Can you solve assessment number 4.22 on the maximum power transfer
@juderavenatta917320 сағат бұрын
Good day, sir You didn't solve the second required in the problem which is the time necessary for the capacitor voltage to decay to one-third of its value at t=0, you only solve the Vo(t).
@Nawaf-cj3jxКүн бұрын
Wonderful, thank you master I was rescued after watching your explanation. you saved a student just remember that!
@blgn_5356Күн бұрын
Thank u for video
@lizbethlizaolaramirez2069Күн бұрын
thanks!!!
@nabhanubaidillahbic14092 күн бұрын
please dont use calculator
@treynoble17602 күн бұрын
Who ya got, LeGoat!!! or Jorbum
@No-name012 күн бұрын
thank you very much it helps a lot
@treynoble17602 күн бұрын
Hello Ardi!
@treynoble17602 күн бұрын
Looking Great as always Ardi!
@IntisarJabir-zj3zl3 күн бұрын
What happened to closing all independent sources except one 😢
@mdsaifuddin62793 күн бұрын
Thank you.
@aguler063 күн бұрын
hi sir why cant we use current divider at current source only? for 8ohm 1A and for 2+6ohm again 1A from node B is it because current can go from node a to b on current source?
@alimran82353 күн бұрын
Thank you
@himalhasan18824 күн бұрын
Please 9.70 math solve video
@marissolfelez80364 күн бұрын
KevinLanda28, the currents that flow to AMP (in two terminals) are zero because the AMP is ideal, right? So, this doesn't mean that there isn't current flowing at the 10k resistor since it is under a DDP = -1-Vo. Therefore, ArdiSatriawan is right. In addition, the result shown by Ardi disagrees with the book solution. However, I agree with Ardi.
@m.ehsaan.4 күн бұрын
sir, what is your reason of choosing the i1 loop instead of 4A loop when voltage source is turned off? I did that, but got a different answer.
@alimran82354 күн бұрын
Thank you
@youssefmohamed36954 күн бұрын
Thank you bro
@zakibedul12685 күн бұрын
Terimakasih mas ardi atas ilmunya,sangat berguna🙏🔥
@onlydebian6 күн бұрын
thank you sir.
@onlydebian7 күн бұрын
thank you sir.
@onlydebian7 күн бұрын
I sincerely thank you sir!
@onlydebian7 күн бұрын
thank you soooo much! You are a great help for college students. Keep making such videos.
@aleeya58727 күн бұрын
sir why at 3:04 we only multiply j2 for impedance,but not multiply with (4+j2)
@anonymousone30247 күн бұрын
Dei mapla pona thadava exam la practice wasnt enough This time,fingers crossed :)
@rashed0757 күн бұрын
Much Appreciated sir ❤
@Hanabiya5117 күн бұрын
You saved me, thx❤❤
@nadimurrahmanpial86708 күн бұрын
😀
@onlydebian8 күн бұрын
thats wrong. node 1 and 8 are the same, hence will be considered a single node. total nodes=7
@queenhang45519 күн бұрын
Excuse me why you didn't consider IE while there is a resistance in its branch?
@arifhasnatemon9 күн бұрын
Thank you, Sir. Your solution video helps me a lot with my study and assignment.
Thank you sir! Your explanation is amazing and outstanding! 👌
@goktug835110 күн бұрын
why we did it mesh analysis cant we do kcl
@goktug835110 күн бұрын
arda adağa selamlar
@salemos41810 күн бұрын
sir, here at 2:46, how do we know that to find Va, it is equal to -j5 / (20-j5) and not 20 / (20-j5)
@Samrawit-2110 күн бұрын
Thank you sir. u're life saver
@stevenrashid581010 күн бұрын
Jazakallah
@NATASHALWATULA-yd9tg10 күн бұрын
I want this calculator It's doing everything
@salemos4188 күн бұрын
i use a casio fx-991ES plus, it has all these commands to help you solve such questions
@inbdprice495811 күн бұрын
V_c should be 300/7, 3v_c = 60 + 2v_d (Slight miscalculation with plus and minus) v_x= -8.571V.
@RobHoopRunning48 минут бұрын
you're right. was partially following this video and got -8.52(rounded some numbers along the way). He made a mistake at 13:45 where 3Vc should equal "60 + 2Vd" rather than "60-2Vd"
@Fjeka-s7m12 күн бұрын
I am Korean. Thank you very much. It helped me a lot
@HermelaMebrate-p3w12 күн бұрын
when u calculate kcl of v3 where did u do it the 3io one
@AbcDef-yj9gj12 күн бұрын
I'm not sure i understand why the i passing in the power is equal to (i1-i2)2
@ozcanyetut3897 күн бұрын
He wrote that because they are the currents passing through 2 and they are in opposite directions.