TRAN01 - The Ideal Transformer

[@rinrin6454]

This is amazing. Thank you for being so thorough with explaining all the little hooks and crannies of this. Also thank you for the high quality audio!! It makes such a huge difference when trying to concentrate.

[@Kmn0219]

Please make a playlist for feedback control systems, or Electrical Machinery Fundamentals, or Electrical Power Systems! OR ALL OF THEM!! You are the KING of teaching Electrical Engineering!

[@rolinychupetin]

You are too kind ... and I like it.

[@djcronin1989]

Dr. Linares, thank you so much for these videos. We are forever grateful for your efforts! I wanted to point out that when you entered the second equation into the HP-50g, you input V2 - V2 instead of V2 - V1. I wanted to point this out in case anyone solved by themselves and got the wrong answer. I look forward to any future videos. :-)

[@rolinychupetin]

Thank you Mr. DJCronin. Both for the praise and for adding that important errata point for all other viewers.

[@hoola_amigos]

Thank you so much for those concepts and problems.. Cleared many of my doubts.

[@saraemanuel1]

su fluidez y su capacidad de explicación son únicas, ,muy buenas, por eso pido que los hagas en español o si puedes grabarle una traducción.

[@crazywatt3888]

thanks alot you've saved my life

[@rolinychupetin]

Glad to hear that!

[@altuber99_athlete]

As always, you explain topics in such an understandable manner. I know this is just the first video, but I have two questions: 1. What does it mean by "power in" and "power out" in the primary and secondary? I mean, transformers only work for changing currents, which are usually alternating sinusoidal current, and in that case current reverses direction regularly and so does power. My point is, what is referred to as "power in" is just for half a cycle, but for the other half it is "power out." Is this just a convention? 2. When using the transformer for matching impedance, why not simply connect other impedances in series to the load in order to achieve the 800 Ω?

[@rolinychupetin]

Firstly, thank you. Now, on (1), when we say 'power' in the context of AC-SS, very often we mean either average power (also known as active power, and sometimes incorrectly named as "real" power), and sometimes we mean apparent power, in volt-apmpers, in VAs or kVAs or MVAs. And that average has a direction of flow, from the generator into the primary, from the secondary into the load. The actual instantaneous power is fluctuating (as in my video on power in ACSS), but the average power is relentlessly flowing in one direction.

[@rolinychupetin]

Oh, I missed your second question. (2) when we "match impedances", the point is not to extract maximum power out of the generator, or amplifier, or source in general (which would be achieved by the addition of resistors that you propose), but the delivery of ALL of that maximum power to the load, to the customer or to the speaker. If we add those resistors, they will dissipate part of the power delivered by the source and turn it into heat. The customer, the load, would see none of the extra power. Please do the math, set up the circuit and compute the current and with that current compute what is the power actually absorbed by the load. I hope that helps.

[@altuber99_athlete]

@@rolinychupetin Oh, thanks for the explanation!!!! Got it ;)

[@jaysinghchauhan35]

Most useful good concept.

[@10turnerj]

Great video!

[@AR_BSME]

Using ur notes I cannot proof why or how you multiplied and divided to make one circuit..........??.

[@rolinychupetin]

Somewhere in my notes, I show that one can "refer" (move) an impedance from the LVS to the HVS if we multiply its value in ohms by the transformation ratio squared. Going in the other direction (from HVS to LVS) is the opposite.

[@saraemanuel1]

lastima que no los tengas los vídeos en español.