Capacitor Voltage Over Time: Formula and Examples

Capacitor voltage over time depends on how the capacitor is connected and on the current flowing into or out of it. For a capacitor charging or discharging through a resistor, the voltage follows an exponential curve; enter capacitance, resistance, supply voltage, and elapsed time to calculate it.

Formula

For an initially uncharged capacitor charging from a constant supply voltage through a resistor, the voltage is:

V_C(t) = V_S [1 − e^−t/(RC)]

• V_C(t) is the capacitor voltage at elapsed time t, in volts (V).
• V_S is the constant supply voltage, in volts (V).
• R is the series resistance, in ohms (Ω).
• C is capacitance, in farads (F).
• t is elapsed time, in seconds (s), measured from the start of charging.
• e is the mathematical constant approximately equal to 2.71828.

The calculator accepts capacitance in microfarads (µF), so it converts to farads by multiplying by 0.000001. The time constant is τ = RC, in seconds. The voltage as a fraction of its final value is 1 − e^−t/τ; multiplying that fraction by 100 gives the percentage output.

Worked Example

Using the default values—5 V supply, 1,000 Ω resistance, 1,000 µF capacitance, and 1 second elapsed—the calculation is:

1. Convert capacitance: 1,000 µF × 0.000001 = 0.001 F.
2. Calculate the time constant: τ = 1,000 Ω × 0.001 F = 1 s.
3. Substitute into the charging equation: V_C(1) = 5[1 − e^−1/1] V.
4. Evaluate: 5(1 − 0.367879) = 3.1606 V, or 3.161 V to three decimal places.

The time constant is 1.000 s, and the capacitor reaches about 63.2% of its final voltage after one time constant. The percentage output rounds to 63.2%.

How the Charging Curve Works

A capacitor stores electrical charge. Its voltage cannot jump instantly in response to a finite current: current adds or removes charge over time. In a simple resistor-capacitor (RC) circuit, the resistor limits charging current. At first, the difference between the supply voltage and capacitor voltage is large, so current is relatively high. As the capacitor voltage approaches the supply, that difference and the current decrease.

This produces an exponential rise, not a straight-line ramp. The voltage gets increasingly close to the supply but, in the ideal mathematical model, never reaches it at a finite time. A real circuit has tolerances and leakage, so its behavior will not match the ideal curve perfectly.

For a capacitor that starts at a nonzero voltage V₀, the more general form is:

V_C(t) = V_F + (V₀ − V_F)e^−t/τ

Here V_F is the final steady voltage and τ is the time constant. This form covers charging or discharging toward a final voltage. For discharge through a resistor to 0 V, it simplifies to V_C(t) = V₀e^−t/(RC).

Reading the Time Constant

The time constant provides a quick way to estimate how far a capacitor has moved toward its final voltage. For an ideal first-order RC circuit:

• After 1τ, charging reaches about 63.2% of the voltage change; discharging has about 36.8% of its initial voltage left.
• After 2τ, charging reaches about 86.5%; discharging has about 13.5% left.
• After 3τ, charging reaches about 95.0%; discharging has about 5.0% left.
• After 5τ, charging is about 99.3% complete; discharging has about 0.7% left.

These are percentages of the change between the initial and final voltage, not necessarily percentages of a supply if the capacitor starts somewhere other than 0 V. For example, if it starts at 2 V and is moving toward 10 V, one time constant covers about 63.2% of the 8 V difference.

When This Formula Applies

The simple formula is useful for estimating timing in an RC circuit, such as a basic delay, a filter’s transient response, or the time a capacitor takes to approach a steady voltage. It assumes a constant resistance, a constant final voltage, and an ideal capacitor. The resistance in the equation is the total effective series resistance seen by the capacitor, not automatically the value of one component if other resistances matter.

For more complicated circuits, the voltage may not follow a single exponential. Multiple capacitors and resistors can create several time constants; active components, changing sources, or nonlinear components can also alter the response. If the capacitor current is known instead, the underlying relationship is i = C(dV/dt). With constant current, voltage changes linearly: ΔV = I·t/C. That is a different condition from a resistor charging a capacitor from a fixed supply.

Common Mistakes and Limits

• Mixing units: Use ohms, farads, and seconds in RC. This calculator takes capacitance in µF and converts it internally. A value entered as farads when the field expects µF will be off by a factor of one million.
• Using the wrong starting condition: The calculator assumes the capacitor begins at 0 V. Use the general equation when it has an initial charge.
• Confusing resistance with reactance: The formula describes a transient through a resistor; it is not by itself an AC capacitor-impedance formula.
• Expecting an exact final value at a finite time: The ideal exponential approaches its final value asymptotically. A practical design may treat a chosen percentage as “close enough,” but that threshold depends on the application.
• Ignoring real component variation: Capacitance tolerance, resistor tolerance, leakage, and measurement loading can change observed timing.

Calculated values are ideal estimates, not a guarantee of circuit performance. Capacitors can retain charge after power is removed. Follow appropriate circuit procedures and the component’s documentation; do not assume a capacitor is discharged just because a predicted time has elapsed.

Frequently Asked Questions

What is the formula for capacitor voltage over time?
For charging from 0 V through a resistor toward a constant supply, V_C(t) = V_S[1 − e^−t/(RC)]. For discharge to 0 V from an initial voltage, V_C(t) = V₀e^−t/(RC).

How long does it take a capacitor to charge?
There is no finite time at which the ideal equation reaches exactly 100% of its final voltage. It reaches about 95% after 3τ and 99.3% after 5τ, where τ = RC.

What does RC mean in the equation?
RC is the time constant, τ. With resistance in ohms and capacitance in farads, its unit is seconds. A larger resistance or capacitance makes the voltage change more slowly.

Does a capacitor charge linearly?
Not in a resistor-fed RC circuit with a constant supply: the voltage rises exponentially. A constant-current source can produce a linear voltage ramp, according to ΔV = I·t/C.

Can I use this formula if the capacitor is already charged?
Use V_C(t) = V_F + (V₀ − V_F)e^−t/(RC), with the actual initial voltage V₀ and final voltage V_F. The calculator’s charging result specifically assumes V₀ = 0.

Related

• Make Rugs Friday Patch Schedule
• Convert WordPress Multisite to Single Site
• Dig and Clean Codes: Meaning & How to Use

in v | Supply voltage | V | 5 | 0
in r | Resistance | ohm | 1000 | 0
in c | Capacitance | uF | 1000 | 0
in t | Elapsed time | s | 1 | 0
out vc | Capacitor voltage | V | v * (1 - exp(-t / (r * c * 0.000001))) | 3
out tau | Time constant | s | r * c * 0.000001 | 3
out pct | Percent of final voltage | % | 100 * (1 - exp(-t / (r * c * 0.000001))) | 1