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.