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

Computers & Technology

Kristian Leon

30 July, 16:16

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

+1

Answers (1)

Know the Answer?

Cedric Buckley · 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