This was a popular question and it was well attempted. Some candidates had difficulties in the determination of d, A and A-1. Those that were able to determine A1 correctly have difficulties handling the exponential factors of the evaluated quantity A-1. For instance a value of 2.705 x 10-7, the value of 2.7 was used and the 10-7 was not taken into account in evaluating the slope.
The (b) part was poorly attempted.
In part (a), candidates were required to:
- read and record at least 1 decimal place, five values of V, I and d
(ii) convert using the given scale the five values of d
(iii) evaluate correctly and record to at least 3 d.p or s s.f as appropriate five values of R, A and A-1
(iv) record data in composite table showing v, d, R, A and A-1
(v) plot five points correctly on a distinguished axes using reasonable scales with a line of best fit through the points.
(vi) State any two of the following precautions:
- Tight connections were ensured;
- Avoid parallax errors in reading volt-meter/ammeter/meter rules;
- Key opened when readings are not taken
- Ensure clean terminals
- Zero error noted and corrected in reading voltmeter/ammeter/metre rule.
(b)(i) Pt = energy
Pt = QV
t = QV = 30 x 12
p 6
= 60 hours
(ii) The internal resistance of the cell drops a certain amount of voltage across it [lost volts] whenever the cell is delivering current but this is not so when it is not delivering current.
OR
Terminal p.d (v) = I R
Emf (E) = iR + ir
∴ E > V because of ir (lost volts)
