Concept 3 :
Maximum Power Transfer Theorem
%
Sr.
No.
Z s (Source
Impedance)
1.
Rs  j 0
RL  Rs
2.
Rs  jX s
RL  Rs2  ( X s  X L )2
3.
Rs  jX s
XL  Xs  0
4.
Rs  jX s
ZL  Zs *
5.
Rs  jX s
RL  Rs2  X s2
I 2 RL
< 50%
6.
Rs  j 0
RL  Rs2  X L2
I 2 RL
< 50%
Z L (Load
Impedance)
Condition for MPT
PL (max) [maximum
power delivered to load]
I 2 RL 
Vs2
4 Rs
I 2 RL
I 2 RL 
(efficiency)
50%
< 50%
Vs2 RL
( Rs  RL )2
< 50%
Vs2
4 Rs
50%
I 2 RL 
Example 1 :
A source Vs (t )  V cos100t has an internal impedance of (4  j 3)  . If a purely resistive load is
connected to this source to extract the maximum power out of the source, then its value in  should
be
[GATE EC/EE/IN 2013, IIT-Bombay]
(A) 3
(B) 4
(C) 5
(D) 7
Ans. (C)
Sol.
In accordance with case 2 as given in table
RL  Rs2  ( X s  X L )2
Where Rs  4 , X s  3 , X L  0 
RL  42  32  5 
Ans.
Example 2 :
[IES EE 2010]
Consider the network shown in below figure.
The max power that can be transferred in the load Z L is
(A) 12.25 W
(B) 62.5 W
(C) 24.5 W
Page 1 of 2
(D) 500 W
Page 2 of 2
Ans. (B)
Sol. If there is no information of complex variable load. Then by default we assume case 4.
That means
For MPT,
Z L  Z s *  (10  j 50) 
5000
Applying KVL, I 
 2.500 A
10  j 50  10  j 50
I  2.5 A
2
PL (max)  I  RL  (2.5)2  10  (6.25) 10  62.5 W
Ans.