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Wize University Physics Textbook (Master) > DC Circuits

RC Circuits: Charging and Discharging of Capacitors

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Charging Capacitor


A simple RC circuit contains a resistor and a capacitor.



Assume there is no initial charge on the capacitor(q=0)(q = 0). At t=0t = 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=E−iR−qC0=\mathcal{E}-iR-\dfrac{q}{C} .

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


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The charge is the solution to the first-order differential equation above, and is given by:
 q(t)=CE (1−e−tRC)=Q (1−e−tRC) \boxed{ \ q(t)=C\mathcal{E}\ (1-e^{-\frac{t}{RC}})=Q\ (1-e^{-\frac{t}{RC}}) \ }

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



Wize Concept
As time increases t→∞t\rightarrow\infty, we have e−tRC→0e^{-\frac{t}{RC}}\to0 , and the charge qq increases from zero to the maximum value QQ.





The time constant τ\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/e1/e.

For RC circuits the time constant is:

 τ=RC \boxed{ \ \tau=RC \ }


Exam Tip
The larger the time constant, the more time it takes to charge or discharge, and vice-versa.


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Differentiating charge with respect to time gives the current as:
 i(t)=ER (e−tRC)=I e−tRC \boxed{ \ i(t)=\dfrac{\mathcal{E}}{R}\ (e^{-\frac{t}{RC}})=I\ e^{-\frac{t}{RC}} \ }
  • i(t)i(t) is the current as a function of time
  • II is the initial (maximum) current in the circuit, given by I=ERI=\dfrac{\mathcal{E}}{R}




Wize Concept
As time increases t→∞t\rightarrow\infty, we have e−tRC→0e^{-\frac{t}{RC}}\to0 , and the current ii decreases from the maximum value II to zero.
We can use V=qCV=\dfrac{q}{C} to get the potential difference across the charging capacitor:

 VC(t)=E (1−e−tRC) \boxed{ \ V_C(t)=\mathcal{E}\ (1-e^{-\frac{t}{RC}}) \ }
  • VC(t)V_C(t) is the voltage across the capacitor as a function of time


Wize Concept
As time increases t→∞t\rightarrow\infty, we have e−tRC→0e^{-\frac{t}{RC}}\to0 , and the potential difference VCV_C increases from zero to the maximum value of the E\mathcal{E} supplied by the battery.



Exam Tip
  • An empty capacitor behaves like a simple wire, because the potential difference across it is zero.
  • A full capacitor behaves like an open switch, because the current passing through it is zero.


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Discharging Capacitor


Now assume that the capacitor is charged to a value of q=Q0q=Q_0.

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

According to Kirchhoff’s voltage rule we have 0=−iR+qC0=-iR+\dfrac{q}{C} .

Therefore the current is i(t)=−QRC=dqdti(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+q/C in the voltage rule when you move in the direction of the current.




Wize Concept
  • When a capacitor is discharging (when there is no battery linked to it), it almost acts like a battery. It will start giving off its energy (which is limited, unlike a battery).
  • When the capacitor is linked to a battery, it will store energy. When it's fully charged it will behave like an open switch.


Exam Tip
The direction of the current for discharging is opposite to the direction for charging.

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The charge is the solution to the first-order differential equation above, and is given by:
 q(t)=Q0 e−tRC \boxed{ \ q(t)=Q_0\ e^{-\frac{t}{RC}} \ }

  • q(t)q(t) is the charge on the capacitor as a function of time tt
  • CC is the capacitance (of the capacitor)
  • RR is the resistance (of the resistor)
  • Q0Q_0 is the initial (maximum) charge on the capacitor, given by Q0=CV0Q_0=CV_0 (see below)


The current is also decreasing as:

 i(t)=I0 e−tRC \boxed{ \ i(t)=I_0\ e^{-\frac{t}{RC}} \ }
  • i(t)i(t) is the current as a function of time
  • I0I_0 is the initial (maximum) current in the circuit, given by I0=V0RI_0=\dfrac{V_0}{R}


The potential difference across the discharging capacitor is:

 VC(t)=Q0C e−tRC=V0 e−tRC \boxed{ \ V_C(t)=\dfrac{Q_0}{C}\ e^{-\frac{t}{RC}}=V_0\ e^{-\frac{t}{RC}} \ }

  • VC(t)V_C(t) is the voltage across the capacitor as a function of time
  • V0V_0 is the initial (maximum) voltage across the capacitor








Wize Concept
As time increases t→∞t\to\infin , all the collected charge will be used up, the current and voltage will approach zero, and the capacitor will became like an open switch again.

Exam Tip
For problems involving RC circuits it's important to remember what happens at two key times:
  • immediately after the switch is closed
  • after a very long time

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Example: Maximum Charge


The capacitor in the circuit shown below is initially uncharged and the switch S is closed at time t=0t=0. If E=50\mathcal{E}=50V, R=1R=1 MΩ, and C=100C=100µF, after how many seconds will the energy stored in the capacitor be equal to 6060 mJ? What is the maximum charge sitting on this capacitor?


The energy stored by the capacitor is given by:

U=q22CU=\dfrac{q^2}{2C}

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

=[CE(1−e−tRC)]22C=\dfrac{[C\mathcal{E}(1-e^{-\frac{t}{RC}})]^2}{2C}

=12CE2(1−e−tRC)2=\dfrac{1}{2}C\mathcal{E}^2(1-e^{-\frac{t}{RC}})^2

Rearrange for the time:

1−e−tRC=2UCE21-e^{-\frac{t}{RC}}=\sqrt\dfrac{2U}{C\mathcal{E}^2}


e−tRC=1−2UCE2e^{-\frac{t}{RC}}=1-\sqrt\dfrac{2U}{C\mathcal{E}^2}

Put the numbers in:

=1−2(60×10−3)(100×10−6)(50)2=1-\sqrt\dfrac{2(60\times10^{-3})}{(100\times10^{-6})(50)^2}

=0.307=0.307


Change to log form:

−tRC=ln⁡ 0.307-\dfrac{t}{RC}=\ln \ 0.307

t=−RC ln⁡0.307t=-RC\ \ln0.307

=−1×106×100×10−6ln⁡ 0.307=-1\times10^6\times100\times10^{-6} \ln \ 0.307

=118.03=118.03 (s)


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

Qmax=CEQ_{max}=C\mathcal{E}

=(100×10−6)(50)=0.005=(100\times10^{-6})(50)=0.005

=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=0𝑡 = 0. At the instant the switch is closed, find the power supplied by the battery, the power dissipated by resistors R1R_1 and R2R_2, and the charge QQ stored by the capacitor.
b) Repeat part a) after the switch has remained closed for a long time, i.e. 𝑡→∞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 𝑡 = 0 and 𝑡→∞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/e1/e of its maximum value?