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Thevenin’s Statement: A linear network
consisting of a number of voltage sources and resistances can be replaced by an
equivalent network having a single voltage source called Thevenin’s voltage ( V_{th})
and a single resistance called Thevenin’s resistance (R_{th}) .

Norton's statement

A linear
network consisting of a number of voltage sources and resistances can be
replaced by an equivalent network having a single current source called
Norton current ( IN) and a single resistance called Norton’s
resistance (R

There are three cases in Thevenin (Norton )equivalent depending on the sources present in the circuit

1) Circuit consisting of all independent sources

2) Circuit consisting of atleast on dependent source and atleast one independent source

3) Circuit consisting of only dependent sources and no independent sources

As current through 2

Now redrawing the circuit

Applying nodal analysis

_{n}) .There are three cases in Thevenin (Norton )equivalent depending on the sources present in the circuit

1) Circuit consisting of all independent sources

2) Circuit consisting of atleast on dependent source and atleast one independent source

3) Circuit consisting of only dependent sources and no independent sources

**1) Circuit consisting of all independent sources***Solution*

*We want to create a Thevenin equivalent circuit of the circuit to the left of the terminals a-b.now separate R*

_{L}from the circuit and find R_{th}from terminals a-b*R*

_{th}*For finding R*

_{th}. Short circuit the current sources, open circuit the voltage sources*Now the circuit appears as follows**From the above figure, it can be seen that the Thevenin resistance RTH is a parallel combination of a 3Ω resistor and a 6Ωresistor, in series with a 2Ω resistor.*

*R*

_{th}= ( ( 6 * 3 ) / ( 6 + 3 )) + 2 = 4Ω*Finding V*

_{th}*Ω resistance is zero**V*at terminals at a-b appear across 3_{th }*Ω resistor**Applying source transformation technique*Now redrawing the circuit

Applying nodal analysis

*(V*

_{th}*– 18)/6 +*

*V*

_{th}*/ 3= 0*

*Solving above equation,we get*

*V*

_{th}*= 6 v*

*Now the equivalent circuit is*

Find thevenin equivalent
of above circuit across terminal ab

Solution

For V

_{TH}
Applying nodal analysis at
node V

_{1}__V__+

_{1}-20__V__+

_{1}__V__

_{1}- V_{TH }_{}= 0 ----------------------------- 1

40
200 100

At node V

_{1}
i

_{1}= V_{1}/200 ----------------------------------------------------2
Applying nodal analysis at
node V

_{TH}__V__- 1.5 i

_{TH}–V_{1}_{1}=0 -------------------------------------------3

100

Solving above equation we
get

V

_{TH}= 700 / 18 = 38.89 V
For R

_{TH}
From the above the circuit

I

_{n}= I_{X}+ 1.5 I_{1}------------------------------------------ 1
Applying nodal analysis at
V

_{1}__V__+

_{1}-20__V__+

_{1}__V__

_{1}- 0_{}= 0 ------------------------ 2

40
200
100

Solving above equation we
get

V

_{1}= (100 / 8) V
At node V

_{1}
i

_{1}= V_{1}/200 -------------------------------------------- 3
i

_{1}= 1 / 16
I

_{X}= ( V_{1}– 0 ) / 100
I

_{X}= V_{1}/ 100 = (1 / 8)A
Substituting the values of
I

_{X}and i_{1}we get
I

_{n}= 0.21875 A
R

_{TH}= V_{TH}/ In
Substituting the values of V

_{TH}, In we get
R

_{TH}= 177.78Ω**3) Circuit consisting of only dependent sources and no independent sources**
Find thevenin equivalent
to the above circuit across ab

Solution

Connect a known
current source or voltage source to the above circuit because without a
independent source the circuit becomes inactive. So to make it active we are
connecting a known current source or voltage source

From the above circuit V

_{TH}= V_{1}
Applying nodal analysis at
node V

_{1}__V__+

_{1}__V__- 1 = 0

_{1}– V_{X}
100
50

3V

_{1}– 2V_{X}= 100 ---------------------------------1
Applying nodal analysis at
node V

_{X}__V__+ (

_{X}__V__+ 0.1V

_{X}- V_{1})_{1 }= 0

200
50

16V

_{1}+ 5V_{X}= 0
V

_{X}= (16/5)V_{1}----------------------------------- 2
Substituting equation 2 in
equation 1 we get

V

_{1}= 10.6 V
R

_{TH}= V_{TH}/ 1
R

_{TH}= 10.6Ω
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