Current Division

[@billythesunbeltsamurainapi3670]

you are literally one of the most amazing professors I have ever had the opportunity to listen to. I am literally about to be in tears over how much you've helped me with both my digital systems(digital logic course) and now with electrical networks. why cant every professor be like you. just why .

[@mars8499]

OMG 😭😭😭 thank you so much... You taught me better than my lecturer. I owe you. My lecturer teach fast I can't catch up. Thanks sir. 😢❤❤

[@almarobertson4956]

Thank you so much. I have been struggling with this concept for a week. I can't thank you enough.

[@fordhanson7416]

Oh my goodness I like the way you relate to your students and you have made me to understand more than I thought I could, its one of the best among others wishing you best health and your family because I noticed your ring God bless you and your family, you are a blessing to the world community

[@muhammed.s5468]

The proof of this equation makes fully understand ,thanks for you

[@jameshall8562]

very helpful, simple and well explained. thank you!

[@saarike]

Complete explanation. Nice

[@rasyiqahnabilla6796]

thank u soooo much sir ! u always used the simplest ways to solve the problems

[@-scieng-6592]

Great video!

[@splintcell101]

thank you so much, your explanation was so clear and concise

[@333_studios]

Excellent!

[@raezeik3698]

Please give an example of a ladder network with impedance on each branch. thank you.

[@dirohajer7633]

i thank you ans i appréciate your efforts thanks Thanks si much

[@EvolutionXZful]

THANKS!!!

[@mosesmokgosi555]

please do an example containing more than 2 resistors .

[@johnnolen8338]

To reduce a parallel network to two resistors just so you can find the current through one of them is too complicated. Better to recognize that any given branch current is voltage × conductance of said branch. With that in mind, the total current supplied = the sum of the branch currents. The sum of the branch currents = the voltage × the sum of the conductances. Therefore voltage = total current supplied ÷ the sum of the conductances. And the individual branch current = ( the total current supplied ÷ the sum of the conductances) × the individual conductance of the branch in question. For example i1 in the given problem is 20 A ÷ (1/4Ω + 1/6Ω + 1/10Ω) × 1/4Ω = 300/31 A.

[@hamdakhawaja9436]

thank you ..

[@vinissauro]

wish you were my teacher

[@mosesmokgosi555]

with two of them being in series ,find the current through one of the ones in series

[@rf6965]

the current doesn't change when in series

[@manjo49]

thanks