The analysis for i2 is confusing when explained, but you achieved the correct answer. For those who found it hard to follow, I will try to help explain below... Since we know the current going into the node 1 is 0.5A, we can redraw the circuit using 0.5A as the current source and combine 50R + 10R (series) = 60R. 0.5A (Is = Current Source) --> 20R || 30R || 60R i1 i2 i3 We want to find the the current across the 30R = (i2). **** But first, let's find current across 20R or i1. 30R || 60R = 20R NEW Circuit: 0.5A (Is) --> 20R || 20R i1 (i2 || i3) = ix Current Division: Is * ( ix / ( i1 + ix ) ) i1 = 0.5A ( 20R / (20R + 20R) ) = 0.25A 30R || 60R (w/ i2 across 30R and i3 across 60R) i2 i3 Current Division: Is * ( i3 / ( i2 + i3 ) ) i2 = 0.25A (60R / (30R + 60R) = 0.1667A i3 = 0.5A - 0.25A - 0.1667A = 0.0833 Is i1 i2 i3 Check: KCL Rule 0.5A = 0.25A + 0.1667A + 0.0833A Is = i1 + i2 + i3 0.5A = 0.5A (GooD) Hope this helps :-)
This video was extremely helpful. You should make some on nodal analysis including super nodes, super position, operational amplifiers, and pretty much on the rest of circuits. Thanks for the help.
I am having trouble with you designating V2 as 5V. 5V is the volts DROPPED by the 30 ohm resistor. It seems like the large essential node across the top would be 40V, obtained by 60Vs minus the 20V drop across the 40ohm resistor.
Hi. I combined the R2=30 with the 50+10 (far right) using the parallel resistor reduction technique. This gave me a new R2 which is 30||(50+10)=30||60=20. Now that I have 20 for my R2, I can use the current divider formula to find the current through the other resistor in the current divider.
I apologize I did not always use the same notations in the smaller formulas on the bottom as the main formula on top. Hopefully people can see the strategy I was trying to use.
Hi @RedneckDrillMan, great question. For the current divider equation, you use the opposite* parallel resistor in the numerator when finding the current for the branch you are interested in. I accidentally made the i2 in the original circuit the i1 in my current divider, but nonetheless, R2 becomes the opposite branch (15ohm) from the branch I am solving the current for (in our case the 30ohm branch). I hope this helps.
@@aHydrasa Super late but you can actually use that formula in a purely parallel circuit, but the trick is, the resistors are in conductance or the reciprocal of the resistor wherein instead of R1 you use 1 / R1. So in formula it would be: ix = isource ( 1 / Rx ) / ( 1 / R1 + 1 / R2 + 1 / R3 + 1 / Rx + ... ) TL;DR. Use the same resistor as numerator to find the current across it but all of the resistors are in reciprocal.
When calculating V2 which is 0.167A * 30 ohms = 5V wouldn't this be negative since the voltage is flowing in the opposite direction to the current i2??
For the resistor where you needed to find the power. After finding the voltage, and given the resistance; why didnt you use the voltage and resistance to find the current (i = V/R) then do P = IV?
I thought the formula was V1= (R2/(R1+R2))* Vs Also, what would the formula be if you know V1, Vs and R2 and are solving for R1? And for if you know V1, Vs and R1 and are solving for R2?
Hi. In Part2 we are reusing Vo from Part1, which we have found to be 20V. Since we know 20V is the voltage drop across the 40 ohm resistor, we can use V=IR to find the current in this part of the circuit.
Question: How do you get 15 ohms from 20||60? shouldn't it be a third? Edit: I just realized u were adding the resistance in fractions. 1/20 + 1/60 = 1/15 ohms nvm!
All I need is for someone to correct me, i'm sure i'm just being dumb, but I LOVED the first part solving for Vo, but when you started solving for I2 that means the direction at which the current flows plays a factor in the equation. You have every thing set up as if the current were flowing positive-negative and i'm confused because I thought current flows negative to positive? what is going on, someone please help me lol
Voltage divider formula is Vout =( R2 / (R1+R2))xVin ...why in your care is Vout =(R1/(R1+R2)) xVin?...this is confusing...did you made some different notations?
Hi. The voltage divider formula with R2 in the numerator will find the voltage drop across resistor R2. If you want to find the voltage drop across R1 instead (we call this Vo in the video), you simply put R1 in the numerator instead of R1 and you now have V1 (or Vo in the video).
Since the resistors are in series you can switch positions between R1 and R2 and use the voltage divider equation for whichever resistor you need and apply the formula this way.
@@engineeringmadesimple8359 so just to be clear, with voltage division, you put R for whichever voltage drop you are finding, but with current division, you do the opposite?
Would you be able to share your test question? I would double check to make sure you collapsed all of the circuit resistors properly before applying the voltage divider equation.
Also, I'd like to note that if you have any active elements in the circuit (capacitors or inductors) and not simply a resistor circuit, this method will not work
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
The analysis for i2 is confusing when explained, but you achieved the correct answer. For those who found it hard to follow, I will try to help explain below...
Since we know the current going into the node 1 is 0.5A, we can redraw the circuit using 0.5A as the current source and combine 50R + 10R (series) = 60R.
0.5A (Is = Current Source) --> 20R || 30R || 60R
i1 i2 i3
We want to find the the current across the 30R = (i2).
**** But first, let's find current across 20R or i1.
30R || 60R = 20R
NEW Circuit:
0.5A (Is) --> 20R || 20R
i1 (i2 || i3) = ix
Current Division: Is * ( ix / ( i1 + ix ) )
i1 = 0.5A ( 20R / (20R + 20R) ) = 0.25A 30R || 60R (w/ i2 across 30R and i3 across 60R)
i2 i3
Current Division: Is * ( i3 / ( i2 + i3 ) )
i2 = 0.25A (60R / (30R + 60R) = 0.1667A
i3 = 0.5A - 0.25A - 0.1667A = 0.0833
Is i1 i2 i3
Check: KCL Rule
0.5A = 0.25A + 0.1667A + 0.0833A
Is = i1 + i2 + i3
0.5A = 0.5A (GooD)
Hope this helps :-)
Very helpful, the video confused me on that part but this cleared it up
I’m so glad you took time to clarify this. That part of the video was confusing
but why are the resistors of 40 and 70ohms being ignored? im hella confused by that
@@ath6632 because current is the same in series and 40 ohm , 70 ohm is in series.
@@hanaanasr1064 now this was the anwsers i was looking for thk u brotha
Life saver, found this right before my midterm. Keep doing what your doing 😊👍
This video was extremely helpful. You should make some on nodal analysis including super nodes, super position, operational amplifiers, and pretty much on the rest of circuits. Thanks for the help.
I definitely have those topics on my to-do list, thanks for the feedback!
This is really amazing that i'm starting to love circuit until i solve problems by myself. :(
These videos are extremely helpful to stusents and to teachers also.
Thanks from Germany. It helped a lot!
Thank you so much for this video! I was about to give up on circuit dividers altogether before finding this. You are a legend.
man i can't thanks u enough that was really helpful thanks a lot ^^
Bro, you just make it simple, Thank you so much
What happened to 70 ohm resistor???
Excellent video
I am having trouble with you designating V2 as 5V. 5V is the volts DROPPED by the 30 ohm resistor. It seems like the large essential node across the top would be 40V, obtained by 60Vs minus the 20V drop across the 40ohm resistor.
Really useful. Thank you sir ❤️
What software you are using
I don’t get how you got R2. Where did the 20||20 come from?
The 60 ohm resistor is parallel with the 30 ohm resistor, so we get ((1/60)+(1/30))^-1 as the equation to get their equivalent resistor.
R2 became 20||20 because it was 60||30||20. 60||30 is (60*30)/(60+30)=20 so we now have 20||20.
Thanks... Good tutorial
Thats sick dude nice
I don't understand. R2 = 30 but why did you make it 15?
Hi. I combined the R2=30 with the 50+10 (far right) using the parallel resistor reduction technique. This gave me a new R2 which is 30||(50+10)=30||60=20. Now that I have 20 for my R2, I can use the current divider formula to find the current through the other resistor in the current divider.
sir your lecture is very good & my question is in Video you want to find v (not) but while solving you wright v1
I apologize I did not always use the same notations in the smaller formulas on the bottom as the main formula on top. Hopefully people can see the strategy I was trying to use.
If I have an 2amp 12volt supply how will this affect the amperage output if I take the voltage down to 6volt . All DC of course.
When finding i2, why do you use the resistor R2 in the numerator instead of the 30 ohm resistor that we're finding the current in?
Hi @RedneckDrillMan, great question. For the current divider equation, you use the opposite* parallel resistor in the numerator when finding the current for the branch you are interested in. I accidentally made the i2 in the original circuit the i1 in my current divider, but nonetheless, R2 becomes the opposite branch (15ohm) from the branch I am solving the current for (in our case the 30ohm branch). I hope this helps.
EngineeringMadeSimple Awesome thank you, that helps a lot. The equation in my textbook did not make that clear and I kept getting the wrong answer.
@@aHydrasa Super late but you can actually use that formula in a purely parallel circuit, but the trick is, the resistors are in conductance or the reciprocal of the resistor wherein instead of R1 you use 1 / R1. So in formula it would be: ix = isource ( 1 / Rx ) / ( 1 / R1 + 1 / R2 + 1 / R3 + 1 / Rx + ... )
TL;DR. Use the same resistor as numerator to find the current across it but all of the resistors are in reciprocal.
Good job
When calculating V2 which is 0.167A * 30 ohms = 5V wouldn't this be negative since the voltage is flowing in the opposite direction to the current i2??
Hi sir, can i know what happen to the 70ohm and 40ohm?
Is it possible to use the parallel addition on 60 ohms and 30 ohms instead of 60 and 20 for the current divider part? Thanks
For the resistor where you needed to find the power. After finding the voltage, and given the resistance; why didnt you use the voltage and resistance to find the current (i = V/R) then do P = IV?
Simply amazing!
great video, thank you!
I thought the formula was V1= (R2/(R1+R2))* Vs
Also, what would the formula be if you know V1, Vs and R2 and are solving for R1? And for if you know V1, Vs and R1 and are solving for R2?
Oh, think I figured out those formulas.
R1=((v1/vs)*R2)-R2
R2=((v1/(v1-vs))*R1)-R1
That look about right? Math isn't my specialty...lol.
Sir, my question is, there are many resistors there. How are you going to know that, this one is R1,R2??
why you calculate i1 instead of i2.
In second problem how that v=20 instead of 60 as shown in diagram!
Hi. In Part2 we are reusing Vo from Part1, which we have found to be 20V. Since we know 20V is the voltage drop across the 40 ohm resistor, we can use V=IR to find the current in this part of the circuit.
for finding
for finding v1 we take resistance 80 and for v3 we take 10 ohm why
Question: How do you get 15 ohms from 20||60? shouldn't it be a third?
Edit: I just realized u were adding the resistance in fractions. 1/20 + 1/60 = 1/15 ohms nvm!
All I need is for someone to correct me, i'm sure i'm just being dumb, but I LOVED the first part solving for Vo, but when you started solving for I2 that means the direction at which the current flows plays a factor in the equation. You have every thing set up as if the current were flowing positive-negative and i'm confused because I thought current flows negative to positive? what is going on, someone please help me lol
as long as you're consistent it doesn't matter.
Voltage divider formula is Vout =( R2 / (R1+R2))xVin ...why in your care is Vout =(R1/(R1+R2)) xVin?...this is confusing...did you made some different notations?
Hi. The voltage divider formula with R2 in the numerator will find the voltage drop across resistor R2. If you want to find the voltage drop across R1 instead (we call this Vo in the video), you simply put R1 in the numerator instead of R1 and you now have V1 (or Vo in the video).
Since the resistors are in series you can switch positions between R1 and R2 and use the voltage divider equation for whichever resistor you need and apply the formula this way.
@@engineeringmadesimple8359 so just to be clear, with voltage division, you put R for whichever voltage drop you are finding, but with current division, you do the opposite?
what happens to the 70ohm?
Exactly whydid the 70 ohm resistor was not used in calculating the i1 in the circuit.
@@umangrathore6825 i guess its a superfluous element(?) im confused too lmao
Sorry I don't quite understand why you used the value of current i1 to find the value of the voltage at i2...plz anyone with idea plz reply
Dude no one gets how you got R2 and the 20//20 part
R2 became 20||20 because it was 60||30||20. 60||30 is (60*30)/(60+30)=20 so we now have 20||20.
great vid, thank you
good
Increase Sound volume plz
the way you did the voltage divider didn't work for a test question I had
Would you be able to share your test question? I would double check to make sure you collapsed all of the circuit resistors properly before applying the voltage divider equation.
Also, I'd like to note that if you have any active elements in the circuit (capacitors or inductors) and not simply a resistor circuit, this method will not work
Dude !!
i am lost from first 10 second 😢
Wrong major and class sry to break it to you...
Thank you!
i like this video
thaanks ♥
What a g👏
Hey thx youuuuuuu
horrible, if ur solving for something with given labels (R1, R2) you need to mark that on ur diagram otherwise its confusing af like this vid was