Capacitors - the NEW resistors? More on Page 7!

40. minimum C = 1.36 x 10-9F connected in series, maximum C = 1.95 x 10-8F connected in parallel.

41. 300 pF connected in parallel.

42. C1 + (C2C3/(C2+C3)), Q on C1 = 562.5 microcoulombs, Q on C2 = Q on C3 = 375 microcoulombs

44
(a) For 0.4 microfarad capacitor, Q = 2 microcoulombs, V = 5 V
For 0.5 microfarad capacitor, Q = 2 microcoulombs, V = 4 V
(b) Both capacitors have 9V across them, For 0.4 microfarad capacitor, Q = 3.6 microcoulombs, 0.5 microfarad capacitor, Q = 4.5 microcoulombs

72. (a) When the capacitors are connected in parallel, we find the equivalent capacitance from
Cparallel = C1 + C2 = 0.40 µF + 0.60 µF = 1.00 µF.
The stored energy is
Uparallel = 0.5*Cparallel*V^2 = (1.00 x 10–6 F)(45 V)^2 = 1.0 x 10^–3 J.
(b) When the capacitors are connected in series, we find the equivalent capacitance from
1/Cseries = (1/C1) + (1/C2) = [1/(0.40 µF)] + [1/(0.60 µF)], which gives Cseries = 0.24 µF.
The stored energy is
Useries = 0.5*CseriesV^2 = 0.5*(0.24 x 10–6 F)(45 V)^2 = 2.4 ´ 10^–4 J.
(c) We find the charges from
Q = CeqV;
Qparallel = CparallelV = (1.00 µF)(45 V) = 45 µC.
Qseries = CseriesV = (0.24 µF)(45 V) = 11 µC.