/sci/ - Science & Math » Thread #14150887

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Electrical

Anonymous Wed 26 Jan 12:49:46 2022 No.14150887 View Reply Original Report

Quoted By: >>14150912 >>14151042 >>14151243 >>14151729 >>14151765 >>14151988 >>14152839 >>14154305 >>14154316

How do I find the voltage across the current source?

Anonymous

Anonymous Wed 26 Jan 2022 12:58:32 No.14150912 Report

Quoted By: >>14151025

>>14150887
Find the voltage across the 3 kOhm resistor

Anonymous

Anonymous Wed 26 Jan 2022 13:37:35 No.14151025 Report

Quoted By: >>14151608

>>14150912
Would you happen to know how to do it?

Anonymous

Anonymous Wed 26 Jan 2022 13:44:28 No.14151042 Report

Quoted By:

>>14150887
1mA * Kx = Vx
5V - Vx

Anonymous

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Anonymous Wed 26 Jan 2022 13:50:43 No.14151062 Report

Quoted By: >>14151225 >>14154349

I tried to rearrange the diagram to see if I could understand the problem better. How would I find the ratio of the current that would split?

Anonymous

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Anonymous Wed 26 Jan 2022 14:42:18 No.14151225 Report

Quoted By: >>14151280 >>14151593 >>14151774 >>14154349

>>14151062
you rearranged it more confusedly
you need to be able to add the currents together like in picrel
the key is that the voltage on top and bottom is the same for all four
you do know how to solve a system of equations, right, anon?

Anonymous

Anonymous Wed 26 Jan 2022 14:47:18 No.14151243 Report

Quoted By: >>14151285

>>14150887
Are you playing hide and seek? Have you looked everywhere?

Anonymous

Anonymous Wed 26 Jan 2022 14:59:35 No.14151280 Report

Quoted By:

>>14151225
I know how to uses systems of equations. The problem is that I can't find the voltages, because I am missing resistors (U/R = I).

I tires to mess with my second pic, because I was trying to treat it like a ratio problem. Otherwise I am stuck.

Anonymous

Anonymous Wed 26 Jan 2022 15:01:10 No.14151285 Report

Quoted By:

>>14151243
I've had this up for hours, as well as on work safe requests. I'm also busy solving other problems.

Anonymous

Anonymous Wed 26 Jan 2022 16:42:58 No.14151593 Report

Quoted By: >>14151655 >>14151681

>>14151225
I got help in another thread, but I don't see how this could be done in an exam with a calculator that can only do 4 unknowns. Perhaps I'm too dumb for this course.

Anonymous

Anonymous Wed 26 Jan 2022 16:48:07 No.14151608 Report

Quoted By: >>14151708

>>14151025
3/(1+3) * 5v =3.75V

Anonymous

Anonymous Wed 26 Jan 2022 17:03:51 No.14151655 Report

Quoted By: >>14151708 >>14151719

>>14151593
You could have just said "no, I don't understand how to solve a system of equations"

Anonymous

Anonymous Wed 26 Jan 2022 17:09:53 No.14151681 Report

Quoted By: >>14151708

>>14151593
as a brainlet you can always use helmholtz. get the voltage across r3 while replacing the current source with an open circuit and only the voltage source being active. then get the r3 voltage with replacing the 5V source with a short circuit and having only the current source act. then add the 2 voltages to get the correct result.

Anonymous

Anonymous Wed 26 Jan 2022 17:18:54 No.14151708 Report

Quoted By:

>>14151655
I guess I don't. I've done plenty in the past, but I couldn't abstract enough to be able to do this one.

>>14151608
That simplifies things.

>>14151681
I'm not gonna pretend to understand what's going on here.

Anonymous

Anonymous Wed 26 Jan 2022 17:23:01 No.14151719 Report

Quoted By:

>>14151655
Btw, what answer did you get?
Apparently the answer is meant to be 3.923 V

Anonymous

Anonymous Wed 26 Jan 2022 17:27:44 No.14151729 Report

Quoted By: >>14151756

>>14150887
i don't know. but i have a question: what happens if you replace the current source with a resistor? would the current through it still be 1mA? Or the current source is injecting current into the circuit? Also ask in /ohm/ general on /diy/.

Anonymous

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Anonymous Wed 26 Jan 2022 17:40:25 No.14151756 Report

Quoted By:

>>14151729
The current source is meant to produce the 1mA upwards, no matter what the voltage across it may be.

I never knew about the general on /diy/.

Replacing the current source with a resistor is obviously a much harder problem for my dumb ass to solve, seeing how I've managed in this thread.

Anonymous

Anonymous Wed 26 Jan 2022 17:43:43 No.14151765 Report

Quoted By:

>>14150887
Pro-tip
Thevenin between R=3K and I = 1mA

Anonymous

Anonymous Wed 26 Jan 2022 17:46:07 No.14151774 Report

Quoted By:

>>14151225
Not OP but as someone with no experience, this picture is a good example of how to approach this problem, thanks.

Anonymous

Anonymous Wed 26 Jan 2022 17:47:56 No.14151777 Report

Quoted By:

4.08V

Anonymous

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Anonymous Wed 26 Jan 2022 18:15:43 No.14151863 Report

Quoted By: >>14151991 >>14152847

OP here.
I may be a brainlet, but I'm a stubborn one. I will learn the necessary math to master these problems.

Anonymous

Anonymous Wed 26 Jan 2022 18:58:47 No.14151988 Report

Quoted By: >>14151991 >>14153136

>>14150887
thank god i won't ever have to touch or analyze an analog circuit again

Anonymous

Anonymous Wed 26 Jan 2022 19:00:04 No.14151991 Report

Quoted By:

>>14151988
>he never learned about the impedance matrix

>>14151863
for these basic circuits it's just a matter of applying KVL and KCL over and over again

Anonymous

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Anonymous Wed 26 Jan 2022 21:25:30 No.14152513 Report

Quoted By:

there you go, be glad i'm bored. had to remind myself about parallel resistors

Anonymous

Anonymous Wed 26 Jan 2022 23:22:25 No.14152839 Report

Quoted By:

>>14150887
>The Virgin Calculator
The 1kohm and 2kohm form a parallel resistor network forming the top half of a voltage divider, the two 3kohm and other 1kohm form a parallel resistor network forming the bottom half. Find the equivalent parallel resistances and then apply a voltage divider formula.
>The Chad Constructor
Physically build the circuit and measure it.

Anonymous

Anonymous Wed 26 Jan 2022 23:25:25 No.14152847 Report

Quoted By:

>>14151863
> I will learn the necessary math to master these problems.
basic algebra?

Anonymous

Anonymous Thu 27 Jan 2022 01:13:47 No.14153136 Report

Quoted By:

>>14151988
pussy

Anonymous

Anonymous Thu 27 Jan 2022 10:08:08 No.14154305 Report

Quoted By:

>>14150887
go build it in circuitlab my friend

https://www.circuitlab.com/

Anonymous

Anonymous Thu 27 Jan 2022 10:14:22 No.14154316 Report

Quoted By:

>>14150887
answer is 4.080 volts

Anonymous

Anonymous Thu 27 Jan 2022 10:35:00 No.14154349 Report

Quoted By:

>>14151062
>>14151225
yeah, this arrangement shows that the problem is incomplete, because the branch where the current is measured has no resistance associated with it. if we assume there's no resistance there, then the two branches linked in paralel can be completely ignored, and you're left with V=IR, with the R being 1/(1/1000 + 1/2000).

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