# 10.5 Rc circuits

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By the end of the section, you will be able to:
• Describe the charging process of a capacitor
• Describe the discharging process of a capacitor
• List some applications of RC circuits

When you use a flash camera, it takes a few seconds to charge the capacitor that powers the flash. The light flash discharges the capacitor in a tiny fraction of a second. Why does charging take longer than discharging? This question and several other phenomena that involve charging and discharging capacitors are discussed in this module.

## Circuits with resistance and capacitance

An RC circuit    is a circuit containing resistance and capacitance. As presented in Capacitance , the capacitor is an electrical component that stores electric charge, storing energy in an electric field.

[link] (a) shows a simple RC circuit that employs a dc (direct current) voltage source $\epsilon$ , a resistor R , a capacitor C , and a two-position switch. The circuit allows the capacitor to be charged or discharged, depending on the position of the switch. When the switch is moved to position A , the capacitor charges, resulting in the circuit in part (b). When the switch is moved to position B , the capacitor discharges through the resistor.

## Charging a capacitor

We can use Kirchhoff’s loop rule to understand the charging of the capacitor. This results in the equation $\epsilon -{V}_{R}-{V}_{c}=0.$ This equation can be used to model the charge as a function of time as the capacitor charges. Capacitance is defined as $C=q\text{/}V,$ so the voltage across the capacitor is ${V}_{C}=\frac{q}{C}$ . Using Ohm’s law, the potential drop across the resistor is ${V}_{R}=IR$ , and the current is defined as $I=dq\text{/}dt.$

$\begin{array}{c}\epsilon -{V}_{R}-{V}_{c}=0,\hfill \\ \epsilon -IR-\frac{q}{C}=0,\hfill \\ \epsilon -R\frac{dq}{dt}-\frac{q}{C}=0.\hfill \end{array}$

This differential equation can be integrated to find an equation for the charge on the capacitor as a function of time.

$\begin{array}{}\\ \\ \epsilon -R\frac{dq}{dt}-\frac{q}{C}=0,\hfill \\ \frac{dq}{dt}=\frac{\epsilon C-q}{RC},\hfill \\ \underset{0}{\overset{q}{\int }}\frac{dq}{\epsilon C-q}=\frac{1}{RC}\underset{0}{\overset{t}{\int }}dt.\hfill \end{array}$

Let $u=\epsilon C-q$ , then $du=\text{−}dq.$ The result is

$\begin{array}{}\\ \\ \\ \\ \\ -\underset{0}{\overset{q}{\int }}\frac{du}{u}=\frac{1}{RC}\underset{0}{\overset{t}{\int }}dt,\hfill \\ \text{ln}\left(\frac{\epsilon C-q}{\epsilon C}\right)=-\frac{1}{RC}t,\hfill \\ \frac{\epsilon C-q}{\epsilon C}={e}^{\text{−}t}{RC}}.\hfill \end{array}$

Simplifying results in an equation for the charge on the charging capacitor as a function of time:

$q\left(t\right)=C\epsilon \left(1-{e}^{-\frac{t}{RC}}\right)=Q\left(1-{e}^{-\frac{t}{\tau }}\right).$

A graph of the charge on the capacitor versus time is shown in [link] (a). First note that as time approaches infinity, the exponential goes to zero, so the charge approaches the maximum charge $Q=C\epsilon$ and has units of coulombs. The units of RC are seconds, units of time. This quantity is known as the time constant :

$\tau =RC.$

At time $t=\tau =RC$ , the charge is equal to $1-{e}^{-1}=1-0.368=0.632$ of the maximum charge $Q=C\epsilon$ . Notice that the time rate change of the charge is the slope at a point of the charge versus time plot. The slope of the graph is large at time $t=0.0\phantom{\rule{0.2em}{0ex}}\text{s}$ and approaches zero as time increases.

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nothing
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