Universal Collage Of Engineering And Technology

Subject : Circuit & Network

Aim of Topic: Thevenin’s And Norton’s Theorem

Group 7: Div : C Name En.No.

Jay Pandya. : 130460109034

Jay Bhavsar : 130460109006

Darshan Patel : 130460109043

Yagnik Dudharejiya : 130460109013Guided By: Prof. Naveen Sharma

Flow of presentationStatement of Thevnin’s theoremExamples of Thevnin’s theoremStatement of Norton’s theoremExamples of Norton’s theorem

THEVENIN’S THEOREM:Consider the following:

Network1

Network2•

•A

B

Figure 1: Coupled networks.

For purposes of discussion, at this point, we considerthat both networks are composed of resistors and independent voltage and current sources

THEVENIN’S THEOREM:

Suppose Network 2 is detached from Network 1 andwe focus temporarily only on Network 1.

Network1

•

•

A

B

Figure 2: Network 1, open-circuited.

Network 1 can be as complicated in structure as onecan imagine. Maybe 45 meshes, 387 resistors, 91 voltage sources and 39 current sources.

Network1

•

•

A

B

THEVENIN’S THEOREM:

Now place a voltmeter across terminals A-B andread the voltage. We call this the open-circuit voltage.

No matter how complicated Network 1 is, we read onevoltage. It is either positive at A, (with respect to B)or negative at A.

We call this voltage Vos and we also call it VTHEVENIN = VTH

THEVENIN’S THEOREM:

• We now deactivate all sources of Network 1.

• To deactivate a voltage source, we remove the source and replace it with a short circuit.

• To deactivate a current source, we remove the source.

THEVENIN’S THEOREM:Consider the following circuit.

+_

+

+_ _

A

B

V1

I2

V2

I1

V3

R 1R 2

R 3

R 4

Figure 3: A typical circuit with independent sources

How do we deactivate the sources of this circuit?

THEVENIN’S THEOREM:When the sources are deactivated the circuit appearsas in Figure 4.

R 1

R 2

R 3

R 4

A

B

Figure 4: Circuit of Figure 10.3 with sources deactivated

Now place an ohmmeter across A-B and read the resistance.If R1= R2 = R4= 20 and R3=10 then the meter reads 10 .

THEVENIN’S THEOREM:

We call the ohmmeter reading, under these conditions,RTHEVENIN and shorten this to RTH. Therefore, the important results are that we can replace Network 1with the following network.

V TH

R TH

A

B

+_

Figure 5: The Thevenin equivalent structure.

7

THEVENIN’S THEOREM:We can now tie (reconnect) Network 2 back to terminals A-B. A

B

N e tw o r k2

V TH

R TH

+_

Figure 6 : System of Figure 10.1 with Network 1 replaced by the Thevenin equivalent circuit.

We can now make any calculations we desire within Network 2 and they will give the same results as if westill had Network 1 connected.

THEVENIN’S THEOREM:It follows that we could also replace Network 2 with a Thevenin voltage and Thevenin resistance. The results would be as shown in Figure 6

A

B

+ +_ _

R TH 1 R TH 2

V TH 1 V TH 2

Figure 7:The network system of Figure 1 replaced by Thevenin voltages and resistances.

THEVENIN’S THEOREM: Example :1 Find VX by first finding VTH and RTH to the left of A-B.

1 2 4

6 2 V X3 0 V +_

+

_

A

B

Figure 8: Circuit for Example 1.

First remove everything to the right of A-B.

THEVENIN’S THEOREM: Example 1. continued 1 2 4

6 3 0 V +_

A

B

Figure 9: Circuit for finding VTH for Example 1.

(30)(6)10

6 12ABV V

Notice that there is no current flowing in the 4 resistor(A-B) is open. Thus there can be no voltage across theresistor.

THEVENIN’S THEOREM: Example 1. continued We now deactivate the sources to the left of A-B and findthe resistance seen looking in these terminals.

1 2 4

6

A

B

RTH

Figure 10: Circuit for find RTH for Example 10.10.

We see, RTH = 12||6 + 4 = 8

THEVENIN’S THEOREM: Example 1. continued After having found the Thevenin circuit, we connect thisto the load in order to find VX.

8

1 0 VV TH

R TH

2 V X

+

_

+_

A

B

Figure 11: Circuit of Ex. 1 after connecting Thevenin circuit.

10 22

2 8

( )( )

XV V

THEVENIN’S THEOREM:In some cases it may become tedious to find RTH by reducingthe resistive network with the sources deactivated. Considerthe following:

V TH

R TH

+_

A

B

IS S

Figure 12: A Thevenin circuit with the output shorted.

We see;TH

THSS

VR

I Eq .1

THEVENIN’S THEOREM: Example 2. For the circuit in Figure 13, find RTH by using Eq 1.

1 2 4

6 3 0 V +_

A

B

IS S

C

D

Figure 13 : Given circuit with load shorted

The task now is to find ISS. One way to do this is to replacethe circuit to the left of C-D with a Thevenin voltage andThevenin resistance.

THEVENIN’S THEOREM: Example 2. continued Applying Thevenin’s theorem to the left of terminals C-Dand reconnecting to the load gives,

4 4

1 0 V +_

A

B

IS S

C

D

Figure 14 : Thevenin reduction for Example 2.

108

108

THTH

SS

VR

I

Any linear, active, resistive network containing one or more voltage and/or current sources can be replaced by an equivalent circuit containing a current source called Norton’s equivalent current Isc and an equivalent resistance in parallel.

Norton’s Theorem

NORTON’S THEOREM:Assume that the network enclosed below is composedof independent sources and resistors.

Network

Norton’s Theorem states that this network can bereplaced by a current source shunted by a resistance R.

I R

IS S R N = R TH

NORTON’S THEOREM:In the Norton circuit, the current source is the short circuitcurrent of the network, that is, the current obtained by shorting the output of the network. The resistance is theresistance seen looking into the network with all sourcesdeactivated. This is the same as RTH.

NORTON’S THEOREM:

We recall the following from source transformations.

+_

R

RV I =VR

In view of the above, if we have the Thevenin equivalentcircuit of a network, we can obtain the Norton equivalentby using source transformation.

However, this is not how we normally go about findingthe Norton equivalent circuit.

NORTON’S THEOREM: Example 1.

Find the Norton equivalent circuit to the left of terminals A-Bfor the network shown below. Connect the Norton equivalentcircuit to the load and find the current in the 50 resistor.

+_

2 0

6 0

4 0

5 0

1 0 A

5 0 V

A

B

Figure 15: Circuit for Example 1.

NORTON’S THEOREM: Example 1. continued

+_

2 0

6 0

4 0

1 0 A

5 0 VIS S

Figure 16: Circuit for find INORTON.

It can be shown by standard circuit analysis that

10.7SSI A

NORTON’S THEOREM: Example 1. continued

It can also be shown that by deactivating the sources,We find the resistance looking into terminals A-B is

55NR

RN and RTH will always be the same value for a given circuit.The Norton equivalent circuit tied to the load is shown below.

1 0 .7 A 5 5 5 0

Figure 17: Final circuit for Example 10.6.

NORTON’S THEOREM: Example 2. This exampleillustrates how one might use Norton’s Theorem in electronics.the following circuit comes close to representing the model of a transistor.

For the circuit shown below, find the Norton equivalent circuitto the left of terminals A-B.

+_5 V

1 k

3 VX 2 5 IS

+

_

VX

A

B

IS

4 0

Figure 18: Circuit for Example 1.7.

NORTON’S THEOREM: Example 2. continued

+_5 V

1 k

3 VX 2 5 IS

+

_

VX

A

B

IS

4 0

We first find;

SS

OSN I

VR

We first find VOS:

SSXOS IIVV 1000)40)(25(

NORTON’S THEOREM: Example 2. continued

+_5 V

1 k

3 VX 2 5 IS

+

_

VX

A

B

IS

4 0 IS S

Figure 18: Circuit for find ISS, Example 1.7.

We note that ISS = - 25IS. Thus,

40

25

1000

S

S

SS

OSN I

I

I

VR

NORTON’S THEOREM: Example 2 continued

+_5 V

1 k

3 VX 2 5 IS

+

_

VX

A

B

IS

4 0

Figure 19: Circuit for find VOS, Example 1.7.

From the mesh on the left we have;

0)1000(310005 SS II

From which,

mAI S 5.2

NORTON’S THEOREM: Example 2. continuedWe saw earlier that,

SSS II 25

Therefore;

mAI SS 5.62

The Norton equivalent circuit is shown below.

IN = 6 2 .5 m A R N = 4 0

A

B

Norton Circuit for Example 2

Extension of Example 2.

Using source transformations we know that the Thevenin equivalent circuit is as follows:

+_ 2 .5 V

4 0

Figure 20: Thevenin equivalent for Example 2

Reference

www.google.comwww. Wikipidia.comBy U.A Patel (Mahajan Publication)

By U.A Bakshi (Technical Publication)

By chakratwati

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