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