A DC voltage is applied through a resistor R = 10 kΩ to an uncharged capacitor C = 10 µF. Which is closest to the time for the capacitor voltage to reach 50% of its final value?

Answer: B. About 69 ms

The time constant is τ = RC = 10×10³ Ω × 10×10⁻⁶ F = 0.1 s. The charging voltage is v = E(1 − e^(−t/τ)), so v = 0.5E when e^(−t/τ) = 0.5, that is t = τ ln 2 ≈ 0.693 × 0.1 s ≈ 69 ms. 100 ms is the time constant τ itself, at which the voltage has reached about 63% of its final value. The voltage does not rise in proportion to time, so half of τ does not give half the voltage.

• A. 50 ms is half of τ, assuming the voltage rises in proportion to time.
• B. Correct. τ ln 2 ≈ 0.693 × 100 ms ≈ 69 ms.
• C. 100 ms is the time constant, when the voltage reaches about 63% of its final value.