hi friends, 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
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
hi ratch, yes i m trying to find Vab for thevenin equivalent circuit & Rth. pls use nodes or mesh to solve. thanks
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..the answers r correct. when u said nuke the R2, u mean take it out so as to open it? would there any current flow at R2? if so how to obtain the value? as stated in -5+ Vx/4 + 3Vx/6 = 0 it is clear that 5A is divided between R1 & R4 & but excluding R2. thanks
stupid, 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. 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 . 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. Let's use R2 in the supernode current calculations. On the left side of R2 is voltage Vx and the right side is 3*Vx. So the current leaving the left node through R2 is (Vx-3*Vx)/2 , and the current leaving the right node through R2 is (3*Vx-Vx)/2 . Adding the two currents shows that the net current leaving the supernode through R2 is zero. So why bother calculating the current? You know its net value will be zero. Just remove the resistor and be done with it. Ratch
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
hi ratch, thank u so much for yur explanation to my questions. now let say the voltage dependent voltage source is replaced by a independent voltage source, would the voltage across R2 remain the same? what is the difference between dependent source & the independent & whats the impact to the circuit? thanks.
hi, attached are the Pspice simulation for the circuit. can that be used to prove the answers? thanks
well, between points a and b ... you have a current meter and that certainly will prove the current answer. You'd have to replace that with a voltmeter to show the voltage. Or you could do something like ....
stupid, It depends. Yes if the voltage source is constant. No if it is a AC voltage. 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. Ratch
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
thank u ratch, hgmjr & someone whose inputs are so valuable to me.. thanks a million. regards, stupid
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. When calculating Thevenin equivalents, you can't replace dependent sources with shorts or opens. You have to leave the control function in place. This issue came up in a previous thread: http://forum.allaboutcircuits.com/showthread.php?t=25537&highlight="dependent+source"
Thanks all, for getting me back on the straight and narrow path. I was indeed leaping to the erroneous conclusion that dependent sources could be treated as though they were independent sources. That is a mistake I will avoid in the future. In the future, I will be sure to use the more generally applicable method of calculating the open-circuit voltage and then calculate the short circuit current and divide the former by the latter to obtain Rth. It is always refreshing to learn something new from a thread. Thanks, hgmjr