Maximum Power Transfer

[@Jhonz0r]

Thanks! This video was incredibly helpful. You were very clear and concise in your method!

[@ironoside]

at 12:20 u should write 4v not -4 right? cuz u already flipped the polarity

[@ayasera_9501]

hey there , i'm very confused about the load thing , why did you remove the 4 ohms resistor with the load ? can i do that whenever i want ? if i only removed the load and left the 4 ohms resistor it will be wrong ?? please i need help in this ? if i have a voltage source in series in the same branch with the load can i remove it too ??

[@AraujoMatt]

It turns out that maximum power transfer occurs when the resistance of the load is equal to the thevenin resistance! Textbooks will explain why this is....my focus for these videos is how to get homework,quiz and test questions solved rather than theory.

[@jmddddddd1]

Great USM

[@YahyaNaell]

thanks for the video,,, but i have a question if you may: why you neglect the independent current source in the first part when you make mesh analysis at 3:37 .. and you use it in the second part when you use in this equation at 4:43 ?! thanks

[@LeoNogas]

that is how you solve using supermeshes :) you remove the current source in between to main meshes

[@YahyaNaell]

Leo Nogas thanks

[@stevenan93]

so every case of maximum resistance is just summed up to Rload being equal to Rthevenin?

[@abro7827]

Can you please explain why you chose to do +Voc in the KVL at 9:30 ? the polarities in all are opposite so shouldnt it be also -Voc?

[@SolidSnake013Duds]

what happens to the 2 ohm resistor in your first kvl equation?

[@LtLAsian]

if MPT occurs when RL=RTH, and here RTH=-2ohms, why isn't RL also -2ohms? Like at the end of the video, why should RL be = to the RTH(-2) + 4ohms from the original diagram?

[@mitchellgraba7949]

why would you waste your time by applying KVL to find isc, all current will always flow through an ideal conductor when there is a short circuit

[@stevenan93]

I mean maximum power transfer

[@johnnolen8338]

The solution to this problem is incorrect. The proper Thevenin circuit is a -4 v source followed by a Thevenin resistor of -2 ohms ... that part is correct, but the Maximum Power Transfer Theorem says that RL = Rth. (You didn't remove the 4 ohm resistor following the current source, you neglected it. Since no current flows through an open circuit the contribution of the 4 ohm resistor to Voc is zero.) However, when calculating Isc you left the 4 ohm resistor in the circuit (or you should have); hence, the actual, no kidding, value of Rth = -2 ohms. Thus there is no practical way to get the necessary condition that RL = Rth for this specific circuit!

[@sagorlimon]

This is nor correct.