thebattery, meaning it is fully charged, it will notallow any current to pass through it. As acapacitor charges its resistance increases andbecomes effectively infinite (open connection)and all the electricity flows through the resistor.

Chapter 2 77

5. Once the voltage source is disconnected, the capacitor will act as a voltage source itself.

CAPACITOR contd.

6. As time goes on, the capacitor's charge begins to drop, and so does its voltage. This means less current flowing through the resistor.

Chapter 2 78

7. Once the capacitor is fully discharged, no current will flow.

CAPACITOR contd.

Chapter 2 79

CAPACITOR contd.

Chapter 2 80

INDUCTOR

It is a passive element designed to store energy in its magnetic field.

Inductor consists of a coil of conducting wire.

Chapter 2 81

Inductance is measured in henrys (H).

INDUCTOR contd.

Example:

What you see here is a battery, a light bulb, a coil of wire arounda piece of iron (yellow) and a switch. The coil of wire is aninductor.

Chapter 2 82

Without the inductor in this circuit, what you wouldhave is a normal flashlight. You close the switch andthe bulb lights up.

If there is an inductor, when the switch is closed thebulb burns brightly and then gets dimmer. When theswitch is opened, the bulb burns very brightly andthen quickly goes out.

The reason for this strange behavior is the inductor.When current first starts flowing in the coil, the coilwants to build up a magnetic field.

INDUCTOR contd.

Chapter 2 83

While the field is building, the coil inhibits the flow ofcurrent. Once the field is built, current can flownormally through the wire (coil).

A large amount of current will flow through this coil letonly a small amount of current flow to the light bulb.This is why the bulb gets dimmer.

When the switch gets opened, the magnetic fieldaround the coil keeps current flowing in the coil untilthe field collapses. This current keeps the bulb lit for aperiod of time even though the switch is open. Inother words, an inductor can store energy in itsmagnetic field, and an inductor tends to resist anychange in the amount of current flowing through it.

INDUCTOR contd.

Chapter 2 84

Analogy:

One way to visualize the action of an inductor is toimagine a narrow channel with water flowing through it,and a heavy water wheel that has its paddles dippinginto the channel. Imagine that the water in the channel isnot flowing initially. Now you try to start the waterflowing. The paddle wheel will tend to prevent the waterfrom flowing until it has come up to speed with the water.If you then try to stop the flow of water in the channel,the spinning water wheel will try to keep the watermoving until its speed of rotation slows back down to thespeed of the water. An inductor is doing the same thingwith the flow of electrons in a wire - an inductor resists achange in the flow of electrons.

INDUCTOR contd.

Chapter 2 85

INDUCTOR contd.

Chap2 Beta1313 Part i

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