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In the circuit diagram shown below, each cell has an e.m.f of 1.5V and internal resistance of 0.5 ohms. The capacitance of each 1.4 micro faraday. When the switch S is closed determine the:

i).ammeter reading
ii).charge on each capacitor.

Answer Text:
i).Total resistance
= #6 + 5 +(2times 0.5) = 12# ohms
Total e.m.f = 3.0 V.
Current (I) = #3/(12)#= 0.25A
ii).V across each resistor = IR
#0.25 times 11= 2.75v.#
Charge = CV = #1.4 times 2.75#
= 3.85 micro Coulombs.