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# 5.18: Discharging a Capacitor Through a Resistor


$$\text{FIGURE V.24}$$

What you have to be sure of in this section and the following section is to get the signs right. For example, if the charge held in the capacitor at some time is $$Q$$, then the symbol $$\dot Q,\text{ or }dQ/dt$$ means the rate of increase of $$Q$$ with respect to time. If the capacitor is discharging, $$\dot Q$$ is negative. Expressed otherwise, the symbol to be used for the rate at which a capacitor is losing charge is $$-\dot Q$$.

In Figure $$V.$$24 a capacitor is discharging through a resistor, and the current as drawn is given by $$I=-\dot Q$$. The potential difference across the plates of the capacitor is $$Q/C$$, and the potential difference across the resistor is $$IR=-\dot QR$$.

Thus:

$\frac{Q}{C}-IR=\frac{Q}{C}+\dot QR=0\label{5.18.1}$

On separating the variables ($$Q \text{ and }t$$) and integrating we obtain

$\int_{Q_0}^Q \frac{dQ}{Q}=-\frac{1}{RC}\int_0^t dt,\label{5.18.2}$

where $$Q_0$$ is the charge in the capacitor at $$t = 0$$.

Hence

$Q=Q_0 e^{-t/(RC)}.\label{5.18.3}$

Here $$RC$$ is the time constant. (Verify that it has the dimensions of time.) It is the time for the charge to be reduced to $$1/e = 36.8$$% of the initial charge. The half life of the charge is $$RC \ln 2 = 0.6931RC$$.

This page titled 5.18: Discharging a Capacitor Through a Resistor is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.