RC circuits set timing delays and filter signals. Enter the resistance and capacitance to get the time constant, the time to charge almost fully and the filter's cut-off frequency.
Enter the resistance in ohms.
Enter the capacitance in µF.
Read τ, 5τ and fc.
Time constant: τ = R × C
Cut-off: fc = 1 ÷ (2π R C)
When a capacitor charges or discharges through a resistor, it does not change instantly. The voltage follows an exponential curve whose speed is set by one number, the time constant τ = R × C. This calculator takes the resistance in ohms and the capacitance in microfarads, and returns τ in seconds, the time 5τ to reach practically full charge, and the cut-off frequency of the same pair used as a filter.
RC circuits are some of the simplest and most useful in electronics. They produce delays, such as an LED that fades out slowly or a staircase timer; they debounce mechanical switches; they filter noise from sensor signals; and they set the frequency of 555 timer circuits. The topic appears in Class 12 physics and every basic electronics course.
1. Convert the capacitance to farads: 1 µF = 10⁻⁶ F.
2. Multiply by the resistance in ohms: τ = R × C, in seconds.
3. For charging from zero towards a supply V₀: V(t) = V₀(1 − e^(−t/τ)).
4. For discharging from V₀: V(t) = V₀ e^(−t/τ).
5. Practically full charge or discharge takes about 5τ.
6. Cut-off frequency of the RC filter: fc = 1 ÷ (2πRC).
While charging, the voltage across the resistor is the supply minus the capacitor voltage, V₀ − V. So the current is (V₀ − V) ÷ R. That current adds charge to the capacitor, so C × dV/dt = (V₀ − V) ÷ R. The rate of change is proportional to how far the capacitor still has to go, and a quantity that closes a gap at a rate proportional to the gap follows an exponential. Solving gives V = V₀(1 − e^(−t/RC)). As the capacitor fills, the current shrinks, so the approach slows down and never quite finishes.
After one time constant, e^(−1) ≈ 0.368, so the capacitor has reached about 63.2% of the final voltage during charging, or fallen to 36.8% during discharging. After 2τ it is about 86.5%, after 3τ 95%, and after 5τ about 99.3%, which engineers treat as complete. The units work out neatly: ohms times farads is seconds. A 10 kΩ resistor with a 100 µF capacitor gives one second, a handy reference when designing delays by feel.
A capacitor's opposition to AC, its reactance, is Xc = 1 ÷ (2πfC), which falls as frequency rises. In an RC low-pass filter, with the output across the capacitor, low frequencies pass and high frequencies are shorted away. The crossover is where Xc equals R, which gives fc = 1 ÷ (2πRC). At that frequency the output amplitude is 1/√2, about 70.7%, of the input, which is a 3 dB drop in power. Swap R and C and you get a high-pass filter with the same fc.
Nikhil is building a soft-off night lamp, where a 10 µF capacitor discharges through a 47 kΩ resistor to fade an LED driver. He wants to know how quickly the fade happens and what the cut-off frequency of the pair is.
τ = R × C: = 47000 Ω × 10 × 10⁻⁶ F = 0.47 s
Practically full charge (≈99.3%): 5τ = 2.35 s
Cut-off frequency: fc = 1 ÷ (2π τ) = 0.3386 Hz
Answer: Time constant τ 0.47 s; Time to ~99% (5τ) 2.35 s; Cut-off frequency 0.3386 Hz
Entering capacitance in farads or nanofarads when the calculator expects microfarads.
Expecting the capacitor to be fully charged after one time constant; it is only about 63% there.
Forgetting that the charging resistance includes the source's internal resistance and any other series resistance.
Ignoring the wide tolerance of electrolytic capacitors, which makes timing approximate.
Assuming the output voltage stays unaffected when a load draws current from the capacitor.
Designing time delays and soft-start circuits.
Debouncing push buttons and mechanical switches for microcontrollers.
Low-pass filtering of noisy sensor readings and audio signals.
Setting frequencies in 555 timer and oscillator circuits.
Class 12 physics experiments on charging and discharging capacitors.
Why 5τ for full charge?
The charge approaches its final value exponentially; after 5τ it is within 1% for practical purposes.
Is fc the −3 dB point?
Yes, the output is 70.7% of the input amplitude at fc.