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How would I go about solving this? I cannot even begin solving this.

EDIT 1: Thank you all of you I got it:

\$R_{Th}=(\frac{1}{6k}+\frac{1}{12k})^{-1}+2k = 6kΩ\$

Find the potentials at both the nodes connecting the \$12kΩ\$ resistor to the rest of the circuit. \$6V\$ at the top node and \$2V\$ at the bottom. Their difference gives \$V_{Th}\$: \$V_{Th}=6-2=4V\$

The maximum power transfers to the load resistor when \$R_{Th}=R_{L}\$ Using \$P_{L}=\frac{V_{Th}^{2}}{R_{Th}}\$ the power is found to be \$1.5mW\$

Edit 2: Sorry, the power is actually found to be \$0.6667mW\$

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  • \$\begingroup\$ You need to use a Thevenin theorem. So, 1 remove RL and find Vab voltage (Vab = Vth). 2 - Find Rth resistance seen from AB terminal (without Rth) or Short AB terminals and find this short circuit current Isc and calculate Rth = Vth/Isc. 3 - To get max power transfer you need to have Rth = RL electronics.stackexchange.com/questions/377467/… \$\endgroup\$
    – G36
    Commented May 8, 2020 at 22:22
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    \$\begingroup\$ This is not Chegg. We expect you to make a significant effort yourself and to show your work. If you really, truly "cannot even begin" then you might consider changing your major. \$\endgroup\$
    – Elliot Alderson
    Commented May 8, 2020 at 23:28
  • \$\begingroup\$ Hi, Mhf do you know how to figure out the Thevenin equivalent circuit at points A, B? \$\endgroup\$
    – user57037
    Commented May 9, 2020 at 4:25
  • \$\begingroup\$ Any progress in finding Vth and Rth? \$\endgroup\$
    – G36
    Commented May 9, 2020 at 8:46
  • \$\begingroup\$ @G36 Yes, I was on it since the last 6 hours, and I learned and practised Thevenin and Norton circuit equivalent theorems and understood. But don't fully understand why there is a max power transmission when Rth = RL. The link to your answer is extremely useful and I will go through it carefully again. I will post a picture of the calculations I did right now. I can't thank you enough. \$\endgroup\$
    – user226164
    Commented May 9, 2020 at 8:55

3 Answers 3

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Find the Thévenin voltage and resistance as usual (with RL disconnected, looking into A and B).

Then calculate the power dissipation in the load resistor as a function of the load resistance.

Differentiate that equation and set to zero to find the maxima (or minima, but it will be a maxima).

Here is a plot of the function (arbitrary values of source resistance and voltage and range of load resistance)- power on the Y axis, load resistance on the X axis:

enter image description here

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Start with determining the theravin voltage source equivalent to points A and B, with it derived to this point, the maximum power resistance will be the same as the equivalent theravin resistance.

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    \$\begingroup\$ Did you mean Thevenin voltage? \$\endgroup\$
    – JYelton
    Commented May 8, 2020 at 22:04
  • \$\begingroup\$ A theravin voltage source has a voltage and a series resistance, the maximum load will be the same as that resistance \$\endgroup\$
    – Reroute
    Commented May 8, 2020 at 22:06
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    \$\begingroup\$ No, the word is "Thevenin". \$\endgroup\$
    – Elliot Alderson
    Commented May 8, 2020 at 23:25
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My answer based on understanding was too radical for at least one of you. So I'll solve it another way: There is current coming out of that 6V voltage source. 100% of the current flows through the 6k resistor. The current is shared / split between the 12k resistor and the Load-arm.

Making RL = 0 will give maximum current and minimum voltage. (power = 0) Making the RL open circuit will give maximum voltage and no current (power = 0)

So, what's the ideal current /voltage sharing arrangement with this 12k resistor ?. (hint, it's 12k)

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  • \$\begingroup\$ Saying it twice doesn't make it correct. \$\endgroup\$
    – Elliot Alderson
    Commented May 10, 2020 at 12:02
  • \$\begingroup\$ Read the words above Sphero's graph, or ask your mum to do it for you: "arbitrary values of source resistance and voltage" \$\endgroup\$
    – Elliot Alderson
    Commented May 13, 2020 at 13:47

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