Lecture Voltage Divider

8/12/2019 Lecture Voltage Divider

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, 8 2014 Ch. 3 Network Analysis- Part I 1

Example 4

For the given circuit, determine

(a) the value of current I,

(b) the power absorbed by the dependent source, and

(c) the resistance seen by the independent voltage source.

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Solution :

(a) Applying Ohms law to the 4- resistor, gives V1= 4I.

Therefore, the value of dependent voltage source is 4.5V1= 4.5(4I) = 18I. By applying KVL,

24 4 2 18 0I I I I 2 A

(b) For the dependent source, the power absorbed is

1

2 2

(4.5 )( ) 4.5(4 )( )

18 18( 2)

P V I I I

I

72 W

What is the meaning of negative sign ?

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Click

Click


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(c) The resistance seen by the independentvoltage source,

24 24

2R I 12

What is the meaning of negative sign ?

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Click

Ans. :The remainder circuit supplies power tothe independent voltage source.


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4

1.11 Voltage Divider Rule:-

The voltage across one of the series resistors is the total

voltage times the ratio of its resistance to the total

resistance.


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5

1.12 Current Divider Rule:-

Current division is dualof voltage division.

21

12

21

22

21

2

121

1

1

RR

Riior

GG

Gii

RR

R

iiorGG

G

ii


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6


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7


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8


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9


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10

1.13 STAR-DELTA TRASFORMATION:-


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11

If we were to connect a source between terminals a and b

of the Y,the resistance between the terminals would be

Rab=R1+R2 eq.1

But the resistance between terminals a and b of the is

Rab=RC(RA+RB) eq.2Combining eq. 1 & eq. 2 we get,

Using a similar approach between terminals b and c, we get

Eq. 3


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12

and between terminals c and a we get

Eq. 4

Eq. 5

If Equation 4 is subtracted from Equation 3 , then

Eq. 6


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13

Now Adding Equations 5 and 6, we get

Using a similar approach, we obtain

Notice that any resistor connected to a point of the Y is

obtained by finding the product of the resistors connected to

the same point in the and then dividing by the sum of

all the resistances.


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14

If all the resistors in a circuit have the same value, R,

then the resulting resistors in the equivalent Y network will

also be equal and have a value given as,


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, 8 2014 Ch. 2 Ohm's Law 15

(a) Star (Y) connection (b) Delta () connection

Star-Delta Connections :

Two ways of connecting three resistors across three

points.

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, 8 2014 Ch. 2 Ohm's Law 16

Star-to-Delta TransformationSame way, we can derive,

B C1 B C

A

R RR R R

R

C A2 C A

B

R RR R RR

A B3 A B

C

R RR R R

R

Remember

The sum of the two nearest

resistances plus the

product of the same two

resistances divided by the

third resistance.

Next


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17


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18


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20


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21

UNIT-2

NETWORK ANALYSIS


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22

2.1 KIRCHHOFFS CURRENT LAW (KCL):-

It states that the algebraic sum of all currents entering

a node is zero. Mathematically:

Currents are positive if enter ing a nodeCurrents are negative if leaving a node.

Example:


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23

Applying Kirchhoff's current law:

I1+ I2+ I3+ I4= 0

(the negative sign in I2indicates that I2has a magnitude

of 3A and is flowing in the direction opposite to that

indicated by the arrow)

Substituting:

5 - 3 + I3+ 2 = 0 Therefore, I3= - 4A (ie 4A

leaving node)


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24

2.2 KIRCHHOFFS VOLTAGE LAW (KVL):-

It states that the algebraic sum of the voltage drops

around any loop, open or closed, is zero. Mathematically

Example:

Going round the loop in the direction of the current, I,

Kirchhoff's Voltage Law gives:

10- 2I - 3I = 0


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25

- 2Iand - 3Iare negative, since they are voltage drops

i.e. represent a decrease in potential on proceeding

round the loop in the direction of I. For the same reason +

10V is positive as it is a voltage rise or increase in potential.

Concluding:

5 I = 10 Therefore, I = 2A


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, 8 2014 26

Polarity of Voltages

Notethat polarity of the voltage (emf) across abattery does not depend upon the assumed

direction of current.


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Applying KVL

1. Select a closed loop.

2. Mark the voltage polarity (+ and -) across

each element in the closed loop.

3. Go round the selected loop, and add up all

the voltages with + orsigns.

4. Any one of the following two rules can be

followed :

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(i) Rule 1 :While travelling, if you meet a voltage

rise, write the voltage withpositive sign ; if you

meet a voltage drop, write the voltage withnegativesign.

(ii) Rule 2 :While travelling, write the voltage

with positive sign if + is encountered first; writethe voltage with negative sign ifis encountered

first.

We shall be following Rule 1, as it has a stronganalogy with the physical height (altitude)of a

place.

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Example 5 :Use KVL to find vR2and vx.

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For finding vR2, we write KVL eqn. going

around loop abgha clockwise:

If you choose to go around the loopanticlockwise, you get

Giving the same result.

V320436 22 RR vv

V320364 22 RR vv

Next

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Click


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There are two ways to determine vx

1)We can consider this voltage as the voltage

across the gap from dtof. Writing KVL

(habcdfgh) :

2) Knowing vR2, apply a short-cut (bcdfgb) :

V6

x

x

v

v 01412364

V6

x

x

vv 0321412

Next

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Click


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Solution :

We need not find the currentsI1, I2andI3.

Instead, we reduce the network.

Next

E l 7 D i h l f


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Example 7 : Determine the value of

current I .

23I4 = 0 or I = -5 A

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Example 8

Using KCL and KVL, determine the currents ixand iyin the network shown.

Next


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Solution : Using KCL, the currents in other branches are

marked as shown. Writing KVL equations for the loops

1, 2 and 3,

503530)(3502

50227

02)(2505

1001005

0510100

1

1

1

1

1

1

IIIIIII

III

IIII

III

II

yx

yxy

yx

yxx

yx

x

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8 2014 Ch 3 Network Analysis Part I 37

;

50

50100

353

2271005

1

I

II

y

x

Writing the above equations in matrix form,Click

Using Calculator, we solve for Ixand Iy,Click

; andx yI I 3.87 A 0.51 A

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Lecture Voltage Divider