FCC Exam Question: 3-14B6
After two time constants, the capacitor in an RC circuit is discharged to what percentage of the starting voltage?
Explanation: In an RC circuit, a capacitor discharges exponentially. The time constant ($\tau$), calculated as R (resistance in ohms) multiplied by C (capacitance in farads), represents the time it takes for the capacitor's voltage to discharge to approximately 36.8% of its initial value. After one time constant, the capacitor is at 36.8% of the starting voltage. After a second time constant, it discharges by another 36.8% of the *remaining* voltage. So, $36.8\% \times 36.8\% \approx 0.1353$, which means the capacitor is discharged to approximately 13.5% of its starting voltage. Option D (36.8%) represents the remaining voltage after *one* time constant. Option C (63.2%) is how much voltage is *lost* after one time constant during discharge, or how much it *gains* during charging. Option A (86.5%) is how much it *gains* during charging after two time constants, not the remaining voltage during discharge.
3-66J2
3-92O1
3-65J5
3-71K5
3-56G5
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Includes Elements 1, 3, 6, 7R, 8, and 9.