How Does the Time Constant Relate to Charging and Discharging in Capacitors?

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Discussion Overview

The discussion revolves around the relationship between the time constant and the charging and discharging processes of capacitors in RC circuits. Participants explore the mathematical implications of the time constant, particularly how it relates to the exponential behavior of charge over time during both charging and discharging phases.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant states that the time constant is equal to the product of resistance (R) and capacitance (C), and describes its significance in the context of charging and discharging capacitors.
  • Another participant notes that the time constant represents the time taken for the charge on a discharging capacitor to decrease to 37% of its initial value (Qo), while also questioning how this relates to the charging process where the charge increases by 63% of Qo.
  • Some participants discuss the mathematical basis of the time constant, mentioning the exponential functions derived from the differential equation governing RC circuits.
  • There is a clarification that the time taken to gain a charge of 0.63Qo during charging is equal to the time taken to lose the same charge during discharging, which some participants find obvious upon reflection.
  • One participant explains that the 63% figure arises from the mathematical constant 'e-1' in the solution to the differential equation.
  • Another participant emphasizes that in decay, the final value is 0%, while in growth, it is 100%, which is relevant to understanding the time constant's implications.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical relationships involved in the charging and discharging of capacitors, but there is some uncertainty regarding the implications of the time constant and its interpretation in different contexts. The discussion does not reach a consensus on all aspects, particularly regarding the clarity of the mathematical explanations.

Contextual Notes

Some participants express limitations in their understanding of calculus, which may affect their grasp of the mathematical explanations provided. The discussion also highlights the dependence on the definitions of terms like "final value" in the context of charging and discharging.

Hannah7h
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So the rate at which a capacitor charges and discharges is dependent on resistance in a circuit and the magnitude of capacitance of the capacitor? So the time constant is equal to RC. So using this equation where Q=Qoe-t/RC ,time constant is the time taken (when the capacitor is discharging) for charge on a capacitor (Q) to decrease to 37% of Qo ( i.e. charge on the capacitor when it is is fully charged). But I've been reading around and what I don't get is how the time constant is also equal to the time taken for the charge (Q) on a charging capacitor to increase by 63% of Qo . If this makes any sense, would be good if someone could maybe explain it mathematically as well i.e. how 'e' is involved?
 
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Hannah7h said:
i.e. how 'e' is involved?
Have you studied calculus?
 
cnh1995 said:
Have you studied calculus?

Um haven't studied it in much detail, but I could give it a go
 
Hannah7h said:
Um haven't studied it in much detail, but I could give it a go
Well, KVL equation for an RC circuit is a differential equation. Solving that differential equation, you get exponential expressions for charge, current and voltage.
 
cnh1995 said:
Well, KVL equation for an RC circuit is a differential equation. Solving that differential equation, you get exponential expressions for charge, current and voltage.

Ok so... how does this relate to the 63% of Qo?
 
Hannah7h said:
time constant is the time taken (when the capacitor is discharging) for charge on a capacitor (Q) to decrease to 37% of Qo
Or in other words, time taken to lose 63% of Q0.
Hannah7h said:
on a charging capacitor to increase by 63% of Qo
You can see that the time taken to gain a charge of 0.63Qo (while charging) is equal to the time taken to lose the same charge of 0.63Q0(while discharging). It's obvious, isn't it?
 
Hannah7h said:
Ok so... how does this relate to the 63% of Qo?
If you are asking where 63% comes from, it comes from the constant 'e-1' in the solution to the differential equation of the RC circuit.
 
the time constant is the time taken to get to within 37% of the final value.
In decay the final value is 0%!
In growth the final value is 100%
 
cnh1995 said:
If you are asking where 63% comes from, it comes from the constant 'e-1' in the solution to the differential equation of the RC circuit.

Ok that was pretty obvious now I look at it, thank you for explaining it otherwise I probably wouldn't have got there lol
 
  • #10
lychette said:
the time constant is the time taken to get to within 37% of the final value.
In decay the final value is 0%!
In growth the final value is 100%

Yep this makes sense, thanks!
 

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