The formula this page uses
τ = R · C charging: V(t) = V₀(1 − e^(−t/τ)) discharging: V(t) = V₀·e^(−t/τ) f_c = 1 / (2πRC)
← Back to calculators
RC Circuit Calculator
Solve rc circuit problems with clear steps, notation, and a final check.
Ω
Calculate without using AI.
Evaluate the governing formula locally in your browser. Define each known quantity once, then change values to test another case instantly.
Choose an example or enter a formula.
Show your work Step-by-step method
- No account, AI request, or paid token is used.
Recent calculations
No calculations yet.
Applied math
RC Circuit Calculator explained
The short version
- An RC circuit is a resistor and a capacitor together, and it charges or discharges on a curve, not a straight line.
- The time constant τ = R × C is the one number that sets the pace. After 1τ the capacitor is 63.2% charged.
- It is practically finished after 5τ, at 99.3%, which is the rule of thumb every datasheet uses.
What each part means
SymbolWhat it means
R — ResistanceIn ohms. 470 kΩ is 470000 Ω.
C — CapacitanceIn farads. 10 µF is 0.00001 F, and forgetting this conversion is the classic error.
τ — Time constantR × C, in seconds. Ohms times farads really does give seconds.
V₀ — Supply or starting voltageThe final voltage when charging, or the initial voltage when discharging.
Show your work: a full example
- The circuita 9 V supply charging a 10 µF capacitor through 470 kΩ
- Convert to base unitsR = 470000 Ω and C = 0.00001 F
- Time constantτ = 470000 × 0.00001 = 4.7 s
- Voltage after one time constant9 × (1 − e⁻¹) = 9 × 0.632 = 5.689 V
- After three time constants, t = 14.1 s9 × (1 − e⁻³) = 9 × 0.950 = 8.552 V
- After five time constants, t = 23.5 s9 × (1 − e⁻⁵) = 9 × 0.993 = 8.939 V
- Read itpractically full at 23.5 s, and mathematically it never quite reaches 9 V
A second, different case
- A different case: discharging the same capacitorstart at 9 V, same R and C, so τ is still 4.7 s
- After one time constant9 × e⁻¹ = 9 × 0.368 = 3.311 V
- Note the patterncharging climbs to 63.2%, discharging falls to 36.8%, and the two always add to 100%
- After 10 seconds, which is 2.128 time constants9 × e^(−10/4.7) = 9 × 0.1191 = 1.072 V
- Now treat the same parts as a low-pass filterf_c = 1 ÷ (2π × 4.7)
- Cutoff frequency0.0339 Hz, which is extremely low
- Read itthis pair passes only very slow signals, which is exactly what a long time constant means in the frequency picture
Copy-ready example
a 9 V supply charging a 10 µF capacitor through 470 kΩ
The circuit
Questions about the RC Circuit Calculator
Why is 63.2% the magic number?
It is 1 − 1/e, and e is baked into the exponential solution of the circuit's differential equation. Nobody chose 63.2%; it falls out of the mathematics, which is why every RC circuit shares it regardless of the component values.
Does a capacitor ever finish charging?
Not exactly. The curve approaches the supply voltage forever without reaching it. Engineers call 5τ finished because 0.7% left over is smaller than the tolerance of the components themselves.
How do R and C each change the timing?
Both stretch it in direct proportion, since τ is their product. Doubling R to 940 kΩ or doubling C to 20 µF each pushes τ from 4.7 s to 9.4 s, and doing both together gives 18.8 s.
What is the connection to filters?
The same pair is a low-pass filter with cutoff f_c = 1/(2πRC). A long time constant means a low cutoff, so the 4.7 s circuit above passes only signals below 0.034 Hz and smooths everything faster away.