Basic Electricity 2, May/June 2015

Question 2

In figure 2 calculate the;

(i) Equivalent capacitor;

(ii) Total energy stored in the circuit;

(iii) Charge in the 6µF capacitor;

(iv) Potential difference across the 6µF capacitor.  


 The expected answers were;

(i) Equivalent capacitor (Ceq) = (6×12)/(6+12 ) μF = 72/18 = 4×〖10〗^(-6) F = 4μF

(ii) Total energy stored = 〖CV〗^2/2 = (4×〖10〗^(-6 )× 〖30〗^2)/2 = 1.8×〖10〗^(-3) joules = 1.8 m joules

(iii) Charge in the 6µF capacitor (Q) =CV Q = 4×〖10〗^(-6)×30 =120μC

(iv) Potential difference across the 6µF Capacitor V = Q/C = (120×〖10〗^(-6))/(6×〖10〗^(-6) ) = 20V

Candidates were expected to calculate the equivalent capacitance of a series- parallel connection of capacitors and the total energy stored in the circuit, the charge and potential difference across the series (6µF) capacitor.

The candidates were reported to have performed very good in this question.