Finding Thevenin's Voltage and Resistance in a circuit - Page 1
Forum section: electronics / beginners

author: _romeo
author_rank: Newbie
timestamp: September 11, 2022, 04:31:53 pm
content: Hello, I've been going through Practical Engineering for Inventors and I have reached the Thevenin's and Norton's Theorem Section. As an exercise I am supposed to find VTHEVand RTHEVfor the following circuit:Now, I might have missed something but in order to find VTHEVdon't you need a load that has a resistance?The answers mention RTHEV=100Ω and VTHEV=2V. I've managed to reach the same value for RTHEVby calculating the parallel resistor equivalent and using that to calculate in parallel instead of series with the remaining resistor. My intuition tells me that in this case the 2 300Ω resistors are the load in this case.Can someone please give me some hints? Thank you!Edit: I realised I can calculate the circuit resistance, calculate the current then use Kirchhoff's current law and calculate the voltage between A and B

author: ledtester
author_rank: Super Contributor
timestamp: September 11, 2022, 04:48:20 pm
content: V_TH is the voltage across A and B with A and B open (that is, with no extra load).R_TH is found by setting all voltage sources to zero volts (and opening any current sources) and then computing the resistance between A and B.So you don't need to introduce a load between A and B.In your problem when the 6V is replaced by a short it is clear there are three 300R resistors in parallel, so R_TH is 100R.

author: _romeo
author_rank: Newbie
timestamp: September 11, 2022, 04:53:41 pm
content: Thanks! I was failing to replace the source with a short, and couldn't see why the left most resistor is in parallel.

author: iMo
author_rank: Super Contributor
timestamp: September 11, 2022, 05:40:36 pm
content: Good to remember is that the internal impedance of an ideal voltage source is zero, the internal impedance of an ideal current source is infinite. That is why you shorts the voltage sources and opens current sources when doing above exercises.

author: MathWizard
author_rank: Super Contributor
timestamp: September 11, 2022, 07:15:42 pm
content: Yeah and in a simulator you just set the voltage or current sources to zero/off. In fact in LTspice a voltage source set to zero acts like an easy to click on amp meter.

author: MrAl
author_rank: Super Contributor
timestamp: September 12, 2022, 11:02:55 am
content: Quote from: _romeo on September 11, 2022, 04:31:53 pmHi,As you know now, you do not need a load to compute these values. However, you can TEST your results with a load of any impedance value (other than zero or infinite perhaps) applied.To do this, first calculate what you think is Rth and Vth. Then using the circuit in your drawing on the left, add a resistive load and try to choose one that makes the calculation easier. In this case 300 Ohms would be a good choice but really any value would work as long as it is not zero or infinite. Once you connect the resistor to the output, calculate the output voltage again. Do the same with the original circuit, adding the load and calculating the output voltage.If you get the same voltage with both tests then you probably have the right equivalent circuit.If you still have doubts though, use a 2nd resistor value (like say 900 Ohms or something, but really anything other than zero or infinite). If you get the same voltage with that 2nd resistor value load as you did with the first resistor load, then it's almost certain that you did the calculations for Rth and Vth right.The reason for a second load resistor value is because with just one load there is a chance that you got Rth and Vth wrong but in a ratio that happens to work with that particular value of a load, but when you do a second value load test since it is a linear circuit if there was something wrong it would not show the same value for the test output voltage.Off the top of my head i can tell you that with a 300 Ohm load the output voltage would be 1/4 of the source voltage (25v with a source voltage of 100 volts) and with a 900 Ohm load the voltage would be 0.3 of the source voltage (30v if the source was 100 volts).If the source was a current however, then Rth would be just 300 in parallel with 300 because you open circuit the current source, and if that current source was 1 amp then the Vth would be 150 volts because 1 amp times 300 in parallel with 300 would be 1 amp times 150 Ohms which is 150 volts.So you open circuit the current source to get Rth, but keep it in to get Vth after you get Rth (and Vth is the output voltage at that point).It's also interesting that there are other consequences of Norton and Thevenin equivalents.For example, in the original circuit because you have a voltage source in series with a resistor (that first one on the left most side of 300 Ohms) you can replace that with a current source of value V/R in parallel with a value of R Ohms. Since R is 300 in this case, the current source would be 6/300 amps and the parallel resistor would be 300 Ohms.That would change the circuit to a current source of 0.002 amps in parallel with 300 Ohms, and since that 300 Ohms is now directly in parallel with the other two 300 Ohm resistors you can see right off that Rth is the three 300 Ohm resistors in parallel (with the new current source open circuited).If you like you can take it even one more step.Now that you have a current source of 0.02 amps in parallel with 100 Ohms, you can convert that back into a voltage source in series with a resistor.The new voltage is 0.02 amps times 100 Ohms which of course is 2 volts, and the series resistance is 100 Ohms. So guess what we end up with? Yes, the Vth and Rth we were after to begin with: 2 volts and 100 Ohms :-)