| Date | May 2011 | Marks available | 2 | Reference code | 11M.2.SL.TZ2.9 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Determine | Question number | 9 | Adapted from | N/A |
Question
Part 2 Electrical resistance
A resistor of resistance 1.5Ω is made from copper wire of radius 0.18mm. The resistivity of copper is 1.7×10–8Ωm. Determine the length of copper wire used to make the resistor.
The manufacturer of the resistor in (a) guarantees that the resistance is within 10% of 1.5Ω, provided that the power dissipation in the resistor does not exceed 1.0W.
(i) Suggest why the resistance of the resistor might be greater than 1.65Ω if the power dissipation in the resistor is greater than 1.0W.
(ii) Show that, for a power dissipation of 1.0W, the current in a resistor of resistance 1.5Ω is 0.82A.
(iii) The 1.5Ω resistor is connected in series with a variable resistor and battery of emf 6.0V and internal resistance 1.8Ω.
Estimate the resistance R of the variable resistor that will limit the current to 0.82A.
Markscheme
use of \(l = \frac{{RA}}{\rho }\); } (allow if correct substitution seen – watch for use of circumference in place of area)
\( = \left( {\frac{{1.5 \times \pi \times {{\left[ {1.8} \right]}^2} \times {{10}^{ - 8}}}}{{1.7 \times {{10}^{ - 8}}}} = } \right)9.0{\rm{m}}\);
(i) the resistance of a conductor/copper/metal increases with increasing temperature;
increased power (dissipation) leads to higher temperature in the resistor/ resistor heating up;
(ii) \(I = \left( {\sqrt {\frac{P}{R}} = } \right)\sqrt {\frac{{1.0}}{{1.5}}} \);
(=0.82A)
Allow working using 0.82A to show that power is 1.0086W, in this case final answer must be to 2 sig fig or better.
(iii) total resistance = [R+3.3];
6.0=0.82[R+3.3];
to give R=4.0Ω; (allow use of 1.65Ω leading to 3.9Ω)
or
total resistance in circuit = \(\frac{{6.0}}{{0.82}} = \left( {7.3\Omega } \right)\);
internal resistance+fixed resistance=3.3Ω;
to give R=4.0Ω;
