In a first-order RC circuit, after one time constant, what fraction of the final voltage is across the capacitor?

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Multiple Choice

In a first-order RC circuit, after one time constant, what fraction of the final voltage is across the capacitor?

Explanation:
In a charging RC circuit, the capacitor voltage rises exponentially toward the supply voltage. It follows v_C(t) = V_s [1 − e^(−t/RC)], with τ = RC being the time constant. After one time constant, t = τ, so v_C(τ) = V_s [1 − e^(−1)] ≈ V_s (1 − 0.3679) ≈ 0.632 V_s. So the capacitor has about 63.2% of the final voltage at that moment. The remaining portion continues to charge toward the full supply voltage as time progresses.

In a charging RC circuit, the capacitor voltage rises exponentially toward the supply voltage. It follows v_C(t) = V_s [1 − e^(−t/RC)], with τ = RC being the time constant. After one time constant, t = τ, so v_C(τ) = V_s [1 − e^(−1)] ≈ V_s (1 − 0.3679) ≈ 0.632 V_s. So the capacitor has about 63.2% of the final voltage at that moment. The remaining portion continues to charge toward the full supply voltage as time progresses.

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