Thevenin maximum power transfer theorem

chuackl · 2010-09-01T23:52:57+00:00

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.

Thevenin maximum power transfer theorem

chuackl

Member

Updated: Oct 23, 2024

Views: 1.2K

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.

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Saandeep Sreerambatla

Member • Sep 2, 2010

yes , I understand in this case you are getting maximum efficiency but not the maximum power transfer is it?

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chuackl

Member • Sep 2, 2010

Not really maximum efficiency because if maximum efficiency we can put a load resistance that is very big compare to internal resistance so almost all voltage drop will drop at the output resistance but getting a very small amount of output power. At here i trying to find the optimum point that can help to save power but at the same time give a considerable high output of power(of course this will be lower a bit than thevenin maximum power transfer). As you can see in thevenin theorem the efficiency is 0.5 and output power is v^2/4Rin. How do i get the optimum load resistance if i considering to save power at the same time giving output power that is not that low? The only method is to form a formula which relate efficiency and amount of output power than take the product of it differentiate it with n correct? with n=2, efficiency is 2/3 and power output is 2/9v^2/Rin the product of this two is bigger compare to thevenin maximum power transfer theorem. Well, maybe i trying to mean how to get maximum "benefit" from the power source.

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