Interpret capacitor discharge and threshold times

Begin with a different initial condition

The discharge history starts with the capacitor at 6 V and then closes a resistor-only loop after the source has been removed. It is not the continuation of whichever charging point you last selected. The two diagrams deliberately show separate circuits and separate time origins.

Current uses the same positive reference into the upper capacitor plate in both diagrams. During discharge, charge leaves that plate through the resistor, so current is negative. The plate’s voltage can remain positive while its charge decreases. Negative current does not mean negative resistor heating: I²R is nonnegative.

Track what remains

For this resistor-only discharge, V capacitor(t) = 6 exp(−t/τ), with τ = RC. Charge magnitude follows the same fraction because Q = CV and C is fixed. Signed current is I(t) = −(6/R) exp(−t/τ).

At R = 2 kΩ, τ = 2 s. Immediately after connection, voltage is 6 V and current is −3 mA. At 2 s, voltage is 2.2073 V and current is −1.1036 mA. At 4 s, voltage is 0.8120 V and current is −0.4060 mA. Capture these points and explain why current approaches zero from below on the signed plot.

The two voltage histories sum to 6 V at the same time when their R and C values match. That is a consequence of these complementary initial conditions; it is not a statement that the two circuits exchange energy with each other.

Half the voltage does not mean half the energy

At fixed capacitance, stored energy is U = C V²/2. At one time constant, voltage is e⁻¹ ≈ 36.8% of its initial value, but energy is e⁻² ≈ 13.53% of its initial value. Half voltage leaves one-quarter energy.

The half-voltage time is τ ln 2 ≈ 0.6931τ, or 1.3863 s when τ = 2 s. The natural logarithm ln reverses the exponential: it is used here to solve exp(−t/τ) = 1/2. Half-energy time is half as long, approximately 0.3466τ = 0.6931 s, because energy contains the square of the voltage fraction.

These exact threshold times need not lie on the slider’s whole-second sample grid. The browser reports the analytic half-voltage time separately rather than snapping it to a misleading integer second.

A finite plot is not an infinite-time claim

Choose R = 5 kΩ and t = 10 s. Only two time constants have passed. Discharge voltage remains 0.8120 V, and charging voltage is only 5.1880 V. The time to reach 90% of the charging limit is τ ln 10 ≈ 11.5129 s, outside the displayed window. Its label says so; the graph does not silently stretch its time scale.

In the ideal exponential model, discharge voltage approaches zero without becoming exactly zero at a finite time. A practical measurement threshold is different from mathematical zero. Real leakage, meter loading and component tolerances can change the behavior, and this browser does not turn a computed threshold into a physical safety assessment.

Estimate a time constant from a reading

An invented discharge observation shows voltage falling from 6 V to 3 V in 1.4 s. If the simple exponential model is appropriate, estimate τ and R for C = 1000 µF. Then predict a voltage at an additional time before inspecting another reading.

Check the discharge estimate

The observed time is a half-voltage time, so τ = 1.4/ln 2 ≈ 2.0198 s. With C = 0.001 F, R ≈ 2019.8 Ω. At 2.8 s, two half-voltage intervals have passed, so the candidate predicts 1.5 V. This estimate uses invented data without stated measurement uncertainty; it is not evidence of an exact physical resistance. A new reading tests the candidate, whereas replaying the same observation does not.