# Chapter 22 - Worked Solutions

### to 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-Ch22-Finch.pdf

Solutions-Ch22B-Finch.pdf

Solutions-Ch22C-Finch.pdf

The rest are by Glenn Rossiter:

Q10 :-

Given EMF = 12 V and source internal resistance r = 0.4W, the total circuit resistance
Rtot = 15 + 15 + 0.4 = 30.4W
Circuit current calculated from I = EMF/Rtot = 12/30.4 = 3.9 A.
Voltmeter will read 12 - (3.9 x 0.4) = 10.4 V and the ammeter will read 3.9 A as the circuit current.

Q34

(a) Graph appear linear over this short interval but it can't be as a zero watt kettle would take an infinite time to boil. Must be an inverse non-linear relationship. However if you assume it is linear then the slope = Dt/DW = -0.06 s/W. Hence for every watt increase in power, the boiling time decreases by about 0.06 seconds. Alternatively (see the graph below), the equation for the straight line is shown as Time = -0.0823 x Power + 352.92; or if the graph is assumed to be inverse proportional Excel calculates the equation to be: Time = 255022 x Power-0.949.

(b) Several methods to work this out. (i) If 1890 W takes 195 s then a 1600 W kettle would be 295 x 0.06 = 17.7 s longer (= 213 s or 3 min 33 s). Like wise a 2400 W kettle would take about 110 x 0.06 = 7 seconds quicker than a 2290 W kettle (= 160 s - 7 s = 153 s = 2 min 33 s). (ii) However, using the relationships developed above, for 1600 W kettle, the times work out to be 221 s (linear) or 232 s (inverse); for 2400 W  the times are 155 s and 158 s respectively.

(c)  There is no relationship between efficiency and the power of the kettle. The data table below has the kettles ranked in order of power and the highest power kettle does have the best efficiency but there is no trend.

Q35 :-

Each headlight dissipates 60 W at 12 V = 120 W, while each parking light dissipates 5 W at 12 V = 20 W. Thus total dissipation = 140 W. As the battery is rated at 60 AH and the total current drawn I = P/V = 11.7 A
Battery will last (60/11.7) = 5.1 hours.

Q36 :-
Section CD has R = 1.8W. and VA = VB for zero current. Using proportional division of voltage we can equate R/0.4 = 5/1.6 which yields R = (0.4 x 5)/1.6 = 1.25W.

Q37 :-
(a) Zero voltage difference between the opposite ends of the bridge containing the switch and galvanometer.
(b) R1-R2 and R3-Rx divide supply voltage V into equal parts, hence
(R1 x V)/(R1 + R2) = (R3 x V)/(R2 + R3)
or R1/(R1 + R2) = R3/(R2 + Rx)
or R1.R3 + R1.Rx = R1.R3 + R3.R2 which rearranges to Rx = R3.R2/R1 as required.
(c) If R1 = 710W R2 = 317W R3 = 2.24W, then Rx by substitution is 1.0 kW.

Q38 :-
The equivalent circuit needs to be drawn with parallel(//) and series (+) sections. When this is done the circuit reduces to a simpler form as :-
(a) The total circuit resistance Rtot = 40 + 40 + (60//(5 + 15//(10 + 20)))W.
Rtot = 40 + 40 + (60//5 + 10) = 80 + 12 = 92W.
(b) Voltage drop across V(10+20) = 1.0A x 30W = 30 V
hence current through 15W resistor must = 2.0A as it also has 30V across it.
while current through resistor 5W resistor = 1.0 + 2.0 = 3.0A from junction law.
which means that voltage drop across 5W resistor = 3.0 x 5 = 15 V
This allows calculation of a total voltage across 60W resistor of 15 + 30 = 45 V.

Q39 :-
Let the circuit current be (I) and the voltage across R2 be (V).
Hence EMF = 120 V = 100.I + V and also V.I = 30 as given from power dissipation. thus solving simultaneous equations created :-
120 = 100.I + 30/I which reduces to 10.I2 - 12.I + 3 = 0
solving the quadratic yields I = 0.84 A or I = 0.36 A as dual solutions.
Hence using R =P/I2 gives R = 232 or 43W.