(i) Define the electrical resistance of a component.

(ii) Explain whether or not the filament obeys Ohm’s law.

(i) Calculate the resistance of the filament lamp when the potential difference across it is 2.8 V.

(ii) The length of the filament in a lamp is 0.40 m. The resistivity of tungsten when the potential difference across it is 2.8 V is 5.8×10^–7Ω m. Calculate the radius of the filament.

Two identical filament lamps are connected in series with a cell of emf 6.0 V and negligible internal resistance. Using the graph on page 26, calculate the total power dissipated in the circuit.

(i) \(\frac{{{\rm{potential difference across the component}}}}{{{\rm{current in the component}}}}\);
Award [0] for simple statement of voltage divided by current

(ii) Ohm’s law states that voltage is (directly) proportional to current or
\(\frac{{{\rm{potential difference}}}}{{{\rm{current}}}}\) /resistance is a constant;
graph not linear/gradient not constant so Ohm’s law not obeyed / calculation of \(\frac{V}{I}\) at two points showing that they are different;
Award [0] for bald statement of Ohm’s law not obeyed.

(i) (from graph, when V = 2.8 V,) I = 0.33 A; (accept answers in range 0.32 to 0.34 A)
\(R = \frac{V}{I} = \frac{{2.8}}{{0.34}} = 8.5\Omega \); (accept answers in range 8.2 to 8.8 Ω)

(ii) \(A = \left( {\frac{{\rho l}}{R} = \frac{{5.8 \times {{10}^{ - 7}} \times 0.40}}{{8.5}} = } \right)2.7 \times {10^{ - 8}}\);
(accept answers in range 2.6 to 2.8×10^–8)

\(r = \sqrt {\frac{A}{\pi }} \) seen/used;
=9.3×10^-5m; (accept answers in range 9.2 to 9.5×10^–5)

each lamp has a potential difference of 3.0 V so current equals 0.35 A;
(accept answers in range 0.34 to 0.35 A)
2.1 W; (accept answers in range 2.0 to 2.1 W)
Award [1] for answers that use voltage 6.0 V with current 0.52 A to get P=3.1W.