prove that maxmum power transfer theorm
Answers
the maximum power transfer theorem states that “the maximum amount of power will be dissipated in the load resistance if it is equal in value to the Thevenin or Norton source resistance of the network supplying the power“.
Replace any two terminal linear network or circuit to the left side of variable load resistor having resistance of RL ohms with a Thevenin’s equivalent circuit. We know that Thevenin’s equivalent circuit resembles a practical voltage source.
This concept is illustrated in following figures.
Maximum Power Transfer
The amount of power dissipated across the load resistor is
PL=I2RL
Substitute I=VThRTh+RL in the above equation.
PL=⟮VTh(RTh+RL)⟯2RL
⇒PL=VTh2{RL(RTh+RL)2} Equation 1
Condition for Maximum Power Transfer
For maximum or minimum, first derivative will be zero. So, differentiate Equation 1 with respect to RL and make it equal to zero.
dPLdRL=VTh2{(RTh+RL)2×1−RL×2(RTh+RL)(RTh+RL)4}=0
⇒(RTh+RL)2−2RL(RTh+RL)=0
⇒(RTh+RL)(RTh+RL−2RL)=0
⇒(RTh−RL)=0
⇒RTh=RLorRL=RTh
Therefore, the condition for maximum power dissipation across the load is RL=RTh. That means, if the value of load resistance is equal to the value of source resistance i.e., Thevenin’s resistance, then the power dissipated across the load will be of maximum value.
The value of Maximum Power Transfer
Substitute RL=RTh&PL=PL,Max in Equation 1.
PL,Max=VTh2{RTh(RTh+RTh)2}
PL,Max=VTh2{RTh4RTh2}
⇒PL,Max=VTh24RTh
⇒PL,Max=VTh24RL,sinceRL=RTh
Therefore, the maximum amount of power transferred to the load is
PL,Max=VTh24RL=VTh24RTh