2
$\begingroup$

The voltage across a capacitor as a function of time is given by: $$V=V_ie^{-t/RC}$$

Therefore, increasing the resistance and capacitance increases the time it takes for the initial voltage to drop to e.g. 63% of the original value, which also means that the exponential decay graph will be less steep with higher resistance and capacitance.

But is there a physical explanation to justify this phenomenon, especially for capacitance? Since capacitance of a parallel plate is given by $C=q/V$ i.e. the charged stored in one of the plate per unit voltage across the plate, how does capacitance affect the rate of discharge or voltage drop?

$\endgroup$
3
$\begingroup$

I'll answer by analogy to give a clearer picture of what's going on. In a sense, a capacitor is like a storage tank for electrons. This means that a capacitor with a larger capacitance can store more charge than a capacitor with smaller capacitance, for a fixed voltage across the capacitor leads.

The voltage across a capacitor leads is very analogous to water pressure in a pipe, as higher voltage leads to a higher flow rate of electrons (electric current) in a wire for a given electrical resistance, per Ohm's Law.

Given the tank analogy, if I increase the diameter of a tank of water, but keep the starting level of water the same as it initially was, I have more water in the tank. When I poke a hole at the bottom of the tank shell, water will flow out and the tank level will drop. For both the small tank (initial conditions) and the large tank, the initial flow rate of water will be the same for a given hole size because the pressure at the bottom of the tank only depends on the height of the water above the hole, but it is obvious that the small tank of water will run dry first because there is less water in the tank. This means that the time constant of the small tank is smaller than it is for the large tank.

If I decrease the hole size (increase the resistance to flow), the time constant for both tanks will increase, but the small tank will always run dry first if both tanks start at the same level.

Regarding the title of this query, the rate of discharge of a capacitor is normally seen to be the rate at which charge is leaving the capacitor plates. This is the current in the associated circuit. How fast the voltage across capacitor plates is decreasing, and how fast the current in the associated circuit is decreasing, is related to the time constant of the circuit, which is NOT the current flowing in the circuit. In other words be careful not to confuse current in the circuit with the time constant of the circuit.

$\endgroup$
4
  • $\begingroup$ To be sure, capacitors do not store electric charge; with charge $Q$ on one plate and charge $-Q$ on the other plate, the net charge stored in a capacitor is zero. A charged capacitor stores energy and not electric charge. $\endgroup$ – Alfred Centauri Sep 30 '18 at 18:49
  • $\begingroup$ @AlfredCentauri, I certainly realize that. I was trying to give the OP a physical representation that was a good analogy, so he could hopefully "connect the dots". $\endgroup$ – David White Oct 1 '18 at 1:02
  • $\begingroup$ @AlfredCentauri Thanks for the answer and comment. You mentioned capacitor stores energy. May I ask to confirm, is it because the charge Q on one plate and the charge -Q on the other plate are energy carriers or carried energy from e.g. a battery? $\endgroup$ – Bøbby Leung Oct 1 '18 at 5:05
  • 1
    $\begingroup$ @BøbbyLeung, to charge a capacitor, the external circuit, e.g, a battery essentially 'pumps' electrons from one plate of he capacitor to the other. This takes work (moving the electrons against the increasing electric field) and the work done is stored in the energy of the electric field between the two plates. $\endgroup$ – Alfred Centauri Oct 1 '18 at 11:35
3
$\begingroup$

A larger capacitor has more energy stored in it for a given voltage than a smaller capacitor does. Adding resistance to the circuit decreases the amount of current that flows through it. Both of these effects act to reduce the rate at which the capacitor's stored energy is dissipated, which increases the value of the circuit's time constant.

$\endgroup$
1
$\begingroup$

Is there a physical explanation for why increasing a capacitor's capacitance and the circuit's resistance decrease the capacitor's rate of discharge?

The equation you give in your question describes the discharge of a capacitor from an initial voltage across of $V_i$ to zero so this will be the context of this answer. On physical grounds, we have the following:

(1) The amount of charge $Q$ that flows through the resistor over the course of the discharge is, from the fundamental capacitor equation $Q = CV_i$

(2) The initial rate at which the charge flows through the resistor is $\frac{dQ}{dt} = \frac{V_i}{R}$ which is independent of the capacitance $C$ and this is the maximum rate.

That is, increasing $C$ increases the amount of charge $Q$ that must flow through the resistor during discharge without changing the maximum rate at which charge flows. Thus, on physical grounds, it will take longer to discharge the capacitor when $C$ is increased.

Conversely, increasing $R$ decreases the initial (and maximum) rate at which charge flows through the resistor. Thus, on physical grounds, it will take longer to discharge the capacitor when $R$ is increased.

$\endgroup$
1
  • $\begingroup$ Appreciate the very good answer, but I ticked another person’s answer as he pointed out I misunderstood the meaning of rate of discharge of a capacitor, which led to so much confusion on my side $\endgroup$ – Bøbby Leung Aug 12 '20 at 22:43
0
$\begingroup$

...how does capacitance affect the rate of discharge or voltage drop?

At any given voltage level, a larger capacitor stores more charge than a smaller capacitor, so, given the same discharge current (which, at any given voltage level, is determined by the value of the resistor), it would take longer to discharge a larger capacitor than a smaller capacitor.

$\endgroup$

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.