This calculator finds the time constant of an RC circuit and shows you how the capacitor voltage changes over time during charging and discharging. An RC circuit consists of a resistor and a capacitor connected in series. When a voltage is applied, the capacitor charges through the resistor and the voltage across it rises exponentially toward the supply voltage. The time constant tau = R x C in seconds tells you how quickly that happens. After one time constant, the capacitor has reached 63.2 percent of the supply voltage. After two time constants it reaches 86.5 percent. After five time constants, at 99.3 percent, the capacitor is considered fully charged for practical purposes. The same exponential relationship applies in reverse during discharge: after one time constant the voltage has fallen to 36.8 percent of its starting value, and after five time constants it is essentially zero. RC circuits appear throughout electronics. They are the heart of simple low-pass and high-pass filters, debounce circuits for mechanical switches, sample-and-hold circuits, peak detectors, integrators and differentiators. The 555 timer IC uses an RC network internally to generate its output pulses. Enter the resistance in kilohms and capacitance in microfarads for typical timing circuits. The calculator returns the time constant, the voltage at each multiple of tau from 1 to 5, and the time to reach any percentage of the supply voltage that you specify.
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kΩ
µF
V
1 s
time constant τ = RC
At 1τ (63.2%)1 s
At 2τ (86.5%)2 s
At 3τ (95.0%)3 s
At 4τ (98.2%)4 s
At 5τ (99.3%)5 s
R input in kilohms, C input in microfarads. At 1 tau the capacitor charges to 63.2% of supply voltage; at 5 tau it is 99.3% (fully charged for practical purposes).
How it works
The time constant is tau = R x C, with R in ohms (kilohm input x 1000) and C in farads (microfarad input divided by 1,000,000). Tau comes out in seconds. The charging voltage at time t is V(t) = Vs x (1 minus e^(-t/tau)). The discharge voltage at time t is V(t) = V0 x e^(-t/tau). The percentage of full charge at each time constant n is (1 minus e^(-n)) x 100: at 1 tau this gives (1 minus 0.368) = 63.2%, at 2 tau 86.5%, at 3 tau 95.0%, at 4 tau 98.2%, at 5 tau 99.3%. To find the time to reach any fraction p of Vs, invert the formula: t = minus tau x ln(1 minus p).
Worked example
R = 10 kilohms (10,000 ohms) and C = 100 microfarads (0.0001 F). The time constant is tau = 10,000 x 0.0001 = 1.0 s. With a 5 V supply, the capacitor charges to 63.2% of 5 V = 3.16 V at 1 tau (1.0 s), to 86.5% = 4.33 V at 2 tau (2.0 s), and so on up to 99.3% = 4.97 V at 5 tau (5.0 s). These match the default values above.
