In an RC charging circuit with R = 1 kΩ and C = 1 μF, what is the capacitor voltage after one time constant τ if the supply is V?

In an RC charging circuit with R = 1 kΩ and C = 1 μF, what is the capacitor voltage after one time constant τ if the supply is V?

• A. 0.5 × V
• B. V
• C. 0.368 × V
• D. 0.632 × V

Answer: (D)

In an RC charging circuit, the capacitor voltage rises according to Vc(t) = V(1 - e^{-t/RC}), with the time constant τ = RC. Here RC = 1 kΩ × 1 μF = 0.001 s, so τ = 1 ms. At one time constant, t = τ, the voltage becomes Vc(τ) = V(1 - e^{-1}) = V(1 - 0.3679…) ≈ 0.632 V. That means the capacitor has charged to about 63.2% of the supply voltage after one τ. The final steady-state voltage is V, so after one time constant you’re at roughly 0.632 of V.

In an RC charging circuit, the capacitor voltage rises according to Vc(t) = V(1 - e^{-t/RC}), with the time constant τ = RC. Here RC = 1 kΩ × 1 μF = 0.001 s, so τ = 1 ms. At one time constant, t = τ, the voltage becomes Vc(τ) = V(1 - e^{-1}) = V(1 - 0.3679…) ≈ 0.632 V. That means the capacitor has charged to about 63.2% of the supply voltage after one τ. The final steady-state voltage is V, so after one time constant you’re at roughly 0.632 of V.