3.7.4.4
Charging equations use (1 - e^(-t/RC)) for Q and V
Capacitance & Charge/Discharge — AQA A-Level Physics
$$Q = Q_0 \left(1 - e^{-\frac{t}{RC}}\right)$$
- $Q$: chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). at time t (C)
- $Q_0$: maximum chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). when fully charged (C)
- $t$: time (s)
- $RC$: 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 charge. Measured in amperes (A).. 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. (s)
$$V = V_0 \left(1 - e^{-\frac{t}{RC}}\right)$$
- $V$: p.d. at time t (V)
- $V_0$: maximum p.d. when fully charged (V)
- Q0 and V0 are now the final (maximum) values, not the initial values.
- The currentThe rate of flow of charge. Measured in amperes (A). during charging still follows the decay equation: $I = I0 e^(-t/RC). Current always decreases exponentially$.
- At t = 0: $Q = 0$ and $V = 0 (capacitor$ is uncharged).
- As t approaches infinity: Q approaches Q0 and V approaches V0 (fully charged).
Worked Example
A capacitor is to be charged to a maximum p.d. of 12 V. The 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. is 0.5 s. Calculate how long it takes to reach 10 V.
Show Solution
1
List known values
- $V_0 = 12$ V, $V = 10$ V, $RC = 0.5$ s
2
Rearrange V = V0(1 - e^(-t/RC)) for t
$$\frac{V}{V_0} = 1 - e^{-t/RC}$$
$$e^{-t/RC} = 1 - \frac{V}{V_0}$$
$$t = -RC \ln\left(1 - \frac{V}{V_0}\right)$$3
Substitute and calculate
$$t = -0.5 \times \ln\left(1 - \frac{10}{12}\right) = -0.5 \times \ln(0.1\overline{6})$$
$$= -0.5 \times (-1.792) = 0.90 \text{ s (2 s.f.)}$$Answer
$t = 0.90$ s