The source transformation was a neat idea. I never would have thought of doing it that way. Though it took less than a minute to solve in my head anyway. Voltage division gives 8 volts at terminal (a) and there is a 4 volt drop across current source on the RHS. Since the positive end of the current source is grounded that implies the node voltage at terminal (b) is - 4 volts. Voc = Va - Vb = 8 -( -4 ) = 12 volts. Rth was easy. 3k // 6k + 2k = 4k. Since the load equalled Rth it was obvious that Vo = Vth/2 = 6 volts.
Im so confused about when you found the Rth... It looks like the 6ohm and 3 ohm resistor are in Series and the 2 ohm is parallel. When you put the voltage source back in, they were for some reason in series (as you used the Voltage division formula which requires a series). Can someone clarify?
Imagine if the 3 ohm resistor was the right side of the square instead of the top. this is the same thing, maybe that will maake it easier to visualize.
You know two resistors are in parallel because both of their terminals share exactly the same electrical point, that means there are no other elements in between them to make them connected in series instead.
I'm assuming you're referring to the last part, where he uses source transformation and has two voltage source, 8v and 4v. Follow the + of the 4v source and you'll see that they actually sum together.
how come u add the two voltages V` and V" together if the independent sources have different polarities. Shouldn't V" be -4 volts,so then V`+(-V")=8v+(-4v)=4v??
Денис Степанов This is one of the most confusing parts sometimes...it has to do with interpreting the passive sign convention. V' and V'' have the same polarity for both situations. For V'', the current source is driving a current into the positive terminal V'' - for this reason the voltage is positive. When it gets confusing, you can look at it from another direction to make sure: if my V'' is correct (+4V), which way would the current go? It would go the direction that the current source is driving it - hence the solution is valid.
When the current comes in to the left box from the right one, it can travel throuh 6k or 3k to come up to the node (+, first picture). The current don't go in a arount in that left box. Hard to explain, in text.
LET ME EXPLAIN RTH, Where we removed 4k Om resistor he left poententials A and B,So A is at junction where 3 6 and 4 Om resistors are,B is at junction where 4 and 2 Om resistors are,So lets say Potencital C is also at junction where 3,6 and 2 Om resistors meat,that means 6 Om resistor is betwean points A and C ,also 3 Om is betwene A and C,that means they are paralel,so RTH=3||6+2
Using mesh analysis, though, Vo = 5v. I think the math I've done is right, but my equations are 9(i1) - 6(i2) = 12, -6(i1) + 12(i2) = 2, and i3 = 2. Solving for i2, you get 1.25mA. So, 1.25mA * 4k ohms = 5v. Edit -- my second equation was wrong. Should have read ... = 4.
Dude that was amazing! Thank-you so much! You made me understand superposition so much more!!!
Thank you for making circuits beautiful.
The source transformation was a neat idea. I never would have thought of doing it that way. Though it took less than a minute to solve in my head anyway. Voltage division gives 8 volts at terminal (a) and there is a 4 volt drop across current source on the RHS. Since the positive end of the current source is grounded that implies the node voltage at terminal (b) is - 4 volts. Voc = Va - Vb = 8 -( -4 ) = 12 volts.
Rth was easy. 3k // 6k + 2k = 4k. Since the load equalled Rth it was obvious that Vo = Vth/2 = 6 volts.
Im so confused about when you found the Rth... It looks like the 6ohm and 3 ohm resistor are in Series and the 2 ohm is parallel. When you put the voltage source back in, they were for some reason in series (as you used the Voltage division formula which requires a series). Can someone clarify?
I also found this difficult Please Reply :)
Imagine if the 3 ohm resistor was the right side of the square instead of the top. this is the same thing, maybe that will maake it easier to visualize.
Thank you Matthew! You're awesome :)
This made much more sense than when my professor tried to explain it
can anyone explain why the current isn't flowing through the branch in 7:17 ?
so confusing
Have you discovered it yet? xD i kinda would like to know xD
@@franciscomendes4163 i switched to computer science, and a kinda forgot about this stuff sorry )=
@@zawette Its okay ahaha :) how you doing tho?
@@franciscomendes4163 i am doing ok , i graduated this summer ! thanks for asking
@@zawette Daamn congratz! Wish me luck!
You know two resistors are in parallel because both of their terminals share exactly the same electrical point, that means there are no other elements in between them to make them connected in series instead.
thanks for explaining so simply
why is the current coming from the right box? dosent current flow from positive to negative?
min 3:00, How can the 3k and 6k be in parallel? Looks like they are in series...
This helps me so much! Thank you
how can we get Rth? I didn't understand that....
current from 2m source is opposite to the voc=8v
why not 8v minus 4v?
I'm assuming you're referring to the last part, where he uses source transformation and has two voltage source, 8v and 4v. Follow the + of the 4v source and you'll see that they actually sum together.
Thank you so much for help
thank you so much man :) awesome video.
how com Voc double prime is equal to 4? plz answer this.
Great explanation thanks
how come u add the two voltages V` and V" together if the independent sources have different polarities. Shouldn't V" be -4 volts,so then V`+(-V")=8v+(-4v)=4v??
Денис Степанов This is one of the most confusing parts sometimes...it has to do with interpreting the passive sign convention. V' and V'' have the same polarity for both situations. For V'', the current source is driving a current into the positive terminal V'' - for this reason the voltage is positive. When it gets confusing, you can look at it from another direction to make sure: if my V'' is correct (+4V), which way would the current go? It would go the direction that the current source is driving it - hence the solution is valid.
what will happen when the independent source (2mA) have its positive direction going upward... or opposite direction ?? o_O
When the current comes in to the left box from the right one, it can travel throuh 6k or 3k to come up to the node (+, first picture). The current don't go in a arount in that left box.
Hard to explain, in text.
Why did the voltage become 4mA instead of 6mA did I miss something.
+A Fiacco 12v/3k resistor gives you the 4mA
LET ME EXPLAIN RTH, Where we removed 4k Om resistor he left poententials A and B,So A is at junction where 3 6 and 4 Om resistors are,B is at junction where 4 and 2 Om resistors are,So lets say Potencital C is also at junction where 3,6 and 2 Om resistors meat,that means 6 Om resistor is betwean points A and C ,also 3 Om is betwene A and C,that means they are paralel,so RTH=3||6+2
This exact problem was on my first circuits exam...
@ricorico33 because they share a node on both sides
thank you
upload more videos, please
Using mesh analysis, though, Vo = 5v. I think the math I've done is right, but my equations are 9(i1) - 6(i2) = 12, -6(i1) + 12(i2) = 2, and i3 = 2. Solving for i2, you get 1.25mA. So, 1.25mA * 4k ohms = 5v.
Edit -- my second equation was wrong. Should have read ... = 4.
did not explain things very well...