| Date | May 2017 | Marks available | 3 | Reference code | 17M.2.SL.TZ2.5 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Discuss | Question number | 5 | Adapted from | N/A |
Question
A cable consisting of many copper wires is used to transfer electrical energy from a generator to an electrical load. The copper wires are protected by an insulator.
The cable consists of 32 copper wires each of length 35 km. Each wire has a resistance of 64 Ω. The resistivity of copper is 1.7 x 10–8 Ω m.
The copper wires and insulator are both exposed to an electric field. Discuss, with reference to charge carriers, why there is a significant electric current only in the copper wires.
Calculate the radius of each wire.
There is a current of 730 A in the cable. Show that the power loss in 1 m of the cable is about 30 W.
When the current is switched on in the cable the initial rate of rise of temperature of the cable is 35 mK s–1. The specific heat capacity of copper is 390 J kg–1 K–1. Determine the mass of a length of one metre of the cable.
Markscheme
when an electric field is applied to any material «using a cell etc» it acts to accelerate any free electrons
electrons are the charge carriers «in copper»
Accept “free/valence/delocalised electrons”.
metals/copper have many free electrons whereas insulators have few/no free electrons/charge carriers
area = \(\frac{{1.7 \times {{10}^{ - 3}} \times 35 \times {{10}^3}}}{{64}}\) «= 9.3 x 10–6 m2»
«resistance of cable = 2Ω»
power dissipated in cable = 7302 x 2 «= 1.07 MW»
power loss per meter \( = \frac{{1.07 \times {{10}^{ - 6}}}}{{35 \times {{10}^3}}}\) or 30.6 «W m–1»
Allow [2] for a solution where the resistance per unit metre is calculated using resistivity and answer to (b)(i) (resistance per unit length of cable =5.7 x 10–5 m)
30 = m x 390 x 3.5 x 10–2
2.2 k«g»
Correct answer only.
