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Please, can you explain this line of working from the maximum power transfer theorem:

For maximum power dissipation in the load, the condition given below must be satisfied $$\left. \frac{d^2P(R_L)}{dR^2_L} \right|_{R_L=R_{Th}} = -\frac{V^2_{Th}}{8R_{Th}} < 0$$

It's from this page: Maximum Power Transfer Theorem Assignment Help.

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  • \$\begingroup\$ Your first link seems to go to an image sharing site with an unhelpful image. Your second link goes to a page that explains the maximum power transfer theorem. What part of their explanation are you having trouble with? \$\endgroup\$ – The Photon Aug 20 '14 at 22:18
  • \$\begingroup\$ @ThePhoton you have to click on the unhelpful image and it expands and highlights the second derivative test, which I think is what the OP is confused about. \$\endgroup\$ – John D Aug 20 '14 at 23:01
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They are writing the power as a function of Rl, taking the first derivative to find the maximum, and then using the second derivative test to ensure that it is a local maximum and not a minimum. Since Vth is squared, and normally resistance is positive (though incremental resistance can often be negative) it's not hard to meet the second derivative test.

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  • \$\begingroup\$ "Since Vth is squared, and normally resistance is positive (though incremental resistance can often be negative) it's not hard to meet the second derivative test." can you please explain this further \$\endgroup\$ – Ashmar Barbour Aug 21 '14 at 0:25
  • \$\begingroup\$ @AshmarBarbour Well, finding where the first derivative is zero tells you that you are either at the minimum or maximum of a function (slope = 0). In order to know if it is a minimum or maximum (because here we are interested in the maximum) we see if the second derivative is <0, and if it is it is a maximum. Since voltage is a real quantity when we square it it's positive. The Thevanin resistance is positive as well, and the expression has a negative sign in front of it so under those conditions it is always negative. \$\endgroup\$ – John D Aug 21 '14 at 14:45
  • \$\begingroup\$ @AshmarBarbour A negative incremental impedance occurs for example in a switching supply where increasing the input voltage causes the input current to go down. (Constant power.) For purposes of this exercise you can ignore this. \$\endgroup\$ – John D Aug 21 '14 at 14:47

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