| Date | May 2017 | Marks available | 2 | Reference code | 17M.2.HL.TZ2.6 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Calculate | Question number | 6 | Adapted from | N/A |
Question
A cable consisting of many copper wires is used to transfer electrical energy from an alternating current (ac) 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 cable is connected to the ac generator which has an output power of 110 MW when the peak potential difference is 150 kV. The resistivity of copper is 1.7 x 10–8 Ω m.
output power = 110 MW
To ensure that the power supply cannot be interrupted, two identical cables are connected in parallel.
The energy output of the ac generator is at a much lower voltage than the 150 kV used for transmission. A step-up transformer is used between the generator and the cables.
Calculate the radius of each wire.
Calculate the peak current in the cable.
Determine the power dissipated in the cable per unit length.
Calculate the root mean square (rms) current in each cable.
The two cables in part (c) are suspended a constant distance apart. Explain how the magnetic forces acting between the cables vary during the course of one cycle of the alternating current (ac).
Suggest the advantage of using a step-up transformer in this way.
The use of alternating current (ac) in a transformer gives rise to energy losses. State how eddy current loss is minimized in the transformer.
Markscheme
area = \(\frac{{1.7 \times {{10}^{ - 3}} \times 35 \times {{10}^3}}}{{64}}\) «= 9.3 x 10–6 m2»
radius = «\(\sqrt {\frac{{9.3 \times {{10}^{ - 6}}}}{\pi }} = \)» 0.00172 m
Ipeak «\( = \frac{{{P_{peak}}}}{{{V_{peak}}}}\)» = 730 « A »
resistance of cable identified as «\(\frac{{64}}{{32}} = \)» 2 Ω
\(\frac{{{\text{a power}}}}{{35000}}\) seen in solution
plausible answer calculated using \(\frac{{{\text{2}}{{\text{I}}^{\text{2}}}}}{{35000}}\) «plausible if in range 10 W m–1 to 150 W m–1 when quoted answers in (b)(ii) used» 31 «W m–1»
Allow [3] for a solution where the resistance per unit metre is calculated using resistivity and answer to (a) (resistance per unit length of cable = 5.7 x 10–5 m )
Award [2 max] if 64 Ω used for resistance (answer x32).
An approach from \(\frac{{{V^2}}}{R}\) or VI using 150 kV is incorrect (award [0]), however allow this approach if the pd across the cable has been calculated (pd dropped across cable is 1.47 kV).
«\(\frac{{{\text{response to (b)(ii)}}}}{{2\sqrt 2 }}\)» = 260 «A»
wires/cable attract whenever current is in same direction
charge flow/current direction in both wires is always same «but reverses every half cycle»
force varies from 0 to maximum
force is a maximum twice in each cycle
Award [1 max] if response suggests that there is repulsion between cables at any stage in cycle.
higher voltage gives lower current
«energy losses depend on current» hence thermal/heating/power losses reduced
laminated core
Do not allow “wires are laminated”.
