Discharge equations: $Q, I, and V all decay as e^(-t/RC)$

Capacitance & Charge/Discharge — AQA A-Level Physics

Q = Q_0 e^{-\frac{t}{RC}}

$Q$: chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). remaining on the plates (C)
$Q_0$: initial chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). (C)
$t$: time (s)
$R$: resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to currentThe rate of flow of charge. Measured in amperes (A).. Measured in ohms (Ω). (ohm)
$C$: capacitanceThe charge stored per unit potential difference across a capacitor. Measured in farads (F). (F)

I = I_0 e^{-\frac{t}{RC}}

$I$: currentThe rate of flow of charge. Measured in amperes (A). at time t (A)
$I_0$: initial current (A)

V = V_0 e^{-\frac{t}{RC}}

$V$: p.d. at time t (V)
$V_0$: initial p.d. (V)

All three quantities follow the same exponential decay shape.
Smaller time constantThe product of resistanceThe opposition to current flow. The ratio of potential difference to current. Measured in ohms (Ω). and capacitanceThe charge stored per unit potential difference across a capacitor. Measured in farads (F). in an RC circuit. The time taken for the charge (or voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference.) to fall to 1/e (about 37%) of its initial value. RC means faster decay.
Larger I0 means the capacitor takes longer to fully discharge.
No matter how much charge is initially on the plates, the time to halve is always the same (exponential decay property).

Worked Example

A 620 uF capacitor with initial current 0.6 A discharges through a 450 ohm resistor. Calculate the time for the current to fall to 0.4 A.

Show Solution

List known values

$I_0 = 0.6$ A, $I = 0.4$ A
$R = 450 \; \Omega$, $C = 620 \times 10^{-6}$ F

Rearrange I = I0 e^(-t/RC) for t

$$\frac{I}{I_0} = e^{-t/RC}$$

Take ln of both sides:

$$t = -RC \ln\left(\frac{I}{I_0}\right)$$

Substitute and calculate

$$t = -(450)(620 \times 10^{-6}) \times \ln\left(\frac{0.4}{0.6}\right)$$ $$= -0.279 \times \ln(0.667) = -0.279 \times (-0.405)$$ $$= 0.11 \text{ s (2 s.f.)}$$

Answer

$t = 0.11$ s

Examiner Tips and Tricks

The Q equation is on the data sheet.
You must remember that the same form applies to I and V.
Practise rearranging with ln.