A defibrillator with a time constant of 8.00 ms starts at 10.0 kV. About how long does it take to fall to roughly 500 V?

1. In each time constant the voltage falls to 0.368 of the value it held at the start of that interval. 2. After the first 8.00 ms, V = (0.368)(10.0 kV) = 3.680 kV. 3. After the second, V = (0.368)(3.680 kV) = 1.354 kV, at t = 16.0 ms. 4. After the third, V = (0.368)(1.354 kV) = 498 V, at t = 24.0 ms. 5. That is just under the 500 V asked for, so three time constants do the job. 6. 498 V is 4.98 percent of the starting voltage, which shows how quickly three time constants strip a capacitor. 7. 8.00 ms and 16.0 ms stop short, at 3.68 kV and 1.35 kV respectively. 8. 40.0 ms is five time constants, well past the moment the voltage first drops below 500 V. _Source: OpenStax College Physics (CC BY 4.0), Ch 21 "Circuits and DC Instruments", section 21.6 DC Circuits Containing Resistors and Capacitors_