My thanks to Peter Finch from St Joseph's College, Gregory Terrace, for these (pdf) solutions.
The rest are by Glenn Rossiter:
The data plot as shown below(Using an Excel chart)
(a) circuit element X is a semiconductor diode.
(b) switch on voltage is 0.4 volts.
(c) To calculate resistance, an extrapolation is needed to find the estimate of current I at 1.0 volt ie I = 80 mA. Hence R = V/I = 12.5W.
(d) Under reverse bias of only 0.8V the current is zero.
Given R = 2.0kW, and C = 10mF then
(a) time constant t = R.C = 2 x 10-2 s when switch is closed.
(b) In series capacitance total 1/Ctot = 1/C1 + 1/C2, calculates at Ctot = 5.0 mF. which would reduce time constant to a value = 1.0 x 10-2 s.
(c) If S2 switch closed but S1 opened, then no current flows.
(d) If S1 and S2 both closed, the battery would be short circuited which could be quite damaging.
When joined in parallel the combination is as follows :-
Charge on each plate C = Q/V yields Q1 = C1.V = 1.0 x 10-9 C. and Q2 = C2.V = 6.0 x 10-9C initially. The total capacitance is 25.0 pF for the combined capacitor.
(b) Total charge of 7.0 x 10-9C with C = 25 pF will result in a voltage across the combined capacitor of 280 volts.
(a) Total charge redistributed according to the ratio of capacitance :-
C1 = (5/25) of total charge = 1.4 x 10-9C. and C2 = (20/25) of total charge = 5.6 x 10-9C.
(c)Hence in each case charge transferred is 0.4 x 10-9C. Gain on C1 and Loss on C2.
Using Energy W = q.DV
For C1 = 0.4 x 10-9C . 80 V = +3.2 x 10-8 J.
For C2 = 0.4 x 10-9C . 20 V = -0.8 x 10-8 J.
Thus energy dissipated during charge movement = 2.4 x 10-8 J.