Wize University Physics Textbook (Master) > DC Circuits

RC Circuits: Charging and Discharging of Capacitors

Charging Capacitor

A simple RC circuit contains a resistor and a capacitor.

Assume there is no initial charge on the capacitor $(q = 0)$ . At $t = 0$ the switch S is at position "a". The capacitor acts like an ideal wire initially, and charge on it starts to increase.

According to Kirchhoff’s voltage rule, we have $0=\mathcal{E}-iR-\dfrac{q}{C}$ .

Therefore the current is $i=\dfrac{1}{R}\left(\mathcal{E}-\dfrac{q}{C}\right)=\dfrac{dq}{dt}$ .

The charge is the solution to the first-order differential equation above, and is given by:

\boxed{ \ q(t)=C\mathcal{E}\ (1-e^{-\frac{t}{RC}})=Q\ (1-e^{-\frac{t}{RC}}) \ }

$q(t)$ is the charge on the capacitor as a function of time $t$
$C$ is the capacitance (of the capacitor)
$\mathcal{E}$ is the EMF supplied by the battery
$R$ is the resistance (of the resistor)
$Q$ is the final (maximum) charge on the capacitor, given by $Q=C\mathcal{E}$

The time constant $\bcth\tau$ is a measure of how fast the charging or discharging happens. It is the time at which the current in the circuit has changed (dropped) by a factor of $1/e$ .

For RC circuits the time constant is:

\boxed{ \ \tau=RC \ }

Differentiating charge with respect to time gives the current as:

\boxed{ \ i(t)=\dfrac{\mathcal{E}}{R}\ (e^{-\frac{t}{RC}})=I\ e^{-\frac{t}{RC}} \ }

$i(t)$ is the current as a function of time
$I$ is the initial (maximum) current in the circuit, given by $I=\dfrac{\mathcal{E}}{R}$

We can use $V=\dfrac{q}{C}$ to get the potential difference across the charging capacitor:

\boxed{ \ V_C(t)=\mathcal{E}\ (1-e^{-\frac{t}{RC}}) \ }

$V_C(t)$ is the voltage across the capacitor as a function of time

Discharging Capacitor

Now assume that the capacitor is charged to a value of $q=Q_0$ .

The switch S is turned to position "b" at $t=0$ . The charge on the capacitor starts to decrease.

According to Kirchhoff’s voltage rule we have $0=-iR+\dfrac{q}{C}$ .

Therefore the current is $i(t)=-\dfrac{Q}{RC}=\dfrac{dq}{dt}$ (the negative sign means the current is decreasing).

NOTE: when the capacitor is the source of the energy, it appears as $+q/C$ in the voltage rule when you move in the direction of the current.

Image illustrating Discharging Capacitor

The charge is the solution to the first-order differential equation above, and is given by:

\boxed{ \ q(t)=Q_0\ e^{-\frac{t}{RC}} \ }

$q(t)$ is the charge on the capacitor as a function of time $t$
$C$ is the capacitance (of the capacitor)
$R$ is the resistance (of the resistor)
$Q_0$ is the initial (maximum) charge on the capacitor, given by $Q_0=CV_0$ (see below)

The current is also decreasing as:

\boxed{ \ i(t)=I_0\ e^{-\frac{t}{RC}} \ }

$i(t)$ is the current as a function of time
$I_0$ is the initial (maximum) current in the circuit, given by $I_0=\dfrac{V_0}{R}$

The potential difference across the discharging capacitor is:

\boxed{ \ V_C(t)=\dfrac{Q_0}{C}\ e^{-\frac{t}{RC}}=V_0\ e^{-\frac{t}{RC}} \ }

$V_C(t)$ is the voltage across the capacitor as a function of time
$V_0$ is the initial (maximum) voltage across the capacitor

Example: Maximum Charge

The capacitor in the circuit shown below is initially uncharged and the switch S is closed at time $t=0$ . If $\mathcal{E}=50$ V, $R=1$ MΩ, and $C=100$ µF, after how many seconds will the energy stored in the capacitor be equal to $60$ mJ? What is the maximum charge sitting on this capacitor?

Image illustrating Example: Charging a Capacitor

The energy stored by the capacitor is given by:

$U=\dfrac{q^2}{2C}$

Let's put the charge $q=C\mathcal{E}(1-e^{-\frac{t}{RC}})$ into this:

$=\dfrac{[C\mathcal{E}(1-e^{-\frac{t}{RC}})]^2}{2C}$

$=\dfrac{1}{2}C\mathcal{E}^2(1-e^{-\frac{t}{RC}})^2$

Rearrange for the time:

$1-e^{-\frac{t}{RC}}=\sqrt\dfrac{2U}{C\mathcal{E}^2}$

$e^{-\frac{t}{RC}}=1-\sqrt\dfrac{2U}{C\mathcal{E}^2}$

Put the numbers in:

$=1-\sqrt\dfrac{2(60\times10^{-3})}{(100\times10^{-6})(50)^2}$

$=0.307$

Change to log form:

$-\dfrac{t}{RC}=\ln \ 0.307$

$t=-RC\ \ln0.307$

$=-1\times10^6\times100\times10^{-6} \ln \ 0.307$

$=118.03$ (s)

If we wait for a very long time, the charge will increase up to the maximum charge given by:

$Q_{max}=C\mathcal{E}$

$=(100\times10^{-6})(50)=0.005$

$=0.005$ (C)

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Practice: Charge & Discharge

In the circuit depicted below, the switch S has been open for a very long time.

a) The switch is closed at time $𝑡 = 0$ . At the instant the switch is closed, find the power supplied by the battery, the power dissipated by resistors $R_1$ and $R_2$ , and the charge $Q$ stored by the capacitor.

b) Repeat part a) after the switch has remained closed for a long time, i.e. $𝑡 → ∞$ .

c) Derive the expression for the power that's supplied to the capacitor at any time during the charging process. Evaluate the derived expression at $𝑡 = 0$ and $𝑡 → ∞$ .

d) The switch is now opened again after the capacitor has been fully charged. After how many seconds will the energy stored in the capacitor drop to $1/e$ of its maximum value?

Illustration for practice question: Charging a Capacitor

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RC Circuits: Charging and Discharging of Capacitors

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Charging Capacitor

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Discharging Capacitor

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Example: Maximum Charge

Practice: Charge & Discharge

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