TDT06: Initially Charged Transmission Lines

[@Massika2]

This is truly magnificent. Thank you for making this valuable source of information available to students under worse circumstances.

[@FranVarVar]

When I saw TDT02 (Transmission Lines Equations) I thought the TL analysis would be a mathematical nightmare. :) I was wrong, the examples given are very focused on concepts, very useful indeed. The sequence problem at the end of this lecture is just great as a review. The teaching abilities of Grey are outstanding.

[@tohopes]

Greg is such a good teacher, they moved Tuesday and Thursday to be a fraction of a second apart in Georgia so his students wouldn't have to wait for his next class.

[@PauloConstantino167]

Thank you for sharing your wonderful course my friend. Truly appreciate your sharing your knowledge with us.

[@nathan18084]

I just wanted to say thank you for sharing So far I've seen six lectures and I wish my lecturer was as understandable as you are Thanks sir

[@wasimhussain1983]

Dr. Durgin, my humble thanks to you.....

[@Kashif_Javaid]

Very good lectures, I just ordered your book

[@harshvardhansingh4158]

dr greg durgin sir at time 26.53 you said that 2.4 will go back and come back with negative reflection coefficent and erase 4.8 my question is whether it is 2.4 or 4.8 volt that will go back and thank sir for the videos this really helped a lot in understanding of the transmission line

[@jeffsam5495]

its 2.4 volts

[@benjuliebenjulie7414]

great lecture..

[@abdelrahmangamalmahdy]

Does anybody know why 2V+ and 2V- ?! Isn't it supposed to be just V+ and V- ?! Why is it doubled ???

[@rohitn1704]

By now I think you might have figured this out since this was asked long back. If not, see below Assume a Vx at the source in series with Zo and Vy in series with Zo at the load The above model should be equivalent to tx line equation at all times at the source and the load all below variables are function of t at boundary At Load Vload = Vy -Iload *Zo V load = V(+)load + V(-)load and I load = [V(+)load/Z0 ]- [V(-)load / Z0 ] plugging in we get solution for boundary condition at load as Vy = 2*V(+)load At source Vsource = Vx + Isource * Z0 Vsource = V(+)source + V(-)source and Isource = [V(+)source/Z0] - [V(-)source/Z0] plugging in we get solution for boundary condition at source as Vx = 2* V(-)source at that time

[@benjuliebenjulie7414]

defibrillator is about 200 joules...joules (capacitor) = .5*C*V^2. A charged 200V cap can kill, he shouldn't be so flippant about this...

[@TheManzico]

But what if the capacitor is of picofarad value? Then there can't be any killing.....No?