0
\$\begingroup\$

Take a circuit comprised of resistors, a single capacitor, and a single step response voltage source. Then:

  1. Every single voltage and current in the circuit, except at the voltage source, will have a *e^(-t/tau) dependency.

  2. Where tau, the time constant, is the capacitance times equivalent resistance seen by the capacitor.

Why are 1 and 2 true? I understand why it is true for a resistor connected in series with a capacitor but not why this can be generalized, especially since not every RC circuit can be simplified into a resistor in series with a capacitor.

\$\endgroup\$
  • \$\begingroup\$ If a resistor is connected across an ideal voltage source it will have DC flowing through it, regardless of what the rest of the circuit is doing. \$\endgroup\$ – Chu Nov 21 '18 at 7:26
0
\$\begingroup\$

Circuits which have just one capacitor can be divided to 2 parts. First part is the capacitor and the 2nd part is the rest. Thevenin's equivalent of the rest is one resistor and one voltage source. That's the series circuit which you already know.

If we replace the capacitor with a voltage source which has exactly the same in time varying voltage that the capacitor had, the rest of the circuit sees nothing changed.

But for every resistor there's now 2 voltage sources which supply it through a resistor network. Every resistor get a voltage which is a weighted superposition of a step voltage and the exponential charging voltage. That's still an exponential charging voltage with the same tau as the capacitor voltage has, only starting and final values are different.

\$\endgroup\$
  • \$\begingroup\$ Can you explain how you know every voltage difference and current in a purely resistive network is exponential when driven by an exponential voltage source? It has held true for every resistive circuit I can think of but how do I know I didn't just miss the one strange resistive circuit that does not obey it? I know that a resistive network can be modeled as an equivalent resistance but that only shows the voltages at the ports of the resistive network are exponential. \$\endgroup\$ – roobee Nov 20 '18 at 22:47
  • \$\begingroup\$ @roobee Every pair of nodes in a resistive network can be considered to be a port. It's an open port, nothing is connected. You just wrote yourself that ports have exponential voltages and there's no problem in it. When the port happens to be the ends of a resistor, the current in the resistor is exponential by Ohm's law. \$\endgroup\$ – user287001 Nov 20 '18 at 22:57
  • \$\begingroup\$ Nevermind, I just realized that since a purely resistive network has no memory of previous voltages I can use principle of superposition. So at t=1 each voltage will be e-1/tau of the voltage it was at t=0. So that explains why each node is exponential. \$\endgroup\$ – roobee Nov 20 '18 at 23:02
0
\$\begingroup\$

1 and 2 are true because you can derive it from first principles using differential calculus. See:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html

and

https://www.york.cuny.edu/academics/departments/earth-and-physical-sciences/physics-lab-manuals/physics-ii/time-constant-of-an-rc-circuit

The relationship only holds for circuits that can be simplified to an R in series with a C. For other arrangements the equations would be different (tho likely take on a similar form).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.