24b - 🔴MIDSEM EXAMS: Superposition Theorem Problems

[@brytefunky4950]

You are the best teacher I can say😊

[@SkanCityAcademy_SirJohn]

Thanks so much

[@johnnolen8338]

For part b) I got the answer of I1 = 3 A by successive source transformations starting on the left. Superposition is my least favorite circuit theorem but you did it right. 👍

[@SkanCityAcademy_SirJohn]

Thanks so much....

[@Lenga_sa]

When I did current division rule at 7:19 I got 2A for I2. I said I(3ohm)=(6/3+6)×6=4A then I2=0.5×4=2A. where did I go wrong?

[@SkanCityAcademy_SirJohn]

You need to do the division on a step-by-step bases. So starting from the left, (6ohms combines in parallel with another 6) which is 3ohms then in series with another 3, making 6, which later I crossed not to confuse anyone. I1 is 6A, so at the moment, we share I1 between tow 6ohms, which is 3A for each. Moving to the left we try to open up the resistors, .... So kindly visit the video again. I guess you will get it. If anything please let me know.

[@sandilendlovu2513]

Good explanation.. I have a question on the first part of the problem when calculating I2=1.5A. Why did you share the 6A between 6ohm and 6ohm instead of (6:6:3 ohm) resistors

[@SkanCityAcademy_SirJohn]

This is so because all three resistors are not connected in parallel. 6 and 6 are in parallel taking from the left and in series with 3

[@ntwisisochauke3495]

@@SkanCityAcademy_SirJohn Aren't they since the current divides to the 6 and the rest goes to the rest of the circuit meaning all of them are parallel?

[@SkanCityAcademy_SirJohn]

No please

[@paulkumasie]

For the second part, the current is going through the 7ohm resistor so why do you use the 6 ohm resistor as the numerator when calculat ing for the current I prime prime with the current division theorem instead of the one recieving that current as the numerator( the 7ohm instead) ?

[@SkanCityAcademy_SirJohn]

Well, you know resistance offers opposition to the flow of current, so the higher the resistance the smaller the current in it. That's where the idea comes from. I presume you are a little confused about current division and voltage division. For current division, if you want to find current in resistor A, then: = B/(A+B) * I With voltage division, to find voltage across A, it's = A/(A+B) * I

[@rosemarybwalya-cv4st]

I've learn more

[@SkanCityAcademy_SirJohn]

That's great, thanks so much for watching

[@SkanCityAcademy_SirJohn]

Where do you watch from?

[@marktamakloe6119]

Are Kirchhoff's laws applicable in this problem?

[@SkanCityAcademy_SirJohn]

Yes, KCL and KVL can actually be applied in every circuit, at least you can use that concept once or twice, maybe a few times but in this circuit, it's mainly ohms law and current division

[@marktamakloe6119]

@@SkanCityAcademy_SirJohn Oh okay. Thank you

[@hiransirisena4310]

13:50 .... I don't understand how you get minus 13.25

[@SkanCityAcademy_SirJohn]

The initial direction for i1 is to the left, but for simplicity's sake, when 120v was acting alone we changed the direction to face the right. Hence the negative.

[@hiransirisena4310]

thank you. i learned lots of today from you. by the way I'm from Sri Lanka.

[@SkanCityAcademy_SirJohn]

You are most welcome

[@SkanCityAcademy_SirJohn]

Thanks so much

[@imrx2798]

Why you don’t use nodal and mesh analysis

[@SkanCityAcademy_SirJohn]

The question submitted that we solve using superposition

[@KehindeOlatunji-s7d]

Thanks bro

[@SkanCityAcademy_SirJohn]

You are most welcome

[@gulyetik7093]

for find I2 value why we dont do 1.5+(-13.25). ı dont get that

[@SkanCityAcademy_SirJohn]

According to the problem a) we find i1 and i2 when 120v is deactivated, there for i2 we got 1.5A. b) find the value of i1. So no need to find the combined i2 for b

[@gulyetik7093]

@@SkanCityAcademy_SirJohn oh okey thank you ☺️

[@SkanCityAcademy_SirJohn]

you are most welcome@@gulyetik7093

[@EliasKalilo]

😮😮😮

[@kelvinkwawafful4649]

Why is it that in the first question u said the current will split equally but in the second u used current division rule

[@SkanCityAcademy_SirJohn]

Yes, but then it's still current division rule that we need to use to split current. The trick is that if the resistors have the same value, the current will be shared equally.

[@kelvinkwawafful4649]

@@SkanCityAcademy_SirJohn so if the resistors are not of the same value then u apply the current division rule right

[@SkanCityAcademy_SirJohn]

in both cases you use current division rule, I'm only saying that if the resistors going to share the current are the same, you do not bother yourself doing the formula work because they will share the approaching current equally, however if they are not the same, then since we cannot assume we need to solve using the formula@@kelvinkwawafful4649

[@kelvinkwawafful4649]

Ok thank you but when i use the current division formula for the question in which u split the current equally,i had a different answer

[@SkanCityAcademy_SirJohn]

really, kindly check it well okay, because you should same answer@@kelvinkwawafful4649