The candidates were required to:
(i) connect the circuit as shown and obtain voltage and current readings;
(ii) calculate the voltage and current in each branch.
The expected answers were:
Tables 3
Quantity |
Value |
V1 |
5.625V |
V2 |
3.375V |
V3 |
6.907V |
V4 |
2.093V |
I |
490.6mA |
(g) V1 + V2 = (5.625 + 3.375)V = 9v
(h) V3 + V4 = (6.907 + 2.093)V = 9v
(i) Results obtained from steps (g) and (h) were found to be the same.
(j) I3 = V3 = 6.907 = 0.2093A or 209.3mA
R3 33
(k) I1 = V1 = 5.625 = 0.2813A or 281.3m A
R1 20
(l) I3 + I1 = (0.2093 + 0.2813) A = 0.4906A or 490.6mA
(m) On comparing step (l) with the value of I, they were the same.
The candidates understood the task, they connected the resistors as required and obtained readings of the current and voltage drop. However, some of the candidates’ readings were not correct or not expressed in the correct units.