Compact \"ultracapacitors\" with capacitance values up to several thousand farad

Question

Compact "ultracapacitors" with capacitance values up to several thousand farads are now commercially available. One applications for ultracapacitance 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 1400-F ultracapacitor is initially charged to 12.0 V by a battery and is then disconnected 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 device, how long(in days) will in take for the potential difference across capacitance to drop to 6.0 V? Express your answer using two significant figures.

Explanation / Answer

Here,

capacitance , C = 1400 F

Vo = 12 V

dQ/dt = 1 mC/s

let the time taken is t

Now , for the potential to drop to V2 = 6 V

dQ/dt * time = C * (Vo - V2)

1 *10^-3 * t = 1400 * (12 - 6)

solving for t

t = 8.4e6 s

t = 97.2 days

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