Is Capacitor Voltage Curved Like a Hyperbola?

  • Context: Undergrad 
  • Thread starter Thread starter vector222
  • Start date Start date
  • Tags Tags
    Curve Hyperbolic
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 1K views
vector222
Messages
8
Reaction score
0
TL;DR
is this a hyperbolic curve?
Ok Hi everyone!

I was working on what would happen if you apply a linear increasing voltage to a series capacitor resistor.
The question is : If the capacitor voltage is plotted, is the cap voltage curve hyperbolic?

I've done some plots on the cap voltage and it sure looks hyperbolic but I can't prove it.

Any thoughts on this?
 
Physics news on Phys.org
If ##Q## is the charge on the capacitor the equation for the circuit reads
$$Q/C+R \dot{Q}=V(t).$$
Taking the time derivative gives
$$i/C+R \dot{i}=\dot{V}=A=\text{const}.$$
The solution of this obviously is
$$i(t)=C A+a \exp[-t/(RC)].$$
With ##i(0)=0## you get
$$i(t)=CA [1-\exp[-t/(RC)]].$$
Integrating, assuming ##Q(t)=0## gives
$$Q(t)=C U_C(t)=CA t + R C^2 A \{\exp[-t/(RC)]-1\}.$$
 
  • Like
Likes   Reactions: etotheipi
Hi Vanhees71 !

I had come up with
Vres(t) = R/rc (1 - exp (-t/rc) where

Vres(t) = voltage across the resistor at time t
R = input volts per second across the series resistor capacitor
r= ohms
c = farads

Vres(t) = R/rc approaches a max value as t goes infinite

voltage across the cap at time t =
Vcap(t) = R(t) - R/rc (1 - exp (-t/rc) )

Have not verified this but it does look close to your math.
Still don't know if the cap volts are hyperbolic.