i m having a very hard time with the circuit as per attached.
i started with converting current source to voltage source, is that correct?
how should i proceed thence?
thank u
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What are you trying find? Is it the voltage across a,b? Why does the title of the thread say Thevenin? Are you required to solve whatever it is using Thevenin's theorem? Do you want the solution done using nodes, supernode, or meshes?i m having a very hard time with the circuit as per attached.
i started with converting current source to voltage source, is that correct?
how should i proceed thence?
stupid,
What are you trying find? Is it the voltage across a,b? Why does the title of the thread say Thevenin? Are you required to solve whatever it is using Thevenin's theorem? Do you want the solution done using nodes, supernode, or meshes?
Ratch
OK, here goes. Ordinarily, you cannot find the Thevenin/Norton equivalent of this circuit by using either Thevenin or Norton alone. That is because your circuit contains a dependent voltage source. But you can use a combination of Thevenin/Norton and get the correct source resistance irregardless of the dependent sources. You first find the open a,b voltage, and then the short a,b current. Dividing the voltage by the current gives the Thevenin impedance. First we nuke the R2 resistor, because the dependent voltage source is going to establish the same value across it regardless of its resistance. Then let's use the supernode method. So setting up the equations with a,b open we get:yes i m trying to find Vab for thevenin equivalent circuit & Rth.
pls use nodes or mesh to solve.
stupid,
OK, here goes. Ordinarily, you cannot find the Thevenin/Norton equivalent of this circuit by using either Thevenin or Norton alone. That is because your circuit contains a dependent voltage source. But you can use a combination of Thevenin/Norton and get the correct source resistance irregardless of the dependent sources. You first find the open a,b voltage, and then the short a,b current. Dividing the voltage by the current gives the Thevenin impedance. First we nuke the R2 resistor, because the dependent voltage source is going to establish the same value across it regardless of its resistance. Then let's use the supernode method. So setting up the equations with a,b open we get:
-5+ Vx/4 + 3Vx/6 = 0 ======> Vx = 6.66666667 and Vab = 3*Vx = 20 volts
Next we find Vx if a,b is shorted:
-5+ Vx/4 + 3Vx/6 + 3Vx/2= 0 ======> Vx = 2.22222222 , so the voltage across R3 = 3*2.2222222 = 6.66666667 volts, and short circuit current existing in R3 is 6.66666667/2 = 3.33333333 amps. The Thevenin impedance is 20/3.333333 = 6 ohms. Therefore a,b sees a 20 volt source in series with a 6 ohm resistance. Or a 3.33333333 amp source in parallel with a 6 ohm resistance. Take your choice.
Ratch
Yes, that is what I meant. It is a idomatic expression used frequently in the US and perhaps elsewhere. As I said, the voltages at each end do not change at any resistance level, so it need not be considered in voltage calculations.when u said nuke the R2, u mean take it out so as to open it?
You mean current existence through R2? Certainly there is, but it is not necesary to know how much. The voltages at each node remain the same no matter what the current value is. If you want to know what is is, divide the voltage difference across the resistor by the resistance to get its current, which 2*Vx/2 = Vx amps .would there any current flow at R2?
if so how to obtain the value?
Current does exist through R2, but we don't care. I used the supernode method, so the current leaves and enters the same node and cancels out.as stated in -5+ Vx/4 + 3Vx/6 = 0
it is clear that 5A is divided between R1 & R4 & but excluding R2.
It depends. Yes if the voltage source is constant. No if it is a AC voltage.now let say the voltage dependent voltage source is replaced by a independent voltage source, would the voltage across R2 remain the same?
The value of a independent source does not depend on the value of a voltage or current in another part of the circuit or externally. An dependent source is controlled by voltage or current in another part of the circuit or externally. The dynamics of the circuit are completely different depending on whether the source is dependent or independent.what is the difference between dependent source & the independent & whats the impact to the circuit?
Clearly my method of attack in determining Rth is not appropriate for this problem type. It is back to the drawing board for me to determine where I went off the rail.hgmjr,
OK, I see what you did, but I don't think it is correct. You did what I warned stupid against doing in post #5 of this thread. You cannot use Thevenin's theorem or Norton's theorem alone when a dependent source is present. You have to use a hybrid Thevenin-Norton procedure as I have explained above. To check your answer, check the open voltage and short current at a,b. The open voltage is 20 volts which is good. The shorted current is 20/4.4=4.55 amps which is not. It should be 3.3333 amps which was calculated by node analysis.
Ratch
thank u ratch, hgmjr & someone whose inputs are so valuable to me..Clearly my method of attack in determining Rth is not appropriate for this problem type. It is back to the drawing board for me to determine where I went off the rail.
hgmjr
Ratch,
I get the same Vth as you did, 20V. However, I calculated 4.4 ohms for Rth. My approach for calculating the Rth was to replace the voltage source with a short circuit and the current source with an open circuit and then calculate the resulting resistance. The 2 ohm resistor is shorted out by the voltage source so it falls out of the calculation. That leaves 2 Ohms in series with the parallel combination of 4 ohms and 6 ohms. The parallel resistance computes to 24/10 plus 20/10 (for the 2 Ohm resistor) yields 44/10 Ohms. That comes to an Rth of 4.4 Ohms.
hgmjr
hgmjr,
OK, I see what you did, but I don't think it is correct. You did what I warned stupid against doing in post #5 of this thread. You cannot use Thevenin's theorem or Norton's theorem alone when a dependent source is present. You have to use a hybrid Thevenin-Norton procedure as I have explained above. To check your answer, check the open voltage and short current at a,b. The open voltage is 20 volts which is good. The shorted current is 20/4.4=4.55 amps which is not. It should be 3.3333 amps which was calculated by node analysis.
Ratch
Your mistake wasn't so much that you used Thevenin's theorem or Norton's theorem alone, but that you replaced the dependent voltage source with a short circuit.Clearly my method of attack in determining Rth is not appropriate for this problem type. It is back to the drawing board for me to determine where I went off the rail.
hgmjr
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