Hey Sup Forums i was wondering if any of you knew anything about circuits

Hey Sup Forums i was wondering if any of you knew anything about circuits
I need help: how do you calculate the evolution of the voltage decrease in a condenser? (image related, thats the circuit)

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Do you have the capacitance?

t= 0.7 CxR where C is in MFD and R is in MOhms. Theoretically, it's simply CxR but in practice 0.7 is used.
The curve is logarithmic.

You seem onto it, not op but I have an exam in a month or so. Can you please explain overload abit more? Is there a difference between overload and overcurrent or are they the same?

Sorry OP I can't help you
I know very little about circuits but want to learn. Aside from sticking my dick in a socket where do I start?

actually 1-(1/e) ~= 0.632

A static D.C. Circuit
You have an induced voltage stored on the capacitor from some previously connected source. It Pushed so many Coulumbs of electrons on one plate and drew the same number away from the other side.


The previous source placed an induced voltage that still remains on the capacitor.

The new circuit will discharge the excess electrons from one side when the switch is thrown and they will first have to pass through the diode. The resistance to the current will slow the process.

The length of time it takes is based on a function NOT of whatever voltage is on the capacitor- it IS only based on the product of Resistance in Ohms, and the Capacitance in Farads. This can be tweaked to use Non standard inputs by using Ohms in the millions scale (Mega Ohms or Megohms) and Capacitance in millionths of a Farad or microFarad.

Get it so far?

This is part one.

Calculating the Tau in seconds.

Is done by R x C

Once you have the time constant of the circuit calculated you can use the formula that follows.

It takes the Initial voltage on the capacitor and the length of time after the switch is thrown, and the time constant, and gives you the voltage at that later time.

Charging
V
(
t
)
=
V
0
(
1
−
e
−
t
/
τ
)
V(t)=V_{0}(1-e^{{-t/\tau }})

You made me remember why I fucking hated theoric electricity back in technical high-school.

Going for the comp science engineer title now.

V(time) = V(initial_voltage) * ( 1 - e ^ (-Time / Tau) )

It takes roughly 5 or for extreme safety 7 time constants to discharge fully.

check

It's simple as fuck!

You're trying to get a point on a curve


It starts high and goes low but not in a straight line


It's the same curve as mortgages follow, it's the same curve as bacterial growth. You shouldn't be able to be called a scientist Anything if you can't understand a natural logarithmic curve

You meant mFD, not MFD, right?

Not so, good sir, you shouldn't be called a scientist if you aren't interested in finding the truth. Not understanding a natural logarithmic curve? Try not understanding most of the stuff in the Universe, but trying to anyway. That's real science.

Computer science engineer

That's bullshit !

In Asia they take computer engineering Seriously because they use Maxwells Equations (circa 1860) to make sure the traces on the circuit boards aren't Too tiny or Too close together and the Shapes they make don't interfere either


If you can't handle a fuckinh logarithmic curve I say kill yourself before you pay money to idiots to ejumacate yourself in software bullshit and windows bullshit network settings before you DARE let them Hand you a degree with the word Science in it.

I'm 52 fucking years old and I can't believe how fucking retarded you and most graduates are.

Well they call a degree a bachelor of science when most of them aren't involved in SCIENCE or the truth


Grok or Quine.
Are there mathematical truths that are universal- that aliens understand and know or are seeking? Are we playing numbers games with numbers and it's not necessarily the actual mathematical truth?

You do sound like a butt-hurt 52 year old who had to study stuff that nowadays are not necessary for every comp science engineer

,Welcome to best sex dating site >>> v.ht/yKrKm

the circuit loses voltage

It's not linear

Calculate R times C in either Actual Farad Units and Ohms

Or

Calculate R times C in MegaOhms (MΩ) times microFarads (μF)
Using this method the odd scales cancel each other out.
Any capacitor will Not be in the farad range at your retardededed school

Ok

The result you plug into this formula


Voltage Now= Initial Voltage times something

The Something is a thing that describes a curve dropping fast but slowing down

How fast it slows down uses RC
(R times C)

Why?

A big resistor slows it down a lot

A little resistor slows it down a little


The logarithmic function of E to a power subtracted from 1 gives you the shape

What does the 1 do in the formula?
It causes the voltage at time zero to equal the initial voltage.

Gradually later the thing subtracted from the 1 increases and the Time later in seconds goes into the formula

...

I'm Not a computer science engineer

You said you were abandoning something to Fall Back on computer science because you can't hack high school physics. This is not AC circuits with complex numbers. This is a Boy Scout or Cub Scout demonstration showing how to discharge a fucking dangerous capacitor safely.

I am current on current computer languages and cross platform LLVM baded software development . No degree needed.

You- asked for help but can't read a Wikipedia article on RC time circuit?

I didn't ask for anything, I just stated that I find theoretical electricity very thedious and boring. Never said it was hard.

I use schematic capture, a Rip and retry Autorouter to create PCB designs for custom secure banking links. They use three hundred bit long Cross linked LFSRs with a custom ASIC state machine inside to create a seemingly endless supply of random bits that repeat over a hundred quadrillion years later.

I understand both comp sci and electronics out the fuckin wazoo. My company is private and caters to banks and small Non-American countries because the USA prohibits exporting cryptography software or hardware larger than 512 bits.

>Asked for help on Sup Forums
>help me lern bee I r dum

I'm drink and being n asshole
My apologies.

Example 1:

R = 4.7k ohms
C = 750μF

The capacitor is charged to 280 volts. That's D.C. By the way - you can't place an AC charge on something. You CAN however unplug something that's running and be left with a D.C. Charge on some of the capacitors. If they're not clamped with some very large value resistor to discharge them then these highly non conductive IGBT's will prolly not discharge the caps. If you play with a 30hp servo drive without knowing what you're doing you gonna get sparks.

Multiply 4.7e3 times 750e-6

Give me your results so far

I get 3.525

This is in seconds

In 7 time constants the circuit will have like No measurable voltage on it.

How do we tell what it has at 5 seconds?

We plug in 3.525 5 and 280 volts into the formula

It gives us a value

Easy

Voltage at 5 seconds =

280 * ( 1 - e ( -5 * 3.525 ) )

212.2 volts.

Put that in Excel, Numbers, or the LibreOffice spreadsheet.

Make a chart .

C*dV/dt=-V/R