After two time constants, the capacitor in an RC circuit is charged to what percentage of the supply voltage?

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Multiple Choice

After two time constants, the capacitor in an RC circuit is charged to what percentage of the supply voltage?

In an RC charging circuit, the capacitor voltage climbs toward the supply voltage following an exponential curve with a time constant RC. The exact relationship is Vc(t) = Vs [1 − e^(−t/RC)]. The time constant is the step that sets how fast the charging happens. Two time constants means t = 2RC. Plugging into the formula gives Vc = Vs [1 − e^(−2)]. Since e^(−2) ≈ 0.1353, this is about 1 − 0.1353 ≈ 0.8647 of the supply voltage. That is roughly 86.5%. So the capacitor voltage is about 86.5% of the supply voltage after two time constants. For context, after one time constant it’s about 63.2%, and after three time constants it’s about 95%.

In an RC charging circuit, the capacitor voltage climbs toward the supply voltage following an exponential curve with a time constant RC. The exact relationship is Vc(t) = Vs [1 − e^(−t/RC)]. The time constant is the step that sets how fast the charging happens.

Two time constants means t = 2RC. Plugging into the formula gives Vc = Vs [1 − e^(−2)]. Since e^(−2) ≈ 0.1353, this is about 1 − 0.1353 ≈ 0.8647 of the supply voltage. That is roughly 86.5%.

So the capacitor voltage is about 86.5% of the supply voltage after two time constants. For context, after one time constant it’s about 63.2%, and after three time constants it’s about 95%.

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