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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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zubam2002

Answer:

Explanation:

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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