Chapter 23 - Worked Solutions

to selected extension questions from the OUP text Senior Physics - Concepts in Context by Walding, Rapkins and Rossiter

My thanks to Peter Finch from St Joseph's College, Gregory Terrace, for these (pdf) solutions.


The rest are by Glenn Rossiter:

Q27 :-

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.

Q29 :-
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.

Q31 :-
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.

New Edition Q32 :- The plot of the Data provided using Excel appears as follows. Also using the Excel spreadsheet to plot a trendline and deduce a formula results in the equation y = -0.63 X + 602.
This means that the conversion would be LDR-Resistance in Megohms = (-0.63 x Light Intensity candelas) + 602.

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