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The figure below shows a circuit of e.m.f 4.5V, switches A and B, two capacitors C1 = 0.3 micro Faraday and C2 = 0.5 micro Faraday and a voltmeter are connected

a).Determine the charge on C1 when switch A is closed and switch B is open.
b) What is the effective capacitance CT when both switches A and B are closed?
(c) State what is observed on the voltmeter when
(i) Switch A is closed and switch B is open.
(ii) Switch A is closed and opened, and then B is closed.
(iii) Explain the observation made in c (ii) above.

Answer Text:
a).Charge on C1 = C1V
= #0.3 times 4.5 = 1.35 micro C = 1.35 times 10^-6#C
b).CT = C1 +C2
=#(0.3 +0.5)micro Faraday= 8.0 times 10^-7# F
c).i).Voltmeter reads 4.5v
ii).Reads less than 4.5v
iii).The drop is due fact that the charge initially stored in C1 is distributed to C2.Since the
value of C1 and C2 remains constant when the charge reduces the p.d drops.