NETWORK THEOREMS
1. Thevenin’s theorem
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 (VTh )
and a single resistance called Thevenin’s resistance ( RTh ) .
Explanation:
Consider a network or a circuit as shown. Let E be the emf of the cell having its internal
resistance r = 0 . RL → load resistance across AB .
To find VTh :
The load resistance RL is removed. The current I in the circuit is I =
E
.
R1 + R2
The voltage across AB = Thevenin’s voltage VTh .
VTh = I R2 ⇒
VTh =
E R2
R1 + R2
To find RTh :
The load resistance RL is removed. The cell is disconnected and the wires are short as
shown.
The effective resistance across AB = Thevenin’s resistance RTh .
RTh = R3 +
R1 R2
[ R1 is parallel to R2 and this combination in series with R3 ]
R1 + R2
If the cell has internal resistance r , then VTh =
E R2
( R + r ) R2
and RTh = R3 + 1
.
R1 + R2 + r
R1 + r + R2
NETWORK THEOREMS
Proof of Thevenin’s theorem:
Consider the network as shown below
The equivalent circuit is given by
The effective resistance of the network in (1) is R3 and RL in series and this
combination is parallel to R2 which in turn is in series with R1 .
Thus, Reff = R1 +
R2 ( R3 + RL )
-------------- (1)
R2 + R3 + RL
The current I in the circuit is I =
or I =
E
=
Reff
E
R ( R + RL )
R1 + 2 3
R2 + R3 + RL
E ( R2 + R3 + RL )
----------- (2)
R1 R2 + R1 R3 + R1 RL + R2 R3 + R2 RL
The current through the load resistance ( I ′ ) is found using branch current method.
I′ =
I R2
----------- (3)
R2 + R3 + RL
Substituting for I from (2) in (3)
I′ =
E ( R2 + R3 + RL ) R2
( R2 + R3 + RL )( R1 R2 + R1 R3 + R1 RL + R2 R3 + R2 RL )
or I ′ =
ER2
------------ (4)
R1 R2 + R1 R3 + R1 RL + R2 R3 + R2 RL
Thevenin’s voltage VTh =
ER2
----------- (5)
R1 + R2
Thevenin’s resistance RTh = R3 +
R1 R2
---------- (6)
R1 + R2
Consider the equivalent circuit (circuit (2))
The current I ′′ in the equivalent circuit is I ′′ =
VIJAYA COLLEGE
VTh
----------- (7)
RTh + RL
Page 2
NETWORK THEOREMS
Substituting for VTh and RTh from (5) and (6) in (7)
I ′′ =
E R2
E R2
1
×
=
R R +R R +R R +R R +R R
R1 + R2 R + R1 R2 + R
( R1 + R2 ) 3 1 3 2 1 L 2 L 1 2
3
L
R1 + R2
( R1 + R2 )
I ′′ =
or
E R2
------------- (8)
R1 R2 + R1 R3 + R1 RL + R2 R3 + R2 RL
From equations (4) and (8), it is observed that I ′ = I ′′ .
Hence Thevenin’s theorem is verified.
2. Maximum Power Transfer Theorem
Statement: The power transferred by a source to the load resistance in a network is
maximum when the load resistance is equal to the internal
resistance of the source.
Proof of Maximum power transfer theorem:
Consider a network with a source of emf E and internal
resistance r connected to a load resistance RL . The current I in the
circuit is
I=
E
----------- (1)
RL + r
2
 E 
The power delivered to load resistance RL is PL = I 2 RL or PL = 
 RL
 RL + r 
(from equation (1))
PL =
E2
( RL + r )
2
RL ---------- (2)
The variation of PL with RL is as shown.
PL is found to be maximum for a particular value of RL when
dPL
PL is maximum,
=0
dRL
[Q No variation of PL with RL at PL max ]
i.e.,
d  E2 RL

dRL  ( RL + r )2
−3
−2
E2  RL ( −2 ) ( RL + r ) + ( RL + r )  = 0


Differentiating
2 RL
( RL + r )
Thus,
3
=

d  2
−2
E RL ( RL + r )  = 0
 = 0 or



dRL

or
−2 RL
( RL + r )
3
+
1
( RL + r )
2
=0
or
1
( RL + r )
2 RL
=1
RL + r
2
⇒
2 RL = RL + r or RL = r
Thus the power delivered to the load resistance is maximum when the load resistance is
equal to the internal resistance of the source.
VIJAYA COLLEGE
Page 3
NETWORK THEOREMS
To show that the maximum power transfer efficiency of a circuit is 50%:
The power across the load PL = I 2 RL =
E2
( RL + r )
2
RL ----------- (1)
From the maximum power transfer theorem, PL is maximum when RL = r . Putting this
condition in equation (1),
PL max =
E2
( 2 RL )
2
RL ⇒ PL max =
E2
------------- (2)
4 RL
The power that is taken from the voltage source is (or power generated by the source),
E2
E2
E2
P = I 2 ( RL + r ) =
R + r ) or P =
. When RL = r , P =
-------- (3)
2 ( L
RL + r
2 RL
( RL + r )
Dividing equation (2) by (3)
PL max
P
=
E2 2 RL 1
P
× 2 = or PL max =
4 RL E
2
2
Thus the maximum power delivered to the load is only half the power generated by the
source or the maximum power transfer efficiency is 50%. The remaining 50% power is lost
across the internal resistance of the source.
3. Superposition theorem
Statement: In a linear network having number of voltage or current sources and
resistances, the current through any branch of the network is the algebraic sum of the
currents due to each of the sources when acting independently.
Explanation: By mesh current analysis.
1. Consider the network as shown. The currents in different branches of the network are
I 1 , I 2 and I as shown. Also I 1 + I 2 = I .
2. [Let the internal resistance r of the cells be negligible].
The cell E2 is removed and the terminals are short as shown. Now the currents in the
branches are I 1′ , I 2′ and I ′ . Also I ′ = I 1′ + I 2′ .
VIJAYA COLLEGE
Page 4
NETWORK THEOREMS
3. The E1 is removed and the terminals are short as shown. The currents are I 1′′ , I 2′′ and
I ′′ . Also I ′′ = I 1′′ + I 2′′ .
According to superposition theorem I 1 = I 1′ + I 1′′ . I 2 = I 2′ + I 2′′ and I = I ′ + I ′′
I = I1 + I2
Verification of superposition theorem:
1. Consider the network shown. Applying Kirchhoff’s voltage to the loop 1.
[Q I 1 R1 + I ( R3 ) = E1
I 1 R1 + I 1 R3 + I 2 R3 = E1
or I 1 =
I = I1 + I2 ]
E1 − I 2 R3
----------- (1)
R1 + R3
Considering loop 2, I 2 R2 + I R3 = E2 .
I 2 R2 + I 1 R3 + I 2 R3 = E2
I2 =
E2 − I 1 R3
---------- (2)
R2 + R3
Thus, I = I 1 + I 2
I=
E1 − I 2 R3 E2 − I 1 R3
----------- (3)
+
R1 + R3
R2 + R3
2. Consider the circuit shown with E2 removed and
terminals short. Applying Kirchhoff’s law to loop 1.
I 1′ R1 + I ′R3 = E1
As I ′ = I 1′ + I 2′ ,
E − I ′R
I 1 R1 + I 1′ R3 + I 2′ R3 = E1 ⇒ I 1′ = 1 2 3 ---------- (4)
R1 + R3
Similarly for loop 2,
I 2′ R2 + I ′R3 = 0 ⇒
I 2′ = −
I 2′ R2 + I 1′ R3 + I 2′ R3 = 0
I 1′ R3
----------- (5)
R2 + R3
E − I ′R
I ′R
I ′ = I 1′ + I 2′ = 1 2 3 − 1 3 …….(6)
R1 + R3
R2 + R3
3. Consider the circuit with E1 removed and terminals short.
For loop (1) I 1′′ R1 + I ′′R3 = 0
VIJAYA COLLEGE
Page 5
NETWORK THEOREMS
As I ′′ = I 1′′ + I 2′′
− I 2′′ R3
⇒ I 1′′ =
------------ (7)
R1 + R3
I 1′′ R1 + I 1′′ R3 + I 2′′ R3 = 0
For loop (2)
I 2′′ R2 + I ′′R3 = E2
or I 2′′ =
⇒ I 2′′ R2 + I 1′′ R3 + I 2′′ R3 = E2
E2 − I 1′′ R3
------------- (8)
R2 + R3
− I 2′′ R3 E2 − I 1′′ R3
I ′′ = I 1′′ + I 2′′ =
+
-------------- (9)
R1 + R3
R2 + R3
Adding equations (6) and (9)
E1 − I 2′ R3
I ′R
I ′′ R
E − I ′′ R
− 1 3 − 2 3 + 2 1 3
R1 + R3
R2 + R3 R1 + R3
R2 + R3
1 
1
 E − I ′′ R − I ′ R 
=
E1 − I 2′ R3 − I 2′′ R3  +
 R2 + R3  2 1 3 1 3 
R1 + R3 
I ′ + I ′′ =
I ′ + I ′′ =
1
R1 + R3
(
)
1
 E − R I ′ + I ′′  +
3
2
2
 1
 R + R
2
3
(
)
 E − R I ′ + I ′′  …….(10)
3
1
1
 2

Comparing equations (3) and (10) it is observed that
I 1 = I 1′ + I 1′′
I 2 = I 2′ + I 2′′
I = I ′ + I ′′
Hence the proof of the theorem.
VIJAYA COLLEGE
Page 6