# 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.

Solutions-Ch23-Finch.pdf

**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/C_{tot} = 1/C_{1} + 1/C_{2}, calculates at C_{tot} = 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 = C_{1}.V = 1.0 x 10^{-9} C. and Q2 = C_{2}.V = 6.0 x 10^{-9}C initially. The total capacitance is 25.0 pF for the combined capacitor.

(b) Total charge of 7.0 x 10^{-9}C 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^{-9}C. and C2 = (20/25) of total charge = 5.6 x 10^{-9}C.

(c)Hence in each case charge transferred is 0.4 x 10^{-9}C. Gain on C1 and Loss on C2.

Using Energy W = q.DV

For C1 = 0.4 x 10^{-9}C . 80 V = +3.2 x 10^{-8} J.

For C2 = 0.4 x 10^{-9}C . 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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