The candidates were required to:
(i)connect the circuit as shown and take voltage readings;
(ii)calculate the current and the total voltage drop across the resistances in the circuit.
The expected answers were:
(e) Table 1
V1 (volts) |
V2(volts) |
V3 (volts) |
6.186 |
2.722 |
3.093 |
(g) Table 2
Current |
Ammeter reading (mA) |
I1 |
226.80 |
I2 |
82.47 |
I3 |
309.30 |
(h)
V1 =
6.186 = 0.3093A or 309.3mA
R1 20
(i) On comparing step (h) with I3, the two were the same.
(j) V1 + V2 + V3 = 6.186 + 2.722 + 3.093 = 12.001V
(k) On dividing the result obtained in step (j) with the result obtained in step (h)
12.001 = 38.8005 (preferred value 39Ω)
0.3093
(l) Based on step (k), the total resistance of fig 1 was 38.80Ω (preferred value i.e. 39Ω)
From the results, it was evident that the supply voltage was equal to the sum total of the voltage drops across the resistances and that the experimented equivalent resistance was a bit different from the calculated value due to the ratings of the components used.
The candidates understood the task, they connected the circuit fairly and obtained readings, but some candidates’ readings and units were not correct.