Maximum power transfer theorem

           In d.c circuits, maximum power is transferred from a source to load when the load resistance is made equal to the internal resistance of the source as viewed from the load terminals with load removed and all emf sources replaced by their internal resistances, this is the statement of this theorem.

           In the above figure, there is a source emf, nothing but Thevenin's voltage, Vs in series with Thevenin's resistance Rs. RL is the load resistance. Rs is the resistance measured between terminals A and B when RL is removed and the emf source being replaced by its internal resistance. According to this theorem, maximum power transfers when Rs is equal to RL. To calculate maximum power to load the ratio of the  (emf)² to (4* RL) has to be taken.
  Pmax=(Vs)²/(4*RL)
Proof:
I=Vs/(RL+Rs)
P=I²*RL...(Power across load)
P=RL*(Vs²/(RL+Rs)²).........(equation 1)
To obtain maximum or minimum value, first order differentiation has to be made equal to zero.
dP/dRL = 0

or, d/dRL ( RL*(Vs²/(RL+Rs)²)) = 0
or, d/dRL ( RL/(RL+Rs)^2) = 0
or, (RL+Rs)² - RL.2.(RL+Rs) = 0
or, RL² +Rs² + 2RLRs - 2RL² - 2RLRs = 0
or, Rs² - RL² = 0
or, RL = Rs..........(equation 2)

Substitute eq.2 in eq.1
then, we get Pmax=(Vs)²/(4*RL)