How many electrons on a Capacitor calculation

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To calculate the number of electrons in a capacitor, use the formula Q = CV, where Q is charge, C is capacitance, and V is voltage. The user correctly calculated Q for a 47 µF capacitor at 12V, resulting in 5.64 x 10^-4 coulombs. The charging of a capacitor is described by the time constant RC, and the voltage across it follows an exponential curve. The discussion also touches on the principle that the sum of voltage drops in a closed circuit is zero, linked to the conservative nature of electric fields. Understanding these concepts is essential for grasping electrical engineering fundamentals.
ENE
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Hello,
How to find number of electronics in capacitor?
as 1 coulomb and 1 Amp has 6.25x10^18
on what they depend?
Is this correct..?
Q=C*V
C=47uF and V=12V
Q=47*10^-6*12=5.64*10^-4 Coulmb?
 
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ENE said:
How to find number of electronics in capacitor?

The number of electrons in a capacitor is given by Q=CV, which you've calculated correctly.
 
Thanks.

I don't understand this graph
How the capacitor will charge with series Resistance?

rc2.gif
 
Hello,
I got
a. The rate of charging is typically described in terms of a time constant RC.
b. The electrical transient phenomena in capacitors and inductors are exponential processes
tcons.gif
voltage across the capacitor
how we get exponential here?
rc15.gif
 
The voltage on a capacitor is Vc = Q / C.

Note: you copied images from two different circuit explanations and that's causing confusion. The "Vb" in the first image is "Vs" in the second image.
 
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i have to learn this equation?
 
ENE said:
i have to learn this equation?

Oh, the horror! The horror!

Yes, you have to learn it.
 
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Likes ProfuselyQuarky
in electrical engineering how man equation are there?
 
  • #10
ENE said:
voltage across the capacitor
how we get exponential here?
proxy.php?image=http%3A%2F%2Fwww.electronics-tutorials.ws%2Frc%2Frc15.gif

We solve the differential equation for the circuit. See for example

http://web.mit.edu/molly/Public/circuits-b.pdf

and scroll down to page W6-6. This uses Q as the variable, not V, but you can change variables using Q = CV if you prefer.
 
  • #11
Hello,
Why The sum of the voltage drops∆Vi , across any circuit elements that form a closed circuit is zero.??
 
  • #12
fizzle said:
The voltage on a capacitor is Vc = Q / C.
Vanadium 50 said:
Oh, the horror! The horror!

Yes, you have to learn it.
ENE, that's an awfully basic equation, you needn't worry :woot:
 
  • #13
ENE said:
Hello,
Why The sum of the voltage drops∆Vi , across any circuit elements that form a closed circuit is zero.??

The answer to this is kind of in-depth and I don't feel you're ready for it at the moment.
 
  • #14
ENE said:
Hello,
Why The sum of the voltage drops∆Vi , across any circuit elements that form a closed circuit is zero.??
The static electric field is a conservative field. So the work done on a closed path is zero.
Do you understand the relationship between work and potential difference?
 

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