A 470 μF capacitor is charged to 12 V and then discharged through a 22 kΩ resistor.
(a) Calculate the initial charge on the capacitor. [1]
(b) Calculate the initial energy stored by the capacitor. [1]
(c) Calculate the time constant of the circuit. [1]
(d) Calculate the pd across the capacitor 15 s after the discharge starts. [2]
(e) Calculate the time taken for the pd to fall to 3.0 V. [1]
- (a) $Q = CV = 470\times10^{-6}\times12$ = 5.6 × 10$^{-3}$ C(1)
- (b) $E = \frac{1}{2}CV^2$ = 0.034 J (3.38 × 10$^{-2}$)(1)
- (c) $RC = 22\times10^3 \times 470\times10^{-6}$ = 10.3 s (allow 10 s)(1)
- (d) $V = V_0 e^{-t/RC} = 12e^{-15/10.34}$(1)
- (d) $V$ = 2.8 V (allow 2.7 V from $RC$ = 10 s; ecf from (c))(1)
- (e) $t = RC\ln\frac{12}{3.0} = 10.34 \ln 4$ = 14 s (14.3) (accept two half-lives with $T_{1/2} = 0.69RC$)(1)
- Answer: (a) 5.6 × 10$^{-3}$ C (b) 0.034 J (c) 10.3 s (d) 2.8 V (e) 14 s
- Examiner tip: Check your powers of ten: μF is 10⁻⁶ and kΩ is 10³ — the time constant should come out in seconds.