I Is Capacitor Voltage Curved Like a Hyperbola?

[vector222]

TL;DR Summary

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?

 

[vanhees71]

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
$$Q(t)=C U_C(t)=CA t + R C^2 A \{\exp[-t/(RC)]-1\}.$$

 

[vector222]

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.

 

[vanhees71]

This looks pretty similar. It's not a hyperbola.