Unit 2 DC Circuits

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Electrical Engineering IElectrotechnology I

SECTION 2

D.C. CIRCUITS

AND NETWORK ANALYSIS

(A) Series Circuits
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BASIC COMPONENTS

OF AN ELECTRIC CIRCUIT


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SERIES CIRCUIT

Two elements are in series if:

They have only one terminal in common;

The common point between the two elements

is not connected to another current-carryingelement


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R1 and R2 are

connected inseries

R1 and R2 are not

connected in

series as R3 is

also connected tothe common point


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SERIES CIRCUIT

The current through series elements is the

same.

The total resistance of a series circuit is

the sum of the resistances in the circuitwhen looking into the circuit.

T

S

N321T

R

EIthisFrom

R...RRRR


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R1 is in series with R2


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SERIES CIRCUIT

Both of these are

series circuits


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SERIES CIRCUIT

Voltage across each resistor is

determined by the current flow.

Power dissipated by each resistor is thendetermined by the voltage across and

the currentflow through each resistor

1

2

11

2

1111

NN332211

R

VRIIVPthisFrom

IRV,....,IRV,IRV,IRV


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SERIES CIRCUIT

EXAMPLE

Determine:

Total resistance Source current

Voltages across

each resistor

Power

dissipated in

each resistor


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SERIES CIRCUIT

SOLUTION

V5,125x5,2VV5,21x5,2V

V52x5,2V

A5,28

20I

8512R

5

1

2

S

T

W25,315

5,12P

W25,61x5,2P

W5,125x5,2P

2

5

21

2


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VOLTAGE SOURCES

IN SERIES When

connected

in series,

considerpolarity of

individual

sources


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KIRCHHOFFS VOLTAGE LAW

Kirchhoffs voltage law (KVL) states that

algebraic sum of potential rises and drops

around a closed loop / path is zero OR the

algebraic sum of potential rises equals thedrops around a closed loop / path

DROPSRISES

LOOP

VV

0V


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Use assumed direction of current flow to determine

potential rise or drop

When applying KVL, a clockwise (CW) orcounterclockwise (CCW) direction can be used

when following the closed loop

If summing the voltage rises and drops in a loop

and equating them to zero:

When following the loop direction, the polarity (+ or -)

used will be that of the terminal that you leave when

passing through a component.


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0VVE

0V

21

LOOP


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Use assumed direction of current flow to determinepotential rise or drop

When applying KVL, a clockwise (CW) or

counterclockwise (CCW) direction can be usedwhen following the closed loop

If equating the voltage rises to the drops in a loop: When following the loop direction, the polarity (+ or -)

used with a voltage rise will be that of the terminal thatyou leave when passing through a component

When following the loop direction, the polarity (+ or -)used with a voltage drop will be that of the terminal thatyou enter when passing through a component


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21

DROPSRISES

VVE

VV


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Can apply KVL even if no current-carrying

element included in circuit

V6Vx

0V6VxV12

0VLOOP


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EXAMPLE

Determine the voltage V1

V8,2V

V92,4V16EVEV

0EVVE

1

2211

2211


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EXAMPLE

Determine the voltage VX in each case


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SOLUTION


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EXAMPLE

Determine the following

V2

I

R1 and R3


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SOLUTION


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INTERCHANGING

SERIES ELEMENTS Elements (voltage sources or resistors) can

be interchanged without affecting total

resistance, current or power to each element


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INTERCHANGING

SERIES ELEMENTS


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INTERCHANGING

SERIES ELEMENTS


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INTERCHANGING

SERIES ELEMENTS EXAMPLE:

Determine the

current I andthe voltage

across the 7

resistor


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INTERCHANGING

SERIES ELEMENTS SOLUTION:

First interchange the 4 resistor and the 12,5 V battery

Then combine the power sources and solve


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VOLTAGE DIVIDER RULE

Voltage across any resistive

element (RX) will be divided

in the ratio of the magnitude

of the resistances.

The largest resistance will

have the highest voltage

difference across it.

T

XX

R

ERV


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Voltage divides in the ratio of the resistance, not the values of the resistors


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Voltage across anyresistive element willbe divided in the ratioof the magnitude ofthe resistances.

The largestresistance will havethe highest voltagedifference across it.


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EXAMPLE

Using the voltage

divider rule,

determine thevoltages across each

resistor


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SOLUTION

V33,13

10x852

100x10x2

R

ERV

V33,3310x852

100x10x5

R

ERV

V33,5310x852

100x10x8

R

ER

V

3

3

T

33R

3

3

T

22R

3

3

T

1

1R


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Most electrical / electronic systems are

grounded / earthed for reference or safety

purposes

Ground = zero volts Following examples indicates three ways

of sketching the same circuit

NOTATION


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3 different ways of drawing the same circuit


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NOTATION

Two ground symbols indicate same

potential at ve terminal of battery and

one terminal of resistor R2

Note: Point a is positive w.r.t. ground,

Point b is also positive w.r.t. ground but

negative w.r.t. point a


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NOTATION

Replace the special notation for a positive dc voltage

source with a standard symbol


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NOTATION

Replace the special notation for a negative dc voltage

source with a standard symbol


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DOUBLE-SUBSCRIPT

NOTATION A voltage can be measured between two points

orfrom a point to ground

When measured between two points, apply adouble-subscript notation

First letter of the subscript indicates the point ofthe higherpotential the voltage VAB indicatesthat the voltage being measured between pointsA and B and that point A is at the higherpotential, ie the voltage at point A w.r.t. point B


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DOUBLE-SUBSCRIPT

NOTATION Defining the sign for double-subscript notation


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SINGLE-SUBSCRIPT

NOTATION When measured from a point to ground,

the single-subscript notation is used

The letter of the subscript indicates the

point at which the potential is beingmeasured the voltage VA indicates the

voltage at point A relative to ground while

the voltage VB indicates the voltage at

point B relative to ground


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SINGLE-SUBSCRIPT

NOTATION


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NOTATION

EXAMPLE

Determine Vab


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NOTATION

SOLUTION

V4

)V20(V16

VVV baab


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NOTATION

EXAMPLE

Determine Va


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NOTATION

SOLUTION

V9

V4V5

VVV baba


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NOTATION

EXAMPLE

Determine Vab


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NOTATION

EXAMPLE

Determine VAB, VCBand VC for the

network


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NOTATION

SOLUTION

One approach is redraw

circuit as shown

Determine total volt drop

across R1 and R2

Apply Ohms law and

calculate voltages


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NOTATION

SOLUTION

V19EV

V2420A2,1IRV

V3025A2,1IRV

A2,145

V54I

C

1CB

2AB


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NOTATION

SOLUTION Another approach

is to redraw thenetwork

Establish the effectsof E1 and E2

Solve the resultingseries circuit


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NOTATION

SOLUTION

V19EV

V2420A2,1IRV

V3025A2,1IRV

A2,145

V5445

V35V19R

EEI

C

1CB

2AB

T

21


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NOTATION

EXAMPLE

Determine V1 and V2


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NOTATION

SOLUTION


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NOTATION

EXAMPLE

Determine Va, Vb, Vcand Vab


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NOTATION

SOLUTION


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EXAMPLES OF

VOLTAGE SOURCES


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EQUIVALENT OF A

VOLTAGE SOURCE E represents the emf which

can be from a chemical,

solar or electromagnetic

origin RINT represents the internal

resistance


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VOLTAGE SOURCES

INTERNAL RESISTANCES The 3 diagrams indicate:

Ideal voltage source (RINT = 0)

Practical voltage source without any load

Practical voltage source with a load


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Determine value of RINT by:

First measuring the no-load voltage (VNL)

Then connecting a known load (RL) and

measuring the load current (IL) Finally calculating internal resistance (RINT)

LL

NLINT RI

V

R

VOLTAGE SOURCES

INTERNAL RESISTANCES


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EXAMPLE

A battery has an emf of 12 V and an internal

resistance of 0,5 . Determine the load

voltage and the power lost due to internal

resistance if the load resistance is 5 .

VOLTAGE SOURCES



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SOLUTION

W38,25,0x182,2RIP

V91,105,0x182,212RIVV

A182,25,05

12

RR

VI

2

INT

2

LLOST

INTLNLL

INTL

NLL

VOLTAGE SOURCES



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MEASURING VOLTAGE

Meter will be connected in parallel to

component/circuit

Resistance of meter is high (10M)


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MEASURING CURRENT

Meter will be connected in series with

component/circuit

Resistance of meter is low (5m)

Unit 2 DC Circuits

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