# Calculating load resistor for maximum power 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\$$

• 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/…
– G36
May 8, 2020 at 22:22
• 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. May 8, 2020 at 23:28
• Hi, Mhf do you know how to figure out the Thevenin equivalent circuit at points A, B? May 9, 2020 at 4:25
• Any progress in finding Vth and Rth?
– G36
May 9, 2020 at 8:46
• @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.
– user226164
May 9, 2020 at 8:55

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: 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.

• Did you mean Thevenin voltage? May 8, 2020 at 22:04
• A theravin voltage source has a voltage and a series resistance, the maximum load will be the same as that resistance May 8, 2020 at 22:06
• No, the word is "Thevenin". May 8, 2020 at 23:25

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)

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