Capacitor Discharge | Free Notes & Practice

Discharging a Capacitor

When a charged capacitor with capacitance C is connected to a resistor with resistance R, then the charge stored on the capacitor decreases exponentially.

Discharge graph

$Q=Q_0e^{-\frac{t}{RC}}$
Where $Q_0$ is the initial charge on the capacitor.

Time to halve

The time for the charge to reach half of the initial charge, $T_{1/2}$ , can be found:
$Q=\frac{Q_0}{2}=Q_0e^{-\frac{T_{1/2}}{RC}}$
$0.5=e^{-\frac{T_{1/2}}{RC}}$
$\ln (0.5)=\ln (e^{-\frac{T_{1/2}}{RC}})$
$-\ln(2)=-\frac{T_{1/2}}{RC}$
$T_{1/2}=RC\ln 2$
Because $\ln(2) = 0.69,\space T_{1/2}=0.69RC$ .

Time constant

The product RC is known as the time constant.
It is a property of exponential decay graphs that the curve will decrease by a constant fraction with each time constant.

Potential Difference and Current in a Discharging Capacitor

The potential difference and the current in a discharging capacitor have similar forms.

Potential difference

The potential difference across a capacitor with time constant RC, initial potential difference V₀, and having been discharged for a time t is:
$V=V_0e^{-\frac{t}{RC}}$

Current

The current passing through a capacitor with time constant RC, initial current I₀, and having been discharged for a time t is:
$I=I_0e^{-\frac{t}{RC}}$
R is the resistance across the capacitor
$I_0=\frac{V_0}{R}$

Test yourself on Capacitor Discharge

What is $Q_0$ ?
When a capacitor is discharged through a constant resistance, what relationship can be seen between charge stored and time?
Which of these is the correct formula for the potential difference across a discharging capacitor?
Which of these is the correct formula for the current through a discharging capacitor?
What effect does increasing the capacitance have on the rate at which the potential difference in a discharging capacitor falls to zero?

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