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Problem: Calculate the value of load resistance (Rl) for which maximum transfer occurs.

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Attempt: What I know is that for the maximum transfer to occur, the load resistance must be equal to the source resistance, in our case, the equivalent resistance of the resistors in T network.

Maximum transfer results to 50% efficiency, \$n = 0.5 = \frac{Rl}{Rl + Req.} \$. So If I could get the equivalent resistance of the T network, I can plug it in on the formula and get the load resistance. Problem is I don't know what to do next.

Any tip or help will be appreciated or a thorough / step by step explanation would be great for a newbie like me. Thank you.

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    \$\begingroup\$ Do you know the Thevenin theorem? \$\endgroup\$ – G36 Jun 14 '18 at 15:39
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A little help and not an answer: -

How would you calculate the effective source impedance of this: -

enter image description here

What steps would you go through when both resistors are equal (for instance)?

Looking back into the terminals, and imagining the battery produced 0 volts (still a valid value) what impedance would you see?

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  • \$\begingroup\$ 4? Product sum of the 2 resistors? By going with that, I tried to solve the circuit. Product-sum of 20 and 60 then add it to 15 and got 30 ohms which is the answer according the answer key. I don't know if that is just a coincidence or that's really the solution. \$\endgroup\$ – Jayce Jun 15 '18 at 3:49
  • \$\begingroup\$ Product/sum is the right answer when looking back into a source with a series and parallel resistor and, 30 is the correct answer too. Have you grasped this yet? Do you understand why we short out the source in order to examine its impedance? \$\endgroup\$ – Andy aka Jun 15 '18 at 8:32
  • \$\begingroup\$ Actually, I have very little foundation when it comes to circuit analysis but shorting out the source to examine the impedance and/or equivalent resistance is under thevenin's. Lastly, can I ask how can I get the maximum power of the load for the given problem? The answer is 2.7W, but I can't seem to get it. I used the formulas P = IV and P = \$ Vth^2 / 4Rth \$ but results to 4.8 V. \$\endgroup\$ – Jayce Jun 15 '18 at 9:14
  • \$\begingroup\$ In my circuit, if the battery was (say) 24 volts but you didn't know that and you didn't know anything other than that the source impedance was 1 ohm (2||2) AND I gave you a multimeter what would you tell me was the source voltage? \$\endgroup\$ – Andy aka Jun 15 '18 at 9:28
  • \$\begingroup\$ I'm not sure. But there would be a voltage drop across the resistors, so it would be lower than the source voltage but maybe close to 24V. \$\endgroup\$ – Jayce Jun 15 '18 at 10:47

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