Respected Michel Van Biezen, Your videos never fail to fascinate me and I always used to wonder what did you do for your undergrad! Are you an electrical engineer??
On the left side, the current source (2A) pushes the current upwards. On the middle branch, current flow away from the positive side of the battery and towards the negative side of the battery. And since we found I load to be positive, the current flows away from V1 towards the load.
I just can't figure out where he gets the formula IN = (VTH)/(RTH). Since it's a parallel circuit, IN will split up between the two branches, RN and RL. But the intensity going through RN, call it I1, cannot be equal to IN. Therefore, applying Ohm's Law, we'd get that VTH = I1 * RTH, which is to say: I1 = VTH / RTH, BUT I1, NOT IN. Please helppp.
Alex,Take a look at the next video (Norton's theory summarized) and see if that helps you understand it. If not we have another really good example in the physics videos that may help.
I'm not sure if you're still struggling 4 years later, but Thevenin's and Norton's circuits are essentially the same circuit with a source transformation applied. Thevenin's resistor is in series with Thevenin's voltage, and Norton's resistor is parallel to Norton's current. This relationship between the source and resistor in these circuits is why IN=(VTH)/(RTH).
PLEASE HELP - why did assumed that i3 is also entering node 1? I get that for current i1 is entering CS is directed that way anf i2 because VS + sign in that way but why for i1?
I did that intentionally to illustrate that it doesn't matter which direction you assume (even if they are not correct). If the assumed direction is different from the actual direction you will simply get a negative answer. In this problem it is obvious that I3 is actually in the opposite direction, but again it doesn't matter which direction you assume to get the correct answer.
Since the current flow "upward" through the 8 ohm resistor, we can assume that V1 is at a lower voltage than 8 volts, and therefore 8 - V1 is a positive quantity.
Those current directions are arbitrarily chosen. (It doesn't matter what direction you chose). If the actual current is in the opposite direction, you'll get a negative value for the current.
It helped me alot in my final exam! Thank you so much!
i love ur quote in the end, thats how its done. majestic :D!
And thats how its DONEE XD, your videos really help us ace freshman year courses at college, keep it up professor!
Respected Michel Van Biezen,
Your videos never fail to fascinate me and I always used to wonder what did you do for your undergrad! Are you an electrical engineer??
My degrees are in physics, but I worked in many different fields of engineering.
Awesome video sir! Thank you!
Keep it going......
thanks a lot again!!
You are welcome!
Very helpful!
thank you, you really helped me
We can't see the board when you were writing,sir
fantastic explanation tq
+Neerigatti Somasekhar
Thanks for your feedback.
sir can u give your mail id
Sorry sir I still little bit confuse why current from i3 moving to V1, not leaving v1?
On the left side, the current source (2A) pushes the current upwards. On the middle branch, current flow away from the positive side of the battery and towards the negative side of the battery. And since we found I load to be positive, the current flows away from V1 towards the load.
I just can't figure out where he gets the formula IN = (VTH)/(RTH). Since it's a parallel circuit, IN will split up between the two branches, RN and RL. But the intensity going through RN, call it I1, cannot be equal to IN. Therefore, applying Ohm's Law, we'd get that VTH = I1 * RTH, which is to say: I1 = VTH / RTH, BUT I1, NOT IN. Please helppp.
Alex,Take a look at the next video (Norton's theory summarized) and see if that helps you understand it. If not we have another really good example in the physics videos that may help.
I'm not sure if you're still struggling 4 years later, but Thevenin's and Norton's circuits are essentially the same circuit with a source transformation applied. Thevenin's resistor is in series with Thevenin's voltage, and Norton's resistor is parallel to Norton's current. This relationship between the source and resistor in these circuits is why IN=(VTH)/(RTH).
Sir can you please explain how did you get the formula for computing the current through the load resistor.
It is current divider for parallel circuit
PLEASE HELP - why did assumed that i3 is also entering node 1?
I get that for current i1 is entering CS is directed that way anf i2 because VS + sign in that way but why for i1?
I did that intentionally to illustrate that it doesn't matter which direction you assume (even if they are not correct). If the assumed direction is different from the actual direction you will simply get a negative answer. In this problem it is obvious that I3 is actually in the opposite direction, but again it doesn't matter which direction you assume to get the correct answer.
Sir, why can't I use V1-8 instead of 8-V1 ?
Since the current flow "upward" through the 8 ohm resistor, we can assume that V1 is at a lower voltage than 8 volts, and therefore 8 - V1 is a positive quantity.
sir , why is I3 is entering the node V1??
Those current directions are arbitrarily chosen. (It doesn't matter what direction you chose). If the actual current is in the opposite direction, you'll get a negative value for the current.