Basic Electricity 2, May/June 2015

Question 4

(a) Define coulomb.

(b) State the units of the following quantities;

(i) Resistance;

(ii) Potential difference.

(c) A dc motor supplied from 230V supply for 1hour consumes an energy of 36mJ. Calculate;

(i) Power rating of the motor;

(ii) Current taken from the supply.  


The expected answers were;

(a) Coulomb (Q) =I×t Is the quantity of a steady current of one ampere in one second.





Ohm ()

Potential difference

Volt   (V)

(c) (i) Power rating of the motor = (Electrical energy)/time = (36×〖10〗^(-3))/(1×60×60) =0.01×〖10〗^(-3) OR =0.01mW

(ii) Current taken from the supply = Energy/VI = (36×〖10〗^(-3))/(230×60×60) =0.04348×〖10〗^6 =0.043μA


I=P/V =(1×〖10〗^(-3))/230 =4.3×〖10〗^(-3) A =0.043μA

The candidates were to define Coulomb, state unit of measurement of resistance and potential difference. They were to calculate the power rating and current taken from supply by an excited dc motor.

The performance of the candidates was reported to have been very good in this question.