State what is meant by the emf of a cell.

Show that the resistance of the wire AC is 28 Ω.

Determine E.

the work done per unit charge

in moving charge from one terminal of a cell to the other / all the way round the circuit

Award [1] for “energy per unit charge provided by the cell”/“power per unit current”

Award [1] for “potential difference across the terminals of the cell when no current is flowing” 

Do not accept “potential difference across terminals of cell”

[2 marks]

the resistance is proportional to length / see 0.35 AND 1«.00»

so it equals 0.35 × 80

«= 28 Ω»

[2 marks]

current leaving 12 V cell is \(\frac{{12}}{{80}}\) = 0.15 «A»

OR

E = \(\frac{{12}}{{80}}\) × 28

E = «0.15 × 28 =» 4.2 «V»

Award [2] for a bald correct answer

Allow a 1sf answer of 4 if it comes from a calculation.

Do not allow a bald answer of 4 «V»

Allow ECF from incorrect current

[2 marks]