A capacitor is fully charged after 25 seconds to a battery voltage of 20 Volts. The battery is replaced with a short circuit. What will be the voltage across the capacitor after one time constant?
- 0 volts.
- 7.36 volts.
- 12.64 volts.
Explanation
On discharge the capacitor voltage follows V = V0 * e^(-t/tau). After exactly one time constant (t = tau), the factor is e^(-1) = 0.368, leaving 36.8% of the original voltage. So 20 V * 0.368 ~ 7.36 V. The 12.64 V figure is the charging value (63.2%), which applies to charging, not discharging.
roja90 asking:
how to do this question ?
after one time constant voltage is 32.8%?//
Community Comments (3)
D
dimky Posts: 514 16.02.2012 / 14:24
Hello.
There is mistake in answers.
Multi-choise should be:
- 0 volts.
- 7.36 volts.
- 12.64 volts.
I corrected this question. Thank you for report.
There is mistake in answers.
Multi-choise should be:
- 0 volts.
- 7.36 volts.
- 12.64 volts.
I corrected this question. Thank you for report.
M
manzagip Posts: 3 14.03.2014 / 21:53
Roja90 after one time constant voltage is 36.8 %
H
H9m3d Posts: 1 28.04.2016 / 01:33
After one time constant boltage is 63.2%
Short circuit(discharge): 20-(63*20/100)=7.36
Short circuit(discharge): 20-(63*20/100)=7.36
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