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Lesson 4 of 6 · 15 minutes · Free

Capacitors & RC Timing

Use the RC time constant to predict a capacitor’s response to a voltage step.

Try the experiment ↓Practice challenge ↓

See the idea

A capacitor stores energy in an electric field. Its voltage cannot change instantaneously in an ideal RC circuit with finite current. After a supply step, current is initially largest and then decays as the capacitor voltage rises.

The time constant τ equals resistance times capacitance, in seconds when R is in ohms and C is in farads. One time constant brings charging voltage to about 63.2% of the supply; five time constants brings it to about 99.3%.

The bench shows separate analytical experiments: charging starts at 0 V; discharging starts at the supply voltage with the source disconnected. Changing a setting recomputes that experiment rather than preserving a physical capacitor’s previous state.

τ = RC Charging: Vc = Vs(1 − e^(−t/τ)) Discharging: Vc = Vs e^(−t/τ)

RC sets the response time; supply voltage sets the charging limit.

Work through an example

  1. With R = 1,000 Ω and C = 1,000 µF = 0.001 F, τ = 1 s.
  2. At t = 1 s while charging from 0 V with 12 V: Vc = 7.59 V and I ≈ 4.41 mA.
  3. At t = 5 s, Vc ≈ 11.92 V. It approaches 12 V asymptotically rather than reaching it exactly at a fixed time.

Watch for this

A disconnected capacitor may retain charge. This is a browser experiment, not instructions for handling stored energy in equipment.

Predict, then test

  1. Play the charging trace or move the elapsed-time slider to one time constant.
  2. Double capacitance and compare the curve on the same time scale. Switch to discharge and watch voltage decay.

Your virtual bench

Change a setting. See the circuit respond.

Swipe the diagram sideways to explore the connections.

±12 VR1 1000 ΩC 1000 µFReturn reference: 0 V
Capacitor voltage0 V
Current (charge +)12 mA
Time constant τ1 s
Elapsed time0 s
010 s12 V0 V

Indigo: capacitor voltage · Rose cursor: selected elapsed time. Current sign is positive for charging, negative for discharging.

Model: Ideal single-resistor/single-capacitor step response; each setting starts a new charge or discharge experiment. Values are calculated, not equipment measurements. Browser simulation only—do not connect live equipment.

Check your understanding

If resistance doubles while capacitance stays fixed, the time constant:

Put it to work

Choose R and C for a 2-second time constant, then find charging voltage at t = 2 s with 12 V.

Show the worked solution

One option is 2,000 Ω and 1,000 µF. At one time constant Vc = 12(1 − e⁻¹) ≈ 7.59 V.

Try the experiment and answer the check, then record your completion.

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