Basic Electronics, Chapter 4: Challenge 1 solution (capacitor sums)

Formulas: Q = C V    E = 1/2 C V^2    parallel: C1 + C2    series: C1 C2 / (C1 + C2)
          series voltage share: V1 = Vtotal x C2 / (C1 + C2)
          tau = R C    discharge: v = V0 e^(-t/tau)    Xc = 1 / (2 pi f C)

(a) 47 uF charged to 12 V
  Charge:  Q = C V = 47e-6 x 12 = 564e-6 C = 564 uC
  Energy:  E = 1/2 C V^2 = 0.5 x 47e-6 x 12^2 = 0.5 x 47e-6 x 144
             = 3.384e-3 J = 3.38 mJ
  Check:   E = Q^2 / (2C) = (564e-6)^2 / (94e-6) = 3.18e-7 / 94e-6 = 3.38 mJ  (same)

(b) 22 uF and 47 uF
  (i) Parallel:  C = 22 + 47 = 69 uF
  (ii) Series:   C = 22 x 47 / (22 + 47) = 1034 / 69 = 14.99 uF  (less than 22 uF, the smaller)

  Series pair across 24 V (the same charge Q on both):
    V across 22 uF = 24 x 47 / 69 = 1128 / 69 = 16.35 V
    V across 47 uF = 24 x 22 / 69 =  528 / 69 =  7.65 V
    Check: 16.35 + 7.65 = 24.00 V
  The 22 uF part gets 16.35 V, which is ABOVE its 16 V rating: not safe (and in
  practice a part at its limit is not good either). The smaller capacitor takes
  the larger share. Fix: use one capacitor rated for 24 V or more (a 35 V or 50 V
  part), or add balancing resistors.

(c) 220 uF at 9 V discharging through 1 k ohm
  Time constant:  tau = R C = 1000 x 220e-6 = 0.22 s
  Starting current: I = V / R = 9 / 1000 = 9 mA   (decays with the same tau)
  Energy stored:  E = 0.5 x 220e-6 x 9^2 = 0.5 x 220e-6 x 81 = 8.91e-3 J = 8.91 mJ
  Time to fall to 1 V:  1 = 9 e^(-t/tau)  ->  t = tau x ln(9) = 0.22 x 2.1972 = 0.483 s
  Reactance at 50 Hz:  Xc = 1 / (2 pi x 50 x 220e-6) = 1 / 0.06912 = 14.5 ohm

Checklist
  - Units: uF to farads (x 1e-6) before multiplying.
  - Series capacitors: reciprocal formula, and the SMALLER capacitor gets the larger voltage.
  - Energy depends on V squared.