lecture 1-2 meenakshi

8/17/2019 Lecture 1-2 Meenakshi

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BITS PilaniHyderabad Campus

ELECTRICAL SCIENCES

EEE F111

Lectures: 1-2 Meenakshi Viswanathan


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BITS Pilani, Hyderabad Campus


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BASIC CIRCUIT ELEMENTS

Ideal Voltage Source:

A device that produces a voltage or potential difference of V

volts across its terminals regardless of what is connected to

it.


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Ideal Voltage Source


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7/32BITS Pilani, Hyderabad Campus

BASIC CIRCUIT ELEMENTS

Ideal Current Source:

Always supplies the same current I irrespective of

the load connected to it.



CIRCUIT LAWS

KIRCHHOFF’S CURRENT LAW



CIRCUIT LAWS

KIRCHHOFF’S VOLTAGE LAW



Single Loop Circuits



One can represent a “real EMF device” as an ideal one

attached to a resistor, called “internal resistance” of the

EMF device:

Rr iiRir E/E 0



Resistance in Series andParallel



Current Division



Voltage Division



Application of Kirchhoff’s Laws

The first step is to label the unknown currents arbitrarily.

Applying Kirchhoff's current law to node a,

14

14 0

ii

ii

Note that i2 is the current in the 20-V source as well as in the 4Ω resistance.



The next step is to define the unknown element

voltages in terms of the arbitrarily assumed

currents.

2

3

1

4

8

2

iV

iV

iV

cb

bd

ab

Applying Kirchoff 's current law to node b,321 iii

Applying KCL to node d, 0243 iii



Applying KVL to the left hand loop,

032bd ab

V V

Applying KVL to the right hand loop,

020 bd cb

V V

Applying KVL to the outside loop,

03220 bcab

V V

2084

3282

32

31

ii

ii



Substitution method(KCL+ KVL+Ohm’s Law)

321 iii

2084

3282

32

31

ii

ii

Ai

Ai

Ai

3

1

4

3

2

1

20)82(8

328)82(

21

21

I I

I I


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Mesh Analysis / Loop CurrentMethod

1. Reduction in number of unknowns

2. Reduction in number of simultaneous

equations.

3. Can be used only for planar networks.

Loop Current is assumed to circulate around the left loop

Loop Current is assumed to circulate around the right loop

1 I

2 I

By inspection,

213

22

11

I I i

I i

I i


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Mesh Analysis /Loop Current Method

213

22

11

I I i

I i

I i

By applying KVL to the left and right loop,

20)(84

32)(82

020

032

212

211

I I I

I I I

V V

V V

bd cb

bd ab

20)82(8

328)82(

21

21

I I

I I

Physical Interpretation??


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Mesh Analysis : CHECKING

KVL around the outside loop,

Ai

Ai

Ai

3

1

4

3

2

1

03220)1(4)4(203220

bcab V V


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Use Mesh Analysis to find V0

I1=10/3 mA

I2 = 4 mA

V0=4(I2-I1)


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Mesh / Loop Analysis

Calculate the current in the 10Ω resistance usingloop currents.


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I1 = 3 A

TRY USING OTHER LOOPS TO GET SAME ANSWER!


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Find Mesh Currents in thecircuit below

6V

3Ω

7Ω

2Ω 4Ω

+ -

+ - +

-

+

-

V 4

V 1

V 3 V

2

7A

+ -V

i 1 i

2

i 3

Find the clockwise mesh currents in the above circuit.

Example 2.6 from Bobrow.

CURRENT SOURCE COMMON BETWEEN TWO FINITE MESHES


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Nodal Analysis /Node Voltage Method

1. Choose node d as reference and measurevoltage of any node w.r.t node d.

2. Apply KCL to each independent node.

V v

vvv

vvvvv

iiibnode At

b

bbb

bbcba

24

84

20

2

328

0

42

0i, 321 b

Ai Ai

Ai

3

1

4

3

2

1


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Voltage Source Between Nonreference nodes


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Nodal Analysis

Calculate Va

Answer: Va=16V

lecture 1-2 meenakshi

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