3.7.4.4
The time constant tau = RC sets the rate of charge and discharge
Capacitance & Charge/Discharge — AQA A-Level Physics
Key Definition
Time constant (discharging) — The time taken for the charge, current, or voltage of a discharging capacitor to decrease to 37% (1/e) of its original value.
Key Definition
Time constant (charging) — The time taken for the charge or voltage of a charging capacitor to rise to 63% (1 - 1/e) of its maximum value.
$$\tau = RC$$
- $tau$: time constantThe product of 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 chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).. Measured in ohms (Ω). and capacitanceThe chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). 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. (s)
- $R$: resistanceThe opposition to currentThe rate of flow of charge. Measured in amperes (A). flow. The ratio of potential difference to current. Measured in ohms (Ω). of the resistor (ohm)
- $C$: capacitanceThe charge stored per unit potential difference across a capacitor. Measured in farads (F). of the capacitor (F)
- 37% = 1/$e = 0.368. 63% = 1 - 1/e = 0.632$.
- A larger 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. means slower charging/discharging.
- To find tau from a graph: read off the time at which V (or Q or I) has fallen to 0.37 x its initial value (discharging) or risen to 0.63 x its maximum (charging).
Half-lifeThe time taken for half the number of radioactive nuclei in a sample to decay, or for the activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. to halve. of a capacitor
Key Definition
Half-life (capacitor discharge) — The time taken for the charge, current, or voltage to reach half of its initial value.
$$t_{1/2} = 0.69 RC$$
- $t_{1/2}$: half-lifeThe time taken for half the number of radioactive nuclei in a sample to decay, or for the activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. to halve. (s)
- $R$: resistance (ohm)
- $C$: capacitance (F)
- 0.69 = ln 2. The derivation comes from setting $Q = Q0/2$ in the discharge equation and solving for t.
Worked Example
A capacitor of 7 nF is discharged through a resistor. The time constantThe product of resistance and capacitance 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. is 5.6 x 10^-3 s. Calculate the resistance.
Show Solution
1
List known values
- $C = 7 \times 10^{-9}$ F
- $\tau = 5.6 \times 10^{-3}$ s
2
Rearrange tau = RC for R
$$R = \frac{\tau}{C}$$
3
Substitute and calculate
$$R = \frac{5.6 \times 10^{-3}}{7 \times 10^{-9}} = 800 \text{ k}\Omega$$
Answer
$R = 800$ k$\Omega$