CIE AS & A Level Physics 9702: Topic 19: Capacitance- Unit : 19.3 Discharging a capacitor Study Notes

19.2 CHARGING AND DISCHARGING

19.2.1 CAPACITOR DISCHARGE GRAPHS

  • Capacitor Discharge Graphs
  •  So far, only capacitors charged by a battery have been considered
    • This is when the electrons flow from the positive to negative plate
    • At the start, when the capacitor is charging, the current is large and then gradually falls to zero
  • Capacitors are discharged through a resistor
    •  The electrons now flow back from the negative plate to the positive plate until there are equal numbers on each plate
  •  At the start of discharge, the current is large (but in the opposite direction to when it was charging) and gradually falls to zero
  • As a capacitor discharges, the current, p.d and charge all decrease exponentially
  •  The means the rate at which the current, p.d or charge decreases is proportional to the amount of current, p.d or charge it has left
  • The graphs of the variation with time of current, p.d and charge are all identical and represent an exponential decay

The key features of the discharge graphs are:

    • The shape of the current, p.d. and charge against time graphs are identical
    • Each graph shows exponential decay curves with decreasing gradient
    • The initial value starts on the $y$ axis and decreases exponentially
  • The rate at which a capacitor discharges depends on the resistance of the circuit
    •  If the resistance is high, the current will decrease and charge will flow from the capacitor plates more slowly, meaning the capacitor will take longer to discharge
    •  If the resistance is low, the current will increase and charge will flow from the capacitor plates quickly, meaning the capacitor will discharge faster

19.2.2 CAPACITOR DISCHARGE EQUATIONS

The Time Constant

  •  The time constant of a capacitor discharging through a resistor is a measure of how long it takes for the capacitor to discharge
  • The definition of the time constant is:
    The time taken for the charge of a capacitor to decrease to 0.37 of its original value
  • This is represented by the greek letter tau ( $\tau$ ) and measured in units of seconds (s)
  •  The time constant gives an easy way to compare the rate of change of similar quantities eg. charge, current and p.d.
  • The time constant is defined by the equation:

$
\mathbf{T}=\mathbf{R C}
$

  •  Where:
    •  $\tau=$ time constant (s)
    • $\mathrm{R}=$ resistance of the resistor $(\Omega)$
    •  $\mathrm{C}=$ capacitance of the capacitor (F)

Worked example: Time constant

A capacitor of $7 \mathrm{nF}$ is discharged through a resistor of resistance R. The time constant of the discharge is $5.6 \times 10^{-3} \mathrm{~s}$. Calculate the value of $R$.

Answer/Explanation

Step 1: $\quad$ Write out the known quantities

$
\begin{aligned}
& \text { Capacitance, } \mathrm{C}=7 \mathrm{nF}=7 \times 10^{-9} \mathrm{~F} \\
& \text { Time constant, } \mathrm{\tau}=5.6 \times 10^{-3} \mathrm{~s}
\end{aligned}
$

Step 2: $\quad$ Write down the time constant equation
$
\mathbf{T}=\mathbf{R C}
$

Step 3: Rearrange for resistance $R$

$
\mathrm{R}=\frac{\tau}{C}
$
Step 4: Substitute in values and calculate

$
R=\frac{5.6 \times 10^{-3}}{7 \times 10^{-9}}=8 \times 10^5 \Omega=800 \mathbf{k} \Omega
$

Using the Capacitor Discharge Equation

  •  The time constant is used in the exponential decay equations for the current, charge or potential difference (p.d) for a capacitor discharging through a resistor

These can be used to determine the amount of current, charge or p.d left after a certain amount of time when a capacitor is discharging

  •  The exponential decay of current on a discharging capacitor is defined by the equation:

$
\mathrm{I}=\mathrm{I}_0 e^{-\frac{t}{R C}}
$

– Where:

  •  $\mathrm{I}=$ current $(\mathrm{A})$
  • $\mathrm{I}_0=$ initial current before discharge (A)
  • $\mathrm{e}=$ the exponential function
  • $0 \mathrm{t}=$ time $(\mathrm{s})$
  • $\mathrm{RC}=$ resistance $(\Omega) \times$ capacitance $(F)=$ the time constant $\tau$ (s)
  •  This equation shows that the faster the time constant $\tau$, the quicker the exponential decay of the current when discharging
  •  Also, how big the initial current is affects the rate of discharge
    • If $\mathrm{I}_0$ is large, the capacitor will take longer to discharge
  •  Note: during capacitor discharge, $I_0$ is always larger than I, this is because the current I will always be decreasing
  •  The current at any time is directly proportional to the p.d across the capacitor and the charge across the parallel plates
  • Therefore, this equation also describes the change in p.d and charge on the capacitor:

$
Q=Q_0 e^{-\frac{t}{R C}}
$$

  •  Where:
    •  $\mathrm{Q}=$ charge on the capacitor plates (C)
    •  $\mathrm{Q}_0=$ initial charge on the capacitor plates (C)

$
\mathrm{V}=\mathrm{V}_0 e^{-\frac{t}{R C}}
$

  • Where:
    •  $V=$ p.d across the capacitor (C)
    •  $V_0=$ initial p.d across the capacitor (C)

The Exponential Function $e$

  • The symbol e represents the exponential constant, a number which is approximately equal to $\mathrm{e}=2.718 \ldots$
  •  On a calculator it is shown by the button $\mathrm{e}^{\mathrm{x}}$
  • The inverse function of $\mathrm{e}^{\mathrm{x}}$ is $\ln (\mathrm{y})$, known as the natural logarithmic function
    • This is because, if $e^x=y$, then $x=\ln (y)$
  •  The 0.37 in the definition of the time constant arises as a result of the exponential constant, the true definition is:
    The time taken for the charge of a capacitor to decrease to $\frac{1}{e}$ of its original value
  •  Where $\frac{1}{e}=0.3678 \ldots$

The initial current through a circuit with a capacitor of $620 \mu \mathrm{F}$ is $0.6 \mathrm{~A}$. The capacitor is connected across the terminals of a $450 \Omega$ resistor. Calculate the time taken for the current to fall to $0.4 \mathrm{~A}$.

Answer/Explanation

Step 1: Write out the known quantities
Initial current before discharge, $I_0=0.6 \mathrm{~A}$
Current, $I=0.4$
Resistance, $R=450 \Omega$
Capacitance, $\mathrm{C}=620 \mu \mathrm{F}=620 \times 10^{-6} \mathrm{~F}$

Step 2:
Write down the equation for the exponential decay of current

$
\mathrm{I}=\mathrm{I}_0 e^{-\frac{t}{R C}}
$

Step 3:
Calculate the time constant

$
\begin{gathered}
\tau=R C \\
\tau=450 \times\left(620 \times 10^{-6}\right)=0.279 \mathrm{~s}
\end{gathered}
$

Step 4:
Substitute into the current equation

$
0.4=0.6 \times e^{-\frac{t}{0.279}}
$$

Step 5:
Rearrange for the time $t$

$
\frac{0.4}{0.6}=e^{-\frac{t}{0.279}}
$

The exponential can be removed by taking the natural log of both sides:

$
\begin{gathered}
\ln \left(\frac{0.4}{0.6}\right)=-\frac{t}{0.279} \\
t=-0.279 \times \ln \left(\frac{0.4}{0.6}\right)=0.1131=0.1 \mathrm{~s}
\end{gathered}
$

Exam Tip
Make sure you’re confident in rearranging equations with natural logs (In) and the exponential function (e). To refresh your knowledge of this, have a look at the AS Maths revision notes on Exponentials \& Logarithms

Scroll to Top