here a better video on thevenin theorem with dependent sources rth in this video has been calculated using 3 methods (so that you can compare which method takes less time and which is easier to understand) th-cam.com/video/D1QxFA9_U5w/w-d-xo.html by an indian
here a better video on thevenin theorem with dependent sources rth in this video has been calculated using 3 methods (so that you can compare which method takes less time and which is easier to understand) th-cam.com/video/D1QxFA9_U5w/w-d-xo.html
Yes, I chose 1mA because the resistors were in k(ohms). 1 keeps the math simple, but you can choose whatever you want for a test voltage or current. Note if resistors were in ohms(no k) I'd likely choose 1A.
It shouldn't matter. If you choose wrong but keep your signs correct, it works out the same, like a double negative. Worst case, you find a voltage or current is negative, so you account for that in the final ohm's law by checking directions. Current flows from higher voltage.
Vx = 1/2 V1 through voltage division. If you follow v1's path to ground from the left, you get a total resistance of 2k ohms, but Vx is only across one of those 1k ohm resistors, so the splitting factor is 1/2.
Why is there no current source in the equivalent circuit? There is a dependent current source In the original circuit, so I think it must be have current source in the equivalent circuit.
+yosimba2000 Correct me if I'm wrong, but I believe it's because you only have a dependent source. If you had an independent source you might be able to solve for the value of the dependent, but as is you cannot find Rth without introducing a test source. You can't find Rth by deactivating (treat independent voltage sources as short-circuits and independent current sources as open-circuits) independent sources because there is a dependent source. (And you can't 'deactivate' a dependent source.) Ultimately, if you try to solve this without introducing a test source, you'll not be able to simplify the circuit down to Rth because the dependent current source will get in the way.
I think there is a mistake in KCL@V1.[(V2 - V1) / 2k] - [Vx / 1k] - [(V1-Vx)/1k] = 0 must not be true. Because it would be [(V2 - V1) / 2k] - [V1 / 1k] - [(V1 - Vx) / 1k ] = 0 according to me.
That is how it should be. The current isn't changing in nodal analysis so the voltage over that specific resistor is going to stay as (V2-V1)/2k no matter which node you are analyzing.
This is incorrect. Vth should be 0 V for any circuit without an independent source. So Rth is always 0. All of your math looks correct from what I could tell by quickly glancing it over so not sure where you went wrong.
+Chris Sleckman Your statement is not correct. It is true that Vth = 0. But Rth is not always 0. The point of thevenin equivalent circuits is to describe a lumped circuit abstraction which says that you can always express a purely resistive network as a voltage source in series with a resistor. If you put the circuit inside a black box with two terminals, you would be unable to tell an original circuit from its thevenin equivalent. If you put this circuit into a black box and measured its voltage, it would be zero, but if you measured it's resistance, there would be a value.
A fundamental thing to know is that a circuit containing only resistances and controlled sources will appear as a constant value resistance when measured at its input terminals. However, this resistance cannot be measured directly using an ohmmeter (or multimeter) because the dependent current, being an element of an active device (a mosfet for example), only becomes 'alive' when the device is properly biased. So, one can only infer its value by applying a known voltage across the input terminals and measuring the current flowing into it; the Thevenin resistance is obtained by taking the ratio of input voltage to input current. For this particular circuit, analysis will show that the current flowing into the leftmost subcircuit (comprising of the two leftmost parallel branches) is zero for any value of voltage applied across the controlled current source, showing that it behaves as an infinite resistance (open circuit). Hence, when viewed from the right the input resistance of the circuit is simply determined by the 1 kohm shunt resistor since the 2 kohm resistor is open-circuited by the voltage-controlled curent source subcircuit.
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You sure made it easy, thanks.
thank you, your videos really helped a lot.
I love ur videos,please upload more
what about thevenin voltage?
how would you do a nodal analysis if there was a resistor above the dependent source?
this is very helpful.
thanks alot.
thanks sir, it was of gr8 help. respect from india! :)
here a better video on thevenin theorem with dependent sources rth in this video has been calculated using 3 methods
(so that you can compare which method takes less time and which is easier to understand)
th-cam.com/video/D1QxFA9_U5w/w-d-xo.html
by an indian
What if I use redundant sources, for example VA-B is the same as the voltage across 1k resistor =1k*1mA=1v.
So what if you did a norton-to-thevenin source transformation on the test current, then you do mesh analysis on the resulting supermesh?
thank you very much!
i have an exam in 2 hours...oh my god
how did u go?? hope u did well..
here a better video on thevenin theorem with dependent sources rth in this video has been calculated using 3 methods
(so that you can compare which method takes less time and which is easier to understand)
th-cam.com/video/D1QxFA9_U5w/w-d-xo.html
@@SupremeTuber Can't understand a word of that
thanks this helped me a lot
That did help.
How do we know that Vab = Voc? Why doesn't the addition of the current source change the voltage for the open circuit?
Yes, I chose 1mA because the resistors were in k(ohms). 1 keeps the math simple, but you can choose whatever you want for a test voltage or current. Note if resistors were in ohms(no k) I'd likely choose 1A.
How did you determine that the direction of the current in resistor 2 is going on the V1? Why not away?
It shouldn't matter. If you choose wrong but keep your signs correct, it works out the same, like a double negative. Worst case, you find a voltage or current is negative, so you account for that in the final ohm's law by checking directions. Current flows from higher voltage.
thanks
MVP
thank you :))
In summary, in a circuit with only dependent sources, Vth will be 0?
How did you find Vx=1/2 V1 ?
if your confused, try nodal.. you can get it
Vx = 1/2 V1 through voltage division. If you follow v1's path to ground from the left, you get a total resistance of 2k ohms, but Vx is only across one of those 1k ohm resistors, so the splitting factor is 1/2.
Why is there no current source in the equivalent circuit? There is a dependent current source In the original circuit, so I think it must be have current source in the equivalent circuit.
Why is it incorrect to solve for Thevenin without the test voltage/current source?
+yosimba2000 Correct me if I'm wrong, but I believe it's because you only have a dependent source. If you had an independent source you might be able to solve for the value of the dependent, but as is you cannot find Rth without introducing a test source. You can't find Rth by deactivating (treat independent voltage sources as short-circuits and independent current sources as open-circuits) independent sources because there is a dependent source. (And you can't 'deactivate' a dependent source.) Ultimately, if you try to solve this without introducing a test source, you'll not be able to simplify the circuit down to Rth because the dependent current source will get in the way.
+Ezis9 so true. thanks for the explanation
yosimba2000 voltage divider rule
I think there is a mistake in KCL@V1.[(V2 - V1) / 2k] - [Vx / 1k] - [(V1-Vx)/1k] = 0 must not be true. Because it would be [(V2 - V1) / 2k] - [V1 / 1k] - [(V1 - Vx) / 1k ] = 0 according to me.
At node V1, why not V1-V2/2k
You messed up the nodal analysis, you used V2-V1 for both nodes
That is how it should be. The current isn't changing in nodal analysis so the voltage over that specific resistor is going to stay as (V2-V1)/2k no matter which node you are analyzing.
good video but if you're telling us a problem can be solved in two ways then solve it both ways ffs
This is incorrect. Vth should be 0 V for any circuit without an independent source. So Rth is always 0. All of your math looks correct from what I could tell by quickly glancing it over so not sure where you went wrong.
+Chris Sleckman Your statement is not correct. It is true that Vth = 0. But Rth is not always 0. The point of thevenin equivalent circuits is to describe a lumped circuit abstraction which says that you can always express a purely resistive network as a voltage source in series with a resistor. If you put the circuit inside a black box with two terminals, you would be unable to tell an original circuit from its thevenin equivalent. If you put this circuit into a black box and measured its voltage, it would be zero, but if you measured it's resistance, there would be a value.
+Matthew Araujo KCL@v1: the first term should be v1-v2/2k not v2-v1/2k. Check it and see
A fundamental thing to know is that a circuit containing only resistances and controlled sources will appear as a constant value resistance when measured at its input terminals. However, this resistance cannot be measured directly using an ohmmeter (or multimeter) because the dependent current, being an element of an active device (a mosfet for example), only becomes 'alive' when the device is properly biased. So, one can only infer its value by applying a known voltage across the input terminals and measuring the current flowing into it; the Thevenin resistance is obtained by taking the ratio of input voltage to input current. For this particular circuit, analysis will show that the current flowing into the leftmost subcircuit (comprising of the two leftmost parallel branches) is zero for any value of voltage applied across the controlled current source, showing that it behaves as an infinite resistance (open circuit). Hence, when viewed from the right the input resistance of the circuit is simply determined by the 1 kohm shunt resistor since the 2 kohm resistor is open-circuited by the voltage-controlled curent source subcircuit.