Practice Problem 7.12 Fundamental of Electric Circuits (Sadiku) 5th Ed - First Order RL Circuit
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- Опубликовано: 6 окт 2024
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You always manipulate your work making it difficult to understand 👎
Then find other videos
Easier way: as t=infinity the inductor behaves as short circuit branch. 10Ohm and 5 Ohm are in parallel so replace them with a resistor of 3.333333Ohm. Find the voltage between the top and bottom nodes. V=iR. So V=30. Divide the voltage by the 5Ohm resistance to find the current through the inductor to be 6 and plug it back into the formula. 6+(9-6)e^(-t/Tao)
Yeah if you know the formula. I only use the basic formula for each component.
Thanks sir...I can understand easily when you teach❤️❤️
Great!
thak uuu you saved again
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Gracias, buen video
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its (u+v)' so we should get d/dt( e^10t + i) not d/dt( e^10t * i) ?????
No
@@ArdiSatriawan could explain pls
@@aymen2698 already did in the video
On serious note: It should be (uv)'
sir y 2 days not upload any videos? 7.13 and 7.14??
I am busy at the moment
Why can't say (IB-IA)*10=0 , that's mesh right?
Maybe
@@ArdiSatriawan What do you mean "Maybe"? please dont give us vague answer when we seriously nicely askiing
Professor, when the switch is opened at time t=0, the inductor is also under dc condition (current from the current source can still reach the inductor) so why is it not treated as short circuit?
It can't change instantaneously
Sir, IB is 9-IA isnt it?
Don't know
I really confused can you explain it
No