For those concerned below, around 5:40 is not a mistake, just a skipped step. Ik + 2k = 3k, then that's then put in parallel when he converts back to the current source, but at the same time he combines it with the 2k resistor that was in parallel on the far right of the circuit to end up with a single 1.2k resistor. Note: You have to leave the V0 2k resistor uncombined, as it is the load you're doing the calculation upon.
I'm not sure what you mean, but you might be talking about the step I took at 5:20. Notice that I added the two series resistors (The 1k and 2k = 3k). Once the source has been changed to a current source, the 3k is in parallel with the 2k on the far right. This combines to 1.2k. I don't show these steps because I'm trying to keep the videos short.
Seems like a mistake was made when the 3k resistor was combined. It is not in parallel with the 2k resistor to the far right. The two 2k resistors are in parallel and combine to a 1k resistor, which combines to 4k in total.
This was awesome, but I have a question. When you looked at the contribution due to the first source, why did you do a second source transform? I was able to find Vo' and io without doing the second transform. Just used both voltage and current dividers.
The second transform of the 12V source to the 4mA current source, how does he get 1.2k//2k? according to source transformation , I thought the 2k resistors are in parallel which are then parallel to the 3k resistor? As in we have our current source and then three branches in parallel...3k//2k//2k
Dont understand why Vnot is found using the 1.2k resistor for the numerator if Voltage at Vnot is the 2k resistor. Unless he meant to write the Vnot next to the 1.2k instead of the 2k?
what you said 11 month ago was informative but what i dont get is how you got the 4m at 6:20 . 12/3 is 4 but dont you have to take in account the 2 k resister thats in parallel so it would be 12/1.2 not 12/3 right???
drive.google.com/file/d/0B5AeJ-6GS1vxVk5DOUd6U0RMbXM/edit?usp=sharing I hope this helps :) , it's what I know from what I've reached in studying my exam **_** sorry if it's 3 weeks late :v
The step he skipped is the step that caused a little confusion. When he transformed the 12V into 4 mA, the 3k becomes parallel with that other 2k. This two parallel resistors can then be replaced with a 1.2k (1,2k =2*3/(2+3)).
at 6:10 when he said sum of resistors in parallel,,, but he then just add 1.2 & 2 ... like we add in series... We should use formula for resistors in parallel... no?
For those concerned below, around 5:40 is not a mistake, just a skipped step. Ik + 2k = 3k, then that's then put in parallel when he converts back to the current source, but at the same time he combines it with the 2k resistor that was in parallel on the far right of the circuit to end up with a single 1.2k resistor. Note: You have to leave the V0 2k resistor uncombined, as it is the load you're doing the calculation upon.
Thank you for keeping your videos both short and simple!!
I'm not sure what you mean, but you might be talking about the step I took at 5:20. Notice that I added the two series resistors (The 1k and 2k = 3k). Once the source has been changed to a current source, the 3k is in parallel with the 2k on the far right. This combines to 1.2k. I don't show these steps because I'm trying to keep the videos short.
Life Saver! Thanks!
love u bro........u r awsm.....a wndrful way to solve circuits....:)
Thank you so much, that's great
Seems like a mistake was made when the 3k resistor was combined. It is not in parallel with the 2k resistor to the far right. The two 2k resistors are in parallel and combine to a 1k resistor, which combines to 4k in total.
This was awesome, but I have a question. When you looked at the contribution due to the first source, why did you do a second source transform? I was able to find Vo' and io without doing the second transform. Just used both voltage and current dividers.
The second transform of the 12V source to the 4mA current source, how does he get 1.2k//2k? according to source transformation , I thought the 2k resistors are in parallel which are then parallel to the 3k resistor? As in we have our current source and then three branches in parallel...3k//2k//2k
Dont understand why Vnot is found using the 1.2k resistor for the numerator if Voltage at Vnot is the 2k resistor. Unless he meant to write the Vnot next to the 1.2k instead of the 2k?
Thank you! Very simply explained.
what you said 11 month ago was informative but what i dont get is how you got the 4m at 6:20 . 12/3 is 4 but dont you have to take in account the 2 k resister thats in parallel so it would be 12/1.2 not 12/3 right???
drive.google.com/file/d/0B5AeJ-6GS1vxVk5DOUd6U0RMbXM/edit?usp=sharing
I hope this helps :) , it's what I know from what I've reached in studying my exam **_** sorry if it's 3 weeks late :v
According to the rule of Source Transformation, you've got to take the first R that is in series with the voltage source, which is 3 in this case.
good job dude thanks for video.
Dude in 4:23-5:25 what happened to the other 2k resistor?
The step he skipped is the step that caused a little confusion. When he transformed the 12V into 4 mA, the 3k becomes parallel with that other 2k. This two parallel resistors can then be replaced with a 1.2k (1,2k =2*3/(2+3)).
i have these in my 141
at 6:10 when he said sum of resistors in parallel,,, but he then just add 1.2 & 2 ... like we add in series... We should use formula for resistors in parallel... no?
He's applying the current division formula which is i at R2 = R1*Is/(R1+R2). (Is = I source).
oh!!! alright ... thanks :)
Michael Vuthy Sarunn oh!!! alright ... thanks :)
This is very wrong. You messed up on the first source exchange.
You are missing and 2k resistor in your calculations
you fucked up...
very wrong