| Electronics > Beginners |
| Thévenin's theorem, open circuit voltage calculation? |
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| rstofer:
I stuffed the equations I gave above into wxMaxima. The voltage VA is indeed 12.07V. Attached is the program and the output. The output starts where the equations are rewritten at "(eq1)" The last 5 lines of code sort the variables. The program itself was written long ago for another example. I just changed the equations a bit. ETA: You can see in eq1 the one current going into node Vx and the two currents leaving. Similarly, at eq2, you can see the 3 currents entering node Va. Positive currents are entering and negative currents are leaving. The assumed direction is irrelevant and you can just as well have all of the current entering and none leaving. Or you can have them all leaving and none entering. How can that be? In the solution, some of the entering or leaving currents will turn out to be negative. If this happens, your assumed direction was incorrect. No big deal, just realize what the negative result means. But it doesn't affect the solution! There are ways to simplify the circuit by various manipulations (like converting the current source to a voltage source with a series resistor. These shortcuts can be quite handy. But the thing about Kirchhoff's Laws is that they take a methodical approach to solving these circuits. You make some assumptions about direction but even that doesn't have to be right. Two simple rules: The currents entering and leaving a node must add up to zero. Current can not spill out on the floor. Second, the sum of the voltage drops around a mesh must add up to zero. Voltage can not pile up. ETA (again): Only the first two equations have anything to do with the solution. There are two equations (eq1 and eq2) and two unknowns (Vx and Va). The other equations just allow me to name the components and assign values. That keeps numerics out of the first two equations and that allows the equation to look a lot like the circuit. What if I used numerics and all the resistors had the same value. It would be confusing! |
| Zero999:
The voltage is higher because the current source and resistance adds to V1. I solved the first two steps in my head, with no calculator. It should be fairly self-explanatory, just by looking at the schematics. Step 1, a current source in parallel with a resistor, is equivalent to a voltage source in series with a resistor. Step 2, the voltages and resistances of the two sources can be added together to form one voltage source and series resistance, as both sources and resistances are in series. The final step was a bit more tricky. Calculate the current through R2 and R3, which is the same, as they're in series. I = V/R V = V1 - V2 = 11.6V R = R1 + R2 = 5k + 3.2k = 8.2k I = 11.6/8200 = 0.0014146 = 1.4146mA Calculate the voltage across R3: V = I*R V = 0.0014146*5 = 0.007073 = 7.073V Calculate the voltage at node A, which is simply the voltage across R3 added to V2. VA = 5 + 7.073 = 12.073V What's more, the whole circuit can be simplified to a simple voltage source and a series resistor. The voltage is equal to the voltage at node A and the resistance equal to R2 and R3 in parallel. R1 = (3.2k*5k)/(3.2k+5k) = 16k/8.2k = 1.951k |
| rstofer:
--- Quote from: Sudo_apt-get_install_yum on September 18, 2018, 02:22:11 pm --- --- Quote from: kulky64 on September 18, 2018, 02:00:41 pm ---I calculated this equivalent model: V = 12.073170731707 V R = 1.9512195121 kOhm --- End quote --- --- Quote from: oPossum on September 18, 2018, 02:04:46 pm ---2k2 * 3mA = 6.6V 10V + 6.6V - 5V = 11.6V 11.6V / (1k + 2k2 +5k) = 1.4mA 5k * 1.4mA = 7.07V 5V + 7.07V = 12.07V (1k + 2k2) * 1.4mA = 4.53V 10V + 6.6V - 4.53V = 12.07 --- End quote --- Thanks for the help! It seems like the answer is 12.07 just like the simulation said. Could you please explain a little bit more, like how you add the voltages together in step two? I have a hard time imagining it. I’m definitively not used to the kinds of circuits. --- End quote --- As I understand the solution, Step 2: The current source and parallel resistor are now treated as a voltage source of 6.6V. Think about 2 batteries in series, they add. So, 10V + 6.6V => 16.6V equivalent. Then the 5v source is subtracted when a mesh equation is written around the 3 sources and 3 resistors (step 3). The next two steps use the mesh current across the 5k resistor to get the final output voltage. The last 2 steps cross-check that the same voltage is contributed by the 16.6V source as is contributed by the 5V source and the 5k resistor. It balances so the output is indeed 12.07V. |
| rstofer:
Strictly speaking, my solution uses Kirchhoff's Laws, not Thevenin's Theorem. Here is a better description of the process. https://www.electronics-tutorials.ws/dccircuits/dcp_7.html |
| Vtile:
There is many ways to skin this cat, which is still really simple mixed source problem and like said above can be solved in many ways. I just want to point out that remember to simplify the networks. You can even go as far as the attachment 2 (network2) and ask yourself what is the current in "ground leg". Edit2. Better drawn network2. |
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