In a discharge RC circuit with V0 = 5 V, R = 1 kΩ, and τ = 1 ms, what is Vc after t = 1 ms?

In a discharge RC circuit with V0 = 5 V, R = 1 kΩ, and τ = 1 ms, what is Vc after t = 1 ms?

• A. Vc ≈ 1.84 V
• B. Vc ≈ 0.84 V
• C. Vc ≈ 5 V
• D. Vc ≈ 2.5 V

Answer: (A)

The voltage across a discharging capacitor in an RC circuit decays exponentially: Vc(t) = V0 · e^(−t/τ). With V0 = 5 V and τ = 1 ms, at t = 1 ms you get Vc = 5 · e^(−1) ≈ 5 · 0.3679 ≈ 1.84 V. Conceptually, after one time constant the voltage has fallen to about 37% of its initial value, so it’s around 1.8 V here. It can’t be still 5 V because the capacitor is discharging, and it isn’t 2.5 V because the half-life (approx 0.693 ms) would give 2.5 V, and at 1 ms the voltage is already lower than that. The result around 1.84 V matches the exponential decay with the given time constant.

The voltage across a discharging capacitor in an RC circuit decays exponentially: Vc(t) = V0 · e^(−t/τ). With V0 = 5 V and τ = 1 ms, at t = 1 ms you get Vc = 5 · e^(−1) ≈ 5 · 0.3679 ≈ 1.84 V. Conceptually, after one time constant the voltage has fallen to about 37% of its initial value, so it’s around 1.8 V here.

It can’t be still 5 V because the capacitor is discharging, and it isn’t 2.5 V because the half-life (approx 0.693 ms) would give 2.5 V, and at 1 ms the voltage is already lower than that. The result around 1.84 V matches the exponential decay with the given time constant.