Capacitor discharge
Capacitors - OCR A-Level Physics
Worked Example
A 470 \mu F capacitor is charged to 12 V and then discharged through a 10 k\Omega resistor. Calculate: (a) the 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., (b) the 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. after 3.0 s, (c) the time for the 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 2.0 V.
Show Solution
1
$(a) \tau = RC = 10 \times 10^3 \times 470 \times 10^{-6} = 4.7 \text{ s}$
2
$(b) V = V_0 e^{-t/RC} = 12 \times e^{-3.0/4.7} = 12 \times e^{-0.638} = 12 \times 0.528 = 6.3 \text{ V}$
3
(c) Rearrange
$t = -RC \ln\left(\frac{V}{V_0}\right) = -4.7 \times \ln\left(\frac{2.0}{12}\right) = -4.7 \times (-1.792) = 8.4 \text{ s}$
Answer
(a) \tau = 4.7 s, (b) V = 6.3 V, (c) $t = 8.4 s$