Superposition Example-Dependent Source
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- เผยแพร่เมื่อ 10 ม.ค. 2012
- Using superposition to analyze a circuit that has a dependent source. More instructional engineering videos can be found at www.engineeringvideos.org.
This video is licensed under the Creative Commons BY-SA license creativecommons.org/licenses/b....
Superposition works by finding the voltage or current due to each source individually, then adding them together to get the actual voltage or current. So vx prime is the voltage due to the 1V source with the 4V source shorted. vx prime prime is the voltage due to the 4V source with the 1V source shorted.
thank you very much
you are awesome! thanks
How can you do KCL when there is no input current; all output current on Vx''?
When doing the analysis, it does not matter what direction you draw the current as long as you keep track of the direction correctly. With superposition, you do the analysis twice, once for each of the independent sources; each analysis is separate from the other. So I arbitrarily chose the current into the node for the first analysis and out of the node for the second. Sorry for the confusing inconsistency.
Yes, it looks like I am counting current and then voltage. I was not clear that the gain of the controlled source has units of Siemens (1/Ohms). So the units for the current are Volts*Siemens = Volts/Ohms = Amps. The method is correct, but I was not at all clear about what is actually going on.
Hi darryl which software, equipments do you use to make these videos. Thanks for uploading. The videos are great
Yes
i cant understand what you did there at the upper node why is the upper node = vx prime
is it like a ground discharge ?
V(x)" and I(x)" should be negative right because at 10:38 the equation should equal to -4/6.
And an other way to see that you should make a substracion is that the two batteries are facing each other with the same pole (+)(+) and (-)(-).
???
how do we know that the current source is the dependent source? is this something we are told or one just knows by looking at the circuit?
For 4 volts, I make 2 ohm and 3 ohm resistors in parallel (2*3/2+3) I got vx'' = 0.196 volts instead of .22 volts. why ?
Yeah me too
can u plz tell when we open circuit the wire the current flowing in the resistor is zero or not
Ya I=0.in open ckt
Darryl what in the world is that thing in the 3rd branch? Looks like a diamond with an arrow pointing down, is that supposed to be a current source?
+Jorge B yes... That's what we call a "Dependent Current Source"..
i didn't understand why the voltage of the 2 ohm resistor is (1v - Vx')/2 in fact it' the (1v - Vx') part that i don't understand
why is the current through the 6 ohm resistor Vx''- 4V / 6 Ohm. Why isn't this Vx'' + 4V / 6 Ohm?
Going left to right the voltage is going to be higher on the left. You subtract because it's greater potential, the higher voltage, minus the lower potential, lower voltage. That's why it's. Vx-4v/(6).
Because Vx includes all nod and 4V is just a part of it we need other part to calculate i think...
Because he took the current as leaving we ca also write it with + sign if we take it as current entering the node and other as leaving but in that case we have to write it on the other side of the = .So, then end result remains the same.
please talk faster. the videos could be half their size but thank you for these videos. they help tremendously.
i didn't understand why the current of the 2 ohm resistor is (1v - Vx')/2
The current is going into the node so it should be -(Vx' - 1v)/2 which is the same as (1v-Vx')/2
Why current of 2 Ohm resistor is (1V-Vx)/2 Ohm, but Ix is not (1V-Vx)/3 Ohm?
+Александр Михеев because in 3 ohm resistor there is no voltage source(voltage source direction is + to -). in 2 ohm resistor there is a voltage source
whenever there is a diamond kinda shape instead of the normal shape of the current/voltage source, it's a dependent source
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looking jst the diamond shape one can know that the source is a dependent source
you left handed?
Seems to me when you are analyzing the 4v source, you have the 6 ohm resistor in series with the 2 ohm and 3 ohm resistors in parallel with each other. This gives you an equivalent resistor of (2x3/2+3) or 1.2 ohms. Using Vx = Vs(Rx/Rt),` the Voltage across 2 ohm & 3 ohm resistors (4)(1.2/7.2) = .667 volts. Thus current through 3 ohm resistor is .222 Amps.(For the 4v source only. When analyzing the 1v source you have 2 ohms in series with the 3 ohm and 6 ohm in parallel which gives you an equivelant resistance of 2 ohms. So voltage divides evenly. You have 1/2 volt across the 3 ohm resistor and you get .167 amps.
7:15 how can you add volt and current ? They're different huh?
K got it siemens. Everything is fine.
jhund
It is useful, but how could it be entertaining ...