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 Form 3 Electrostatics II Physics Questions and Answers

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:
13.png
i).ammeter reading
ii).charge on each capacitor.

 (4m 0s)
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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.


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