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A 10 Superscript negative 910-9 -F capacitor (11 nanofaradnanofarad ) is charged to 5050 V and then disconnected. One can model the charge leakage of the capacitor with a RC circuit with no voltage source and the resistance of the air between the capacitor plates. On a cold dry day, the resistance of the air gap is 66times*10 Superscript 131013 Upper OmegaΩ ; on a humid day, the resistance is 77times*10 Superscript 6106 Upper OmegaΩ. How long will it take the capacitor voltage to dissipate to half its original value on each day

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  1. 21 April, 11:00
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    i) On a cold day, Capacity = (66 X 910 X 10∧-9) F

    Charge, Q = (66 X 910 X 10∧-9) X 5050 = 0.30 Columb

    Current, I = 5050/131013 = 0.039 Amp (where Resistance = 131013 ohms)

    Time, t = 0.30/0.039 = 7.7 seconds

    ii) On a humid day, Capacity = (77 X 910 X 10∧-9) F

    Charge, Q = (77 X 910 X 10∧-9) X 5050 = 0.35 Columb

    Current, I = 5050/6106 = 0.83 Amp (where Resistance = 6106 ohms)

    Time, t = 0.35/0.83 = 0.42 h or 0.42 X 60 mins = 25 mins
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