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 .
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
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.
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 .
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. 😢❤❤
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
Thank you so much. I have been struggling with this concept for a week. I can't thank you enough.
The proof of this equation makes fully understand ,thanks for you
very helpful, simple and well explained. thank you!
Great video!
thank u soooo much sir ! u always used the simplest ways to solve the problems
Complete explanation. Nice
Please give an example of a ladder network with impedance on each branch. thank you.
thank you so much, your explanation was so clear and concise
please do an example containing more than 2 resistors .
Excellent!
i thank you ans i appréciate your efforts thanks Thanks si much
THANKS!!!
with two of them being in series ,find the current through one of the ones in series
the current doesn't change when in series
thank you ..
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.
wish you were my teacher
thanks