Thevenin maximum power transfer theorem
@chuackl-nMVZ8P
•
Oct 23, 2024
Oct 23, 2024
1.3K
In Thevenin maximum power transfer theorem it state that the load resistance,RL must be equal to internal resistance of the source,Rin. However, the efficiency of the circuit is just 50%. At here, can i say i use power source in a more effective way if be able to have an output of higher efficiency and at the same time considerable high output voltage but this output voltage will be slightly lower abit than when RL=Rin. The efficinecy will be higher alot if using RL>>Rin but the amount of output voltage will be extremely low. So in order to choose a load resistor give optimization to power saving and total output power, can consider it as a product of efficiency and total oputput power?
here i let X=efficiency*output power and consider a voltage source connected in series with an internal resistance and a load resistance,nRin where n is the value to be find.
efficiency=Pout/Pin=V^2[(nRin)/(Rin+nRin)^2]/V^2(nRin+Rin)=n/(1+n)
output power=v^2(nRin)/(n+1)^2Rin^2)=(V^2/Rin)(n/(n+1)^2)
hence X=n^2/(n+1)^3[V^2/Rin]
differentiate it with respect to n using UV method will finally get
dx/dn=[n/(n+1)^3][2-[3n/(n+1)]]
when dx/dn=0 n=0(rejected) and n=2
so at here, can i say if i take into consideration of power saving and the amount of output power dissipated, taking n=2 will give me what i want? a desirable high efficinecy and high output power.
here i let X=efficiency*output power and consider a voltage source connected in series with an internal resistance and a load resistance,nRin where n is the value to be find.
efficiency=Pout/Pin=V^2[(nRin)/(Rin+nRin)^2]/V^2(nRin+Rin)=n/(1+n)
output power=v^2(nRin)/(n+1)^2Rin^2)=(V^2/Rin)(n/(n+1)^2)
hence X=n^2/(n+1)^3[V^2/Rin]
differentiate it with respect to n using UV method will finally get
dx/dn=[n/(n+1)^3][2-[3n/(n+1)]]
when dx/dn=0 n=0(rejected) and n=2
so at here, can i say if i take into consideration of power saving and the amount of output power dissipated, taking n=2 will give me what i want? a desirable high efficinecy and high output power.