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 66times10 Superscript 131013 Upper OmegaΩ ; on a humid day, the resistance is 77times10 Superscript 6106 Upper OmegaΩ. How long will it take the capacitor voltage to dissipate to half its original value on each day

Question

Nickolas Church

Computers & Technology

21 April, 10:51

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 66times10 Superscript 131013 Upper OmegaΩ ; on a humid day, the resistance is 77times10 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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Answers (1)

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Timothy Tran · Answer 1

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