Electronics > Beginners

Thévenin's theorem, open circuit voltage calculation?

(1/3) > >>

Sudo_apt-get_install_yum:
Hi everyone!

So my friend showed me a question he got on a test (see attached image) and I realized that I don’t know how to solve this.
Now I’ve googled around and skimmed through one of my books (Analog electronics) and I am pretty sure that I have to use "Thévenin's theorem" to solve this. Reason why is because it’s a linear circuit and fits the Wikipedia explanation of the theorem https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

So I fired up LTSpice and simulated the circuit, the resulting voltage between AB is ~12V witch confuses me even more since this voltage is higher than both sources.

Could someone please explain to me what’s going on?
I’ve never encountered a real life example of this, where is this used and in what circumstance is it used, my friend didn’t solve it correctly so I’m also indirectly asking on his behalf.

Thanks for any help!

Mr. Scram:
Following for fun.

kulky64:
I calculated this equivalent model:
V = 12.073170731707 V
R = 1.9512195121 kOhm

Mario87:
Ok, I THINK I know how to do this, but bear with me as I'm only just starting to go through the AoE textbook and lab books which funnily enough cover voltage dividers & theveins theorem in the first chapter. Then again I could be TOTALLY wrong in what I'm about to explain, so hopefully someone else can clarify.

The way I looked at it was to take the first part of the circuit as a voltage divider with 1k & 2.2k

10v over those resistors gives you 6.875v across the 2.2k resistor

From there we can calculate the resistance of the current source. So using ohms law, V=IR which goes onto 6.875=0.003R, therefore R=2291

So now our 2.2k resistor is in parallel with a 2291 ohm resistor which is no different than a single resistor of 1122 ohms

So now we have a voltage divider with 3 resistors. First one is 1k, second is made up of the parallel devices which we have just calculated have a resistance of 1122 ohms and the final resistor is 5k

So then, if we have 10v across all 3 resistors, that means the voltage across the final 5k resistors is 7.02v

Take that 7.02v and add on the 5v supply you have in series with it and you get an output of 12.02v

Like I said, I could be TOTALLY wrong and may have just stumbled upon the answer by sheer chance, but that's how I have understood it.

If someone with more knowledge can confirm it would be great, and if I am correct it would be good to know that I am actually understanding what I am reading in these books.

oPossum:
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

Navigation

[0] Message Index

[#] Next page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod