9. (II) Compact “ultracapacitors” with capacitance values up to several thousand

Question

9. (II) Compact “ultracapacitors” with capacitance values up to several thousand farads are now commercially available. One application for ultracapacitors is in providing power for electrical circuits when other sources (such as a battery) are turned off. To get an idea of how much charge can be stored in such a component, assume a 1200-F ultracapacitor is initially charged to 12.0V by a battery and is then discon- nected from the battery. If charge is then drawn off the plates of this capacitor at a rate of 1.0 mc/s, say, to power the backup memory of some electrical gadget, how long (in days) will it take for the potential difference across this capacitor to drop to 6.0 V?

Explanation / Answer

arifkhananswered this

C = 1200 F

Vi = initial Voltage = 12 volts

Qi = initial charge stored = C Vi = 1200 (12) = 14400 C

Vf = Final Voltage = 6 volts

Qf = initial charge stored = C Vf = 1200 (6) = 7200 C

Charge given away = Q = 14400 - 7200 = 7200 C

rate = 0.001 C/s

Time = Charge given away / Rate = 7200 / 0.001 = 7200,000 sec = 7200,000 (1h / 3600 s) (1 day/24) = 83.3days

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