How many electrons on a Capacitor calculation

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SUMMARY

The calculation of the number of electrons on a capacitor is derived from the formula Q = C * V, where Q is the charge, C is the capacitance, and V is the voltage. In this discussion, a capacitor with a capacitance of 47 µF and a voltage of 12 V results in a charge of 5.64 x 10^-4 Coulombs. The number of electrons can be calculated using the conversion factor of 1 Coulomb = 6.25 x 10^18 electrons. Additionally, the charging behavior of capacitors is described by the time constant RC, and the voltage across a capacitor can be expressed as Vc = Q / C.

PREREQUISITES
  • Understanding of basic electrical concepts such as charge (Q), capacitance (C), and voltage (V).
  • Familiarity with the formula Q = C * V for capacitor calculations.
  • Knowledge of the relationship between charge and the number of electrons (1 Coulomb = 6.25 x 10^18 electrons).
  • Basic understanding of exponential functions and their application in electrical circuits.
NEXT STEPS
  • Learn about the time constant in RC circuits and its significance in capacitor charging.
  • Study the differential equations governing capacitor and inductor behavior in electrical circuits.
  • Explore the concept of conservative fields and their relation to voltage drops in closed circuits.
  • Investigate the relationship between work and potential difference in electrical systems.
USEFUL FOR

Electrical engineering students, electronics enthusiasts, and anyone interested in understanding capacitor behavior and calculations in circuits.

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.
 
  • Like
Likes   Reactions: ENE
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.
 
  • Like
Likes   Reactions: 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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