Technically the fraction IS the more exact value, but the main reason is to show how the math occurred. Recall that a resistor pair in parallel is R1*R2/(R1+R2), so for example if R1 is 10 and R2 is 3, you get: 30/13. 30/13 is an exact number, which you can approximate as 2.3077 but that is not the exact number, nor is it clear where 2.3077 comes from without showing the fraction first. You also could have done it using the 1/Rt = 1/R1 + 1/R2... formula, but that also gives you the same fraction, it just happens to work for more than 2 resistors at once. Tip - any parallel combination of resistors will always have LESS resistance than even the smallest resistor. In this case, 10 and 3 were combined to give roughly 2.3. If you get a number larger than the smallest resistor, you did it wrong, and your most likely error is you forgot to flip the number at the end, that is 1/x.
I know im randomly asking but does anybody know of a way to log back into an Instagram account..? I stupidly forgot the password. I love any help you can give me.
@Jett Chaim i really appreciate your reply. I got to the site thru google and I'm waiting for the hacking stuff atm. I see it takes quite some time so I will reply here later with my results.
+Titan Xelon Idk why he's doing that but if you just follow the formula you'll get the answer he got. Just remember that according to the formula the numerator is the OPPOSITE resistor from the resistor we are interested in. I1 = [r2/(r1+r2)] * Is. There's no reason to do reciprocals like him. The formula works
You can use KCL and KVL and set up a system of equations, and solve through algebra. The superposition method allows you to solve more directly, so you don't have to do nearly as much algebra. It's a simple matter of drawing each version of the circuit with the other source nullified, combining resistors, and using voltage dividers and current dividers to isolate the either voltage or current that applies to the resistor of interest.
so if i have r1, r2, r3 in parallel and 2 power sources, if i want to find the current at just r1,do i only need to workout current for each source using this theorem, then add together ?
There is no 10A, it is a voltage source 10V. Current sources are opened to remove their influcence (ie. set current to 0A) and Voltage sources act as a short when their influcence is removed (ie. voltage between two pts is 0).
Yes it is current division. That is why he does (1/8.33)/(1/2 + 1/8.33) * 1A. Notice that he does not do R1/(R1+R2). Instead he does (1/R1)/(1/R1 + 1/R2). Look at SLIDE 3 of this PDF note (tuttle.merc.iastate.edu/ee201/topics/analysis_techniques/dividers.pdf)
It is not a Voltmeter or Ammeter -- the problem shows a VOLTAGE SOURCE and a CURRENT SOURCE. Setting Current source to 0 means no current (aka broken wire)... Setting Voltage source to 0 means your are measuring voltage across the same node (aka same wire).
Hi nice job, Im IC layout designer and I would like to talk to you sometime, I love the way you work in Mentor tools. Let me know of you interested talking sometime. thanks
Above all, can you please use your ORIGINAL accent to explain rather than comparing it with foreigners..i couldn't understand any of your twisted and rolled tongue words. It is not necessary to show off your language or vocabulary while you are teaching. Besides this spoils your teaching lessons... Thankyou.
this means you're just dumb to begin with. if foreigners can understand heavily accented english; in this case non-indians understanding indian accented english, then why can't you?
![“อาร์ต พศุตม์” เกือบเอี่ยวทุกคดีดัง เอ๊ะไว้ก่อน ไม่มีอะไรได้มาง่ายๆ | แฉ 16 ต.ค. 67 [2/3] | GMM25](http://i.ytimg.com/vi/0Gx-0-kUBv8/mqdefault.jpg)
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Studying for the FE and this helped me understand that current sources get opened, while voltage sources get shorted. Thanks for the video. Cheers
great men you deserve two cold beer🍺🍺 .please share star delta and node voltage
th-cam.com/video/YtI294cjKMc/w-d-xo.html
the best to ever do it 🙏
Give that man a cookie! :D
th-cam.com/video/YtI294cjKMc/w-d-xo.html
Any reason for using 1/8.33, 1/8.33, & 1/2? Instead of inserting the exact values 8.33, 8.33, & 2 the when calculating I1? #confused.
that is how we calculate the total resistance in parallel. 1/R(p.total) =1/R1 + 1/R2. @ENGRTUTOR , may you please confirm if l am right.
Technically the fraction IS the more exact value, but the main reason is to show how the math occurred. Recall that a resistor pair in parallel is R1*R2/(R1+R2), so for example if R1 is 10 and R2 is 3, you get: 30/13. 30/13 is an exact number, which you can approximate as 2.3077 but that is not the exact number, nor is it clear where 2.3077 comes from without showing the fraction first. You also could have done it using the 1/Rt = 1/R1 + 1/R2... formula, but that also gives you the same fraction, it just happens to work for more than 2 resistors at once.
Tip - any parallel combination of resistors will always have LESS resistance than even the smallest resistor. In this case, 10 and 3 were combined to give roughly 2.3. If you get a number larger than the smallest resistor, you did it wrong, and your most likely error is you forgot to flip the number at the end, that is 1/x.
something looks off with this guys current division. Am I the only one confused?
NOP
Actually... He used a strange formula for it, but it's the correct answer.
Yeah, I was thrown off by that as well.
yeah,....that's method is not eassy
2:50 “Lets use voltage divider.” Then places a huge ass equation without explaining any of it.
Thank you Sir - you are the best
#RespectFromSouthAfrica
Great video, thanks a lot! Superposition, Thevenin and Norton seemed kind of abstract to me, but you explained it very well :)
I know im randomly asking but does anybody know of a way to log back into an Instagram account..?
I stupidly forgot the password. I love any help you can give me.
@Deangelo Cesar instablaster =)
@Jett Chaim i really appreciate your reply. I got to the site thru google and I'm waiting for the hacking stuff atm.
I see it takes quite some time so I will reply here later with my results.
@Jett Chaim it did the trick and I actually got access to my account again. I am so happy!
Thank you so much you saved my ass!
@Deangelo Cesar happy to help :)
Great vid, I get it now
thank you for learning how to speak good english
Why are you doing !/R for you current divider?
Yeah, at 6:30 I think it's supposed to be ((1/2)/((1/2)+(1/8.33)))*1 since it's current division (which is I1 = (R2/(R1+R2))*Isource)
Yes that's correct but why did he use 1/R instead of just R --> 1/2 instead of 2 and 1/8.33 instead of 8.33 ??
+Titan Xelon
Idk why he's doing that but if you just follow the formula you'll get the answer he got. Just remember that according to the formula the numerator is the OPPOSITE resistor from the resistor we are interested in.
I1 = [r2/(r1+r2)] * Is. There's no reason to do reciprocals like him. The formula works
Good question, that's because they're in parallel and he didn't wanna get the voltage then divide by the resistor. Piece of cake
The RECIPROCAL always works. The simple formula you list is a special case where there are only 2 resistors.
Some heroes don't wear capes...
Thanks😀
Why not use kcl voltage node to slove it ?
Find v2,v1 and then to find IR5 current
There are plenty of ways to solve this problem. I am showing ONE.
Some students for assignments/homework are forced to demonstrate the use of this method. Some problems are harder to solve without it.
You can use KCL and KVL and set up a system of equations, and solve through algebra. The superposition method allows you to solve more directly, so you don't have to do nearly as much algebra. It's a simple matter of drawing each version of the circuit with the other source nullified, combining resistors, and using voltage dividers and current dividers to isolate the either voltage or current that applies to the resistor of interest.
so if i have r1, r2, r3 in parallel and 2 power sources, if i want to find the current at just r1,do i only need to workout current for each source using this theorem, then add together ?
better than my lecturer😂
superb excellent outstanding
Can you use nodal analysis in this question?
Hi why is that you open the 1A then you short the 10A?
There is no 10A, it is a voltage source 10V. Current sources are opened to remove their influcence (ie. set current to 0A) and Voltage sources act as a short when their influcence is removed (ie. voltage between two pts is 0).
In first can we take 2ohm restistor series with 10ohm resistor parrell to 5ohm resistor?
Very Clean!
Did he calculate the current i1 correctly? it's on the beginning of 6:30
Help!
yes
Yes it is current division. That is why he does (1/8.33)/(1/2 + 1/8.33) * 1A. Notice that he does not do R1/(R1+R2). Instead he does (1/R1)/(1/R1 + 1/R2). Look at SLIDE 3 of this PDF note (tuttle.merc.iastate.edu/ee201/topics/analysis_techniques/dividers.pdf)
Yes, you can do it that way as well
Why did he put 1/2 in the resistors when he can just use the formula I1 = 2 / (8.33+2) * 1A ?????
definitely need more questions very nice method
Here ,i have one doubt instead of voltage divider is there any other method..
at 6:15 I still dont understand why the Resistances all had a 1/R in the Current divider equation!
I had the same doubt, I just plugged in some numbers in the calc and realized that (1/r1)/(1/r1+1/r2) = (r2/r1+r2)
th-cam.com/video/YtI294cjKMc/w-d-xo.html
i found one method using current division.. Thank you
Very helpful, thanks a lot :)
Thank you for this☺️
why wounldn't you the mesh or nodal in between this analysis .
or is this the only rule to solve this by superposition theorem
He is using nodal
Pointing v1,V2 in junction
Much more understandable.
I want source transformation technique to the same problem
kinda confusing bc it seemed like you were using nodal analysis with superposition
Got the same answer using mesh analysis.
how did he get 15/50 ohm 5:24
@@jianyanchoong5762 He added the two resistor using parallel addition.
Thanks a bunch ❤️
THANKS MYAN
How to find current by spt? broooo
This is way to Heavy...Man...)...Got a piping sound in my Left ear,trying to follow-Had to give it Up...)... X ...)
good... thanks alot
Ridiculous explanation
6:10 can i bypass all and only left 2 ohm
no . because the equivalent resistance is even less than 2 ohms and we all know how the current likes the path of least resistance
I1 at 6:30 should equal 0.1936 Amps not 0.1935, not that it matters during rounding.
i have a problem solving complex numbers, I need some one to help me over come this situation.
looks like the problem is very complex
@@Shafixy nice
Thanks a lot
pretty helpful.
i regret the time i have spent watching this video
WHY DIDNT YOU REMOVE THE WIRE IN THE VOLTMETER CASE BUT REMOVED IT IN THE AMMMETER CASE
THIS MAKES THINGS DIFFERENT
PLS EXPLAIN
It is not a Voltmeter or Ammeter -- the problem shows a VOLTAGE SOURCE and a CURRENT SOURCE.
Setting Current source to 0 means no current (aka broken wire)... Setting Voltage source to 0 means your are measuring voltage across the same node (aka same wire).
Thanks sir
th-cam.com/video/YtI294cjKMc/w-d-xo.html
Noice
I find this method stupid
Hi nice job, Im IC layout designer and I would like to talk to you sometime, I love the way you work in Mentor tools. Let me know of you interested talking sometime. thanks
th-cam.com/video/YtI294cjKMc/w-d-xo.html
What an awful way of explaining superposition
노무현
just awful explanation.
Above all, can you please use your ORIGINAL accent to explain rather than comparing it with foreigners..i couldn't understand any of your twisted and rolled tongue words. It is not necessary to show off your language or vocabulary while you are teaching. Besides this spoils your teaching lessons... Thankyou.
this means you're just dumb to begin with. if foreigners can understand heavily accented english; in this case non-indians understanding indian accented english, then why can't you?