How do you get the Thevenin equivalents on question 8 of the thevenin worksheet, especially the Thevenin resistance? The Thevenin resistance is 210.53 and the Thevenin voltage is 8.772. The circuit sort of confuses me. http://www.allaboutcircuits.com/worksheets/thev.html I have a test tomorrow night so I'd like to know today or tomorrow.
To find the thevenin resistance you short circuit all the voltage sources and open circuit all the current sources. Here you have a voltage source only, if you short it out the you have: Rth=(300//1K)//(400+2K) To find Vth use the original circuit (voltage source not shorted): The equivalent resistance of 1K-2K-400=R1 equals R1=(2000+400)//1000=705.9 R By using the voltage divider equation the voltage across R1 is: VR1=15*R1/(R1+300)=10.53V Then use again the voltage divider equation to find the voltage across the 2K: V2K=VR1*2K/(2K+400)=8.77V
Mik3, After looking at this problem closer I think there may be an error in the calculation of Rth. Shouldn't Rth be: ((300||1K)+400)||2000 = 479.532 Ohms What do you think? hgmjr
No, its Rth=(300//1K)//(400+2K) because the upper side of the 2K is connected to the upper side of the 1K and 300R resistors, have in mind that the battery is shorted
Ohhh yes, i checked it better and you are right hgmr I apologize for my previous post. Heidegger, the correct Rth=((300||1K)+400)||2000 = 479.532 Ohms as hgmr said, i wasnt very careful before
Mik3, No apology needed. Its a pretty obtuse network. You had me starting to doubt my understanding of Thevenin's Theorem. I feel better now. hgmjr
hjmjr and mik3: You are both forgetting the fundamental princible of Thevenin`s Theorem! The Rth is viewed from the terminals of interest. In this case the terminals are across the two terminals outside the dotted lines (in parallel with the 2k resistor); as the printed answer dictates-displaying the Rth and Vth with the same two terminals. There are different answers-depending on the defined terminals looking into the circuit. [ (2000+400) || (300 || 1000) ] = 210.53 ohms
You told the correct answer by yourself (viewed from the terminals of interest) but you gave the wrong expression. The correct one is ((300||1K)+400)||2000 = 479.532 Ohms Think of it, the terminals are connected across the 2K.
Greetings Alwayslearning, If you look at the problem closely I think you will see that value of 210.53 Ohms is Rth at the junction of the 300, 400, and 1000 Ohm resistor. Not at the terminals indicated by the dashed line. hgmjr
No mik3, I'm not agreeing with Alwayslearning. I intended to point out that the equation he is advocating for Rth is for the node at which the 300, 400, and 1000 Ohm resistor come together. What the problem is asking for the value of Rth at the node where the 400 Ohm and 2000 Ohm come together. Sorry for any confusion. hgmjr
Another way to find Rth is to divide Vth by the short circuit current out of the terminals. If you do it and the calculations you find exactly 479.532 ohms
hjmjr and mik3: The textbook answer is correct! We are asked to Thevenize the circuit; from the terminals outside the dashed lines! As stated previously from the terminals looking in; an electron would first encounter the (2k resistor in series with 400 ohms) in parallel with (1k||300 ohms) = 210.53 ohms. Following the rules of Thevenin Theorem, the calculated solution above, the textbook solution-there is no doubt! I would suggest you both wire up the circuit and measure the resistance?
Mik3, I too simulated the network and I obtained 479.532 as well. Alwayslearning, did your simulation yield 210.53? hgmjr
Here is the simulation that I used to check Rth. The bottom circuit is the network straight from the worksheet. The top circuit is the same network loaded with a 1000 ohm load. Rth can be calculated as follows: The second image is the Vth and Rth only. The voltage across the 1000 load resistance is the same as the original circuit loaded with 1000. This should illustrate clearly that Rth is 479.5 Ohms. hgmjr
I'm getting RTH = (300 // 1k + 400 ) // 2k 630.76923 // 2k = 479.53 ------------------------------------------------------- and ETH= 15V across (300 on neg. and (1k // 2k) ) 15V in series with 300 ohms and (667ohms) = 10.34V across thew output terminals. and 4.65V across 300 ohms. whoops ETH is wrong I left out the 400 ohm resistor. I'm getting ETH=8.77V at the output term. when I included 400ohms. in the circuit. Is that correct??
hjmjr and mik3: Well I am glad we had this discussion! I have learned a valuable lesson on the subtleties of series/parallel circuits. The 400 ohm is in series with the (300||1k)||2k; all I was seeing was the 400 ohm in series with the 2k||(1k+300). Though I dont feel that bad considering that the textbook writers/screeners along with many others missed it also? Thank-you hjmjr and mik3!
Greetings hobbyist, Your updated value for the output voltage of 8.77V is correct. The simulation I have supplied validates my original conclusion concerning the error in the calculation of Rth as indicated in the problem. The incorrect value of 210.53 Ohms as stated in the example needs to be amended to indicate that the actual value for Rth is 479.53 Ohms. hgmjr