Topic 1. Basic Concepts Laws

8/9/2019 Topic 1. Basic Concepts Laws

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1.1 Basic Concepts & Definitions

1.2 Circuit Elements

1.3 Sources1.4 Circuit Notation

1.5 Kirchhoff’s Law1.6 Series and Parallel Resistors Circuit

1.7 Delta-Wye Transformations1.8 Application

Chapter 1: Basic Concepts and Laws


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• Elementary Charge is 1.602176487×10−19 C(1 proton or electron)

• Question: How many protons/electrons have

passed from the positive to negative terminal for

1 Ampere of current flowing for 10 seconds?

&


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DC (Direct current)- current that remains constant

with time- current flows in uni-directional

AC (Alternating current)- current that varies sinusoidallywith time

- current flows in bi-directional

&


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Resistance (R)- Resistance is an opposition to the current flow’s intensity.

- Like the slope that controls the flow of water in pipeline.- Resistance is measured in ohm (Ω).

&

A

l

R ρ =


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Conductance (G)- Conductance is the tendency to conduct current.- It is the reciprocal of resistance, measured in siemens (S),

G = 1/R; [S] = [Ω]-1

- Electronic engineers also use Mho (℧).

Question: Beside metals, can water or other fluids conductelectricity?

&


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• Current & Voltage are two basic variables in electriccircuits, also serve as signals of information

• Voltage / potential difference = move the electrons in aparticular direction, work must be done on every unitcharge, measured in Volts (V)

• (+) & (-) signs defines the direction or voltage polarity.• 1 V = 1 J/C = 1Nm/C

• Energy is capacity to do work, in Joules (J)

&

dQdW V =

baab

baab

V V

V V V V

−=

=−= 9


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Ground Connections- The chassis ground should be earthed for safety

precaution.- The earth is a neutral body with huge electric capacity(electrons and protons). It can accept or release electronsand is always 0 V.

- The following shows a return path between two earthgrounds.

&


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Open circuit (OC)- Two terminals (A & B) are not connected.- No current flows due to ∞ resistance, air is an

insulator.- If the potential difference, V is too high, airmay breakdown, and current would flow.

Short circuit (SC)- Two terminals (A & B) are connected withgood conductor such as copper wire, assume to

be zero resistance.- The assumption of negligible resistance no

longer holds for transmission lines that extendover longer distance, when dealing in powersystem.

A

B

οοοο

οοοο

R = 0 Ω

A

B

οοοο

οοοοR = ∞ Ω

&


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Electric Shock

Human body has a resistance of, R = 10 ~ 50 kΩ

- Sweat can reduce body resistance up to 10 times.- It is possible to reduce this resistance down to 1 kΩ bysweaty hands or holding a conductor tightly.

-The electrocution effect on a human body depends on

the current level:1 mA Feeling sensation

10 mA Immobilizing20 mA Breathing difficulty

100 mA Fatal

- Dangerous voltage level is about 30 V and above. We are safe

as long as insulated from the ground, preventing current to flowacross the heart (wear rubber glove, use wooden stick).

&


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Power – rate of change of energy, either supplying orabsorbing energy, measured in Watts (W)

•if current enters through +ve terminal of an element

Element absorbs Power•if current enters through -ve terminal of an elementElement supplies Power

From P = V I (power equation) and V = I R (Ohm's law),

&

VI dt

dW

dt

dQ

dQ

dW

P ==⋅=

R

V R I P

22

==∴


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&

Voltage measurement Current measurement


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&


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2 types of elements:

Passive elements – absorbs energy

e.g. Resistor (R), Capacitor (C) & Inductor(L), often acts as load that convertselectrical energy to mechanical or thermal

energy.

Active elements – generates energy

e.g. Generator & Battery (voltage source),

operating amplifier (current source)


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(i) Ideal independent sources- completely independent of other circuit elements

(ii) Ideal dependent sources- controlled by another voltage / current source


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Ideal Voltage Source• Its voltage is independent from the magnitude &

direction of its current

• When the current leaves +ve terminal, it delivers powerto external circuit – acts as an e.m.f (electromotiveforce) source

• When the current enters to +ve terminal, it acts as aload

Externalcircuit

+

Vi

_

iVi

i

v


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Ideal Current Source• Its current is constant irrespective of magnitude &

direction of the voltage across its terminals

• The voltage across the terminal depends on theelements that connected at the external circuit

External

circuit

+

Vi

_

i

Vi

i

i


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Practical Voltage Source• Voltage terminal of a practical source usually

decreases as current drawn from it increases. This isdue to the ‘voltage drop’ across internal resistor ofthe voltage source

VL = VS - ILRS

S R


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Practical Current Source• Practical current source has an internal resistance RS.

Hence the supplied current varies.• To compute the Load current & Load voltage:

S

LS

S

L IRR

RI ×

+= S

LS

LS

LLL IRR

RRRIV ×

+==


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Source Conversion• Voltage source, VS with a series resistance RS

may be converted to current source, IS with a parallel

resistance, RS without effecting the rest of the circuit &vice versa.• To convert: voltage source current source

IS = VS /RS• To convert: current source voltage source

VS = ISRS


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Example 1:(a) Determine the current IL(b) Convert the voltage

source to a current source(c) Use the resulting current

source of part (b),

calculate the currentthrough the load resistorand compare your answer

to the result of part (a).


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• Branch a single two-terminal element in anelectric circuit

• Node a junction point connecting two or more

branches• Loop a closed path in the circuit

• Mesh a loop which does not contain any other

loops• Ground at zero potential, where voltage of any

node in the circuit is expressed with

reference to the ground (or in theabsence of ground, one of the nodes is

taken as the reference node)


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Series & Parallel Circuit

• Elements in series – only share a single node & having

the same flow of current.• Elements in parallel – share the same pair of terminals /nodes & have same voltage drop across them. The

current have more alternate paths to flow.

• Series components can be called a string.• Parallel components can be called a bank.

• Series-parallel circuit is the combination of strings andbanks


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String

Bank

Another String



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Another Bank

Bank

+

-



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• Specify reference directions

• +ve values means our reference direction is right• -ve values means the opposite direction is true


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Kirchhoff’s Voltage Law (KVL)• States that the algebraic sum of all voltages around a

closed loop is zero

• Sum of potential drop = sum of e.m.f. (electromotiveforce) around the closed loop

• Procedure:

1. Choose either a clockwise or counterclockwise aroundthe closed loop.

2. Assign +ve & -ve signs to the elements of the circuit

3. Apply KVL & write voltage equation for each closed loop


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Solution:-v1 + v2 + v3 - v4 + v5 = 0 or

v1 - v2 - v3 + v4 - v5 = 0

v2 + v3 + v5 = v1 + v4

1. Go clockwise around loop

2. Assign +ve & -ve signs toeach elements

3. Apply KVL & write voltage

equation

0=

∑ n

v

Example 2: Write a voltage equation


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Solution:

(1a) Apply KVL around a path (that includes the source inthe clockwise direction, – E + 12 + Vx = 0 ⇒ Vx = E – 12 = 32 – 12 = 20 V

or(1b) Alternatively, apply KVL around the clockwise path,including resistor R3 gives

– Vx + 6 + 14 = 0 ⇒ Vx = 20 V (same answer)


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Kirchhoff’s Current Law (KCL)• States that the algebraic sum of all currents entering a

node (or a closed boundary) is zero

• Sum of current entering a node = Sum of current leavingthe node

• Apply KCL to a closed boundary

i1 - i2 + i3 + i4 - i5 = 0i1 + i3 + i4 = i2 + i5

0=

∑ni


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At node p - I T + I 1 - I 2 + I 3 = 0

I T =

I 1 -

I 2 +

I 3

Example 4: Write a current equation.P

oSolution:


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Example 5: (KCL)Determine currents I1, I3, I4 and I5 for the circuit


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Applying KCL: Σ Ii = Σ Io

node a I = I1 + I25 = I1 + 4 ⇒ I1 = 1 A

node b I1 = I3 = 1 A

node c I2 = I4 = 4 Anode d I5 = I3 + I4

I5 = 1 + 4 = 5 A

Solution:


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Series Resistors & Voltage Divider• The two resistors are in series, since the same current i

flows through them.

v1 = iR1 v2 = iR2

apply KVL:

-v + v1 + v2 = 0

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

v = i(Req) where

Req = R1+ R2


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Series Resistors• The equivalent resistance of any number of resistors

connected in series is the sum of individual resistances.

N

n=1Req = R1+ R2 + R3+…+ RN = Σ Rn


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Voltage Divider• To determine the voltage across each resistor:

• Principle of Voltage divider – the larger the resistance,the larger the voltage drop

21

1

1

v

R R

Rv

+

=

21

2

2v

R R

Rv

+=

v R

1+ R

2+ …+ R

N

n R

nv =


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Parallel Resistors & Current Divider• The two resistors are connected in parallel and have the

same voltage v across them.

v1 = i1R1 = i2R2

apply KCL at node a:

i = i1 + i2

i +=

1 R

v

2 R

v+=

1 R

1

2 R

1 v

i =

eq R

v


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Parallel Resistors• The equivalent resistance of two parallel resistors is the

product of individual resistances divided by their sum, .

• In the case of a circuit with N resistors in parallel

+1

R1

2 R1=

eq R1 +

N R1+ … or

Geq= G1+ G2 + G3+…+ G N Conductance

+

1 R

1

2 R

1=

eq R

1

2 R

21eq

R R R +=

1 R

21 // R R=


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Current Divider• Principle of current divider – total i current is shared by

the resistors in inverse proportion to their resistances

• If a current divider has N resistors (or conductors) inparallel with the source current i

or

iGGG

G

i N

n

n +++

=

...21

iGG

Gi

iGG

Gi

i R R

Ri

i R R

R

i

21

22

21

11

21

1

2

21

2

1

+=

+=

+=

+=


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Current Divider(a) Suppose R2 = 0 → R2 is short circuit

ii R R

Ri

i R R

Ri

R R

R R Req

=+

=

=+

=

=

+

=

21

1

2

21

2

1

21

21

0

0

Entire current i flows through

the short circuit


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Example 6(network reduction): Find Req for the circuit.Solution:

6//3=2

(2+2)//6=2.4

(1+5)=6

Ω=++=∴ 4.1484.24eq R


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Example 7(network reduction): Find Rab forthe circuit.

6//3

Solution: 3//6=2

(3//6)+1=3

12//4=3 (1+5)=6

Ω=+=∴ 2.11)3 // 2(10ab R


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• Consider the bridge circuit in the following figure, whereresistors are neither in series nor in parallel

• These network are used in three-phase networks,electrical filters and matching networks

• These are the Wye (Y) and delta (∆) network


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• (i) Wye (Y) or Tee (T) network

• (ii) Delta (∆) or pi (Π) network


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Delta to Wye Conversion• Superimpose a Wye network on the Delta network to

find R eq in the Wye network – to simplify computation

)1.1.....()(

)1.1.....()(

,

)1.1.....()(

) //()(

)(

)()(

3234

2113

3112

12

3112

1212

c R R R

R R R R R R

b R R R

R R R R R R

similarly

a R R R

R R R R R R

R R R R

R RY R

RY R

cba

cba

cba

bac

cba

cab

cab

++

+=+=

++

+=+=

++

+=+=

+=∆

+=

∆=


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)3.1.....(

),1.1()3.1(

)3.1.....(

),1.1()2.1(

)3.1.....(

),1.1()2.1(

)2.1.....()(

),1.1()1.1(

3

2

1

21

b R R R

R R R

a fromaSubtract

b R R R

R R R

b fromSubtract

a

R R R

R R R

bto Add R R R

R R R R R

a fromcSubtract

cba

ba

cba

ca

cba

cb

cba

abc

++

=

++

=

++

=

++

−=−

Each resistor in Y network isthe product of resistors in 2adjacent ∆ branches, anddivided by the sum of three

∆ resistors


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Wye to Delta Conversion

Each resistor in ∆ network isthe sum of all possibleproducts of 2 resistors in Ynetwork, and divided by the

opposite Y resistor

( )

)5.1.....(

),1.1()3.1(

)5.1.....(

),1.1()2.1(

)5.1.....(

),3.1()3.1()4.1(

)4.1...(.

.)(

),1.1()1.1(

3

133221

2

133221

1

133221

133221

2133221

c R

R R R R R R R

a fromaSubtract

b R

R R R R R R R

b fromSubtract

a R

R R R R R R R

ctoa Eqby Eq Divide

R R R

R R R R R R R R R

R R R

R R R R R R R R R R R R

ctoa EqFrom

c

b

a

cba

cba

cba

cbacba

++

=

++=

++=

++=++

++

++=++


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Example 8: convert∆

network to an equivalent Y network

Ω=++

×=++

=

Ω=++

×=

++=

Ω=

++

×=

++

=

3151025

1015

5.7151025

1525

5

151025

1025

3

2

1

cba

ba

cba

ca

cba

cb

R R R R R R

R R R

R R R

R R R

R R R

Solution:


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Example 9: Obtain Rab

for the given circuit & find current i.

Ω=×+×+×

=++

=

Ω=×+×+×

=++

=

Ω=×+×+×=++=

705

)105()520()2010(

5.1720

)105()520()2010(

3510

)105()520()2010(

3

133221

2

133221

1

133221

R

R R R R R R R

R

R R R R R R R

R R R R R R R R

c

b

a

R1

R 2

R 3

Solution:


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Solution(continue…):

Ω=

Ω=

Ω=

5.1035 // 15

29.75.17 // 5.12

2130 // 70

A46.1263.9

120

63.921 // )5.1029.7(

===∴

Ω=+=

ab

s

ab

R

vi

R


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• Resistors are often used to model devices thatconvert electrical energy into heat or other forms ofenergy.

• Such devices include conducting wire, light bulbs,electric heaters, stoves, ovens, and loudspeakers.


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Three light bulbs are connected to a 9-V battery as shown inFigure (a). Calculate:

(a) the total current supplied by the battery(b) the current through each bulb(c) the resistance of each bulb.

Solution

(a) The total power supplied by the battery


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is equal to the total power absorbed bythe bulbs; that is,

P = 15 + 10 + 20 = 45W

Since P = VI , then the total currentsupplied by the battery is

AV P I 5

945 ===

(b) The bulbs can be modeled as resistors as shown in

Figure (b). Since R 1 (20-W bulb) is in parallel with the batteryas well as the series combination of R 2 and R 3

V 1 = V 2 + V 3 = 9 V

Solution

The current through R 1 is


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AV

P I 222.29

201

===

By KCL, the current through the seriescombination of R 2 and R 3 is

A I I I 778.2222.2512

=−=−=

(c) Since R I P 2=

Ω===

Ω===

945.1

778.2

15

05.4222.2

20

22

2

22

221

1

1

I

P R

I

P

RΩ=== 279.1

778.2

1022

2

33

I

P R

Topic 1. Basic Concepts Laws

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