This is another popular question among the candidates and it was well attempted by majority of the responding candidates

However, some candidates missed valuable marks on the observation because their values did not tally with the required trend expected of them. Few other candidates recorded the values of OP without decimal place while some did not begin the graph from the origin (0,0), so they could not determine V when I 0 correctly.

In part (b) most candidates could not compare very well lead aid accumulator with dry lechlanche cell. Most candidates only compared the rechargeable aspect while they left out polarization, local action and internal resistance.

The numerical part was poorly tackled.

The expected answers are:

(t) I0 read and recorded to at at least 1 d.p in ampere

(ii) V„ read and recorded to at least 1 d.p in volts

(iii) Five values of I correctly read and recorded

to at least 1 d.p in ampere and in trend (Trend: As OP increases, I increases)

(iv) Six values of OP correctly recorded to at least to 1 d.p in cm

(v) Six values of V read and recorded to at least I d.p in volts and in trend

Trend: As OP increases V decreases

(vi) Composite table showing at least I, OP and V

Candidates are also expected to;

(i) Plot six points on graph

(ii) Distinguish between the axes

(iii) Choose reasonable scales

(iv) Draw line of best fit

(v) Determine the slope of the graph

(vi) Deduce the value of V when I = 0

(vii) State any two of the following precautions in acceptable language.

-- Open key when not taken readings

-- Avoided parallax error in readings ammeter/voltmeter

-- Zero error was noted and corrected for on ammeter/voltmeter

-- Clean and tight terminals Jockey not slide/drag on potentiometer wire

b (i)Advantages of lead-acid cell over lechlanche cell

-- Low internal resistance Rechargeable

-- Has high voltage (2.0V)

-- Supply higher current

-- No defects such as polarization & local

action.

E = l(R + r)

2= I (9 + 1) = 101 I = 0.2A

V= Ir

= 0.2 xl = 0.2V