RC circuit and time constant

The voltage across a capacitor does not jump to the source EMF because changing that voltage means moving charge, and charge can only flow through the resistor: the current is limited by R, so the voltage can only change at a finite rate. Combining Kirchhoff's voltage law E = iR + u with the capacitor law i = C du/dt gives RC du/dt = E − u, whose solution is u(t) = E + (u₀ − E)e^(−t/τ) with τ = RC. The switch closes at t = 0 and the capacitor starts at u₀; after one time constant τ the gap to E has shrunk to 1/e of its original size, and as t → ∞ the voltage tends to E. It is the same circuit reasoning as Ohm's law and Series and parallel, with a capacitor that stores charge — which is what introduces time.

Questions

Why does the voltage across the capacitor need time to change instead of jumping straight to the source EMF?

If you increase the resistance R or the capacitance C, does the capacitor reach the source EMF sooner or later?

If the capacitor already carries a voltage higher than the source (u₀ > E), does it discharge or keep charging, and where does it finally settle?

Predict

Write down your judgement first: with u₀ = 0 and E = 6 V, after one time constant τ is the capacitor voltage about 63%, 86% or 95% of 6 V? If you make τ four times larger, by what factor does the time to reach the same fraction change? Then set u₀ = 10 V and E = 0 V and predict whether the curve rises or falls.

Try

Keep E = 6 V, u₀ = 0 V, R = 40 Ω and C = 0.05 F, so τ = RC = 2 s. Press play and watch the curve leave u₀ and approach E. Pause and use Step to advance 0.10 s at a time, noting the capacitor-voltage reading at t = 0.10, 1.00 and 2.00 s. Then set C = 0.1 F so τ = 4 s, replay, and compare how much flatter the curve becomes. Finally set u₀ = 10 V and E = 0 V to watch discharging. The switch closes at t = 0; changing any parameter (or pressing reset) re-opens the circuit, sets the time back to zero and restores the capacitor voltage to u₀ — the capacitor's internal state is its own voltage, and the run continues from that voltage.

Observe

The curve leaves u₀ and moves monotonically towards E, never overshooting E or turning back at it. A larger τ gives a flatter curve and a smaller τ a steeper one; changing R or C alters only the speed, never the final E. With u₀ < E the curve rises (charging), with u₀ > E it falls (discharging), and with u₀ = E it is a horizontal line. At the default parameters the curve near t = 0 is slanted rather than vertical, which shows that the rate of change of the voltage is finite; with a very small τ (for example 0.2 s) the same change is squeezed into a narrow strip at the left end of the window.

Explain

Why can the capacitor voltage not jump?

The charge stored on the capacitor and its voltage obey q = C u, and the current is the rate at which charge is carried, i = dq/dt = C du/dt. Charge can only move through the resistor, which limits the current to i = (E − u)/R, so du/dt = (E − u)/(RC) is always finite. A finite rate of change means the voltage varies continuously in time and cannot jump to E in an instant. Only as R → 0 would the rate diverge, which is exactly the ideal short-circuit limit.

Where does the exponential come from?

Apply Kirchhoff's voltage law around the loop: the source EMF E equals the drop iR across the resistor plus the capacitor voltage u, so E = iR + u. Substituting i = C du/dt gives RC du/dt + u = E. This is a first-order linear equation; separating variables as du/(E − u) = dt/(RC) and integrating yields u(t) = E + (u₀ − E)e^(−t/τ) with τ = RC. The two constants give the two limits directly: at t = 0 the voltage is u₀, and as t → ∞ the exponential vanishes and u → E.

What does the time constant τ mean?

τ = RC has units of seconds and is the single time scale of the process. After one τ the gap between the capacitor voltage and E has shrunk to 1/e ≈ 0.368 of its original value; charging from u₀ = 0 gives u(τ) = E(1 − 1/e) ≈ 0.632E. After 5τ the gap is only about 0.7%, which is why the process is usually called finished there. The current i(t) = ((E − u₀)/R)e^(−t/τ) is largest at t = 0 and decays with the same exponential. Note that τ depends only on the product RC: swapping R and C (for example halving R and doubling C) leaves the curve shape completely unchanged.

Check

With the default parameters τ = 40 × 0.05 = 2 s. Stepping to t = 0.10 s should read about u = 6(1 − e^(−0.05)) ≈ 0.293 V; stepping to t = 2.00 s should read about 6(1 − 1/e) ≈ 3.793 V; and stepping to t = 10.00 s should read about 6(1 − e^(−5)) ≈ 5.960 V, leaving a gap of only about 0.7%. Set C = 0.1 F so τ = 4 s; the reading at t = 2.00 s should drop to about 6(1 − e^(−0.5)) ≈ 2.361 V, clearly below 3.793 V. Change R from 40 Ω to 80 Ω and the curve should coincide with the previous one, because τ is 4 s in both cases. Conversely, halve R to 20 Ω while doubling C to 0.1 F and τ stays 2 s, so the curve matches the default exactly. Set u₀ = 10 V and E = 0 V; the reading at t = 2.00 s should be about 10/e ≈ 3.679 V. Set u₀ equal to E and the current reading is 0 A while the curve stays flat.

Explore further

Real circuits also have source resistance, capacitor leakage and the parasitic capacitance and inductance of the wires, all of which bend the curve away from a pure exponential, and the capacitor's equivalent series resistance adds a small voltage step at the instant of switching. What happens to the capacitor voltage if the source is not a constant EMF but changes in time, such as a square wave? What curve appears if you cascade two RC circuits? You can also revisit Ohm's law for the instantaneous response of a purely resistive loop, or Series and parallel to compare the rules for combining resistors with the energy storage of the capacitor here.