Voltage Across a Capacitor Equation: RC Charging & Discharging Formulas

The fundamental voltage across a capacitor equation in a static DC state is V = Q / C (Voltage = Charge / Capacitance). However, in practical timing, filtering, and power-supply circuits, you almost always need the transient RC charging equation: V_c(t) = V_s(1 - e^-t/RC). This tells you the exact voltage across the capacitor at any given time t as it charges through a resistor from a source voltage. Below, we break down every symbol, provide real-world component data, and walk through step-by-step calculations with strict unit tracking to ensure your designs work on the bench, not just in theory.

The Core Equations and Symbol Definitions

When a capacitor charges through a resistor, the voltage across its terminals follows an exponential curve. It starts at 0V and asymptotically approaches the source voltage. The governing equation for the charging phase is:

V_c(t) = V_s (1 - e^{-t / RC})

For the discharging phase (when the source is removed and the capacitor bleeds through the resistor), the equation shifts to:

V_c(t) = V₀ (e^{-t / RC})

Table 1: Symbol Definitions and Standard SI Units

Symbol	Definition	Standard Unit	Unit Abbreviation
Vc(t)	Voltage across the capacitor at time t	Volts	V
Vs	Source voltage (charging phase)	Volts	V
V0	Initial voltage (discharging phase)	Volts	V
t	Time elapsed since charging/discharging began	Seconds	s
R	Resistance in the circuit path	Ohms	Ω
C	Capacitance value	Farads	F
e	Euler's number (mathematical constant)	Dimensionless	≈ 2.71828
τ (tau)	Time constant (R × C)	Seconds	s

Real-World Component Data & Magnitude Check

Textbook problems assume ideal capacitors. On the bench, parasitic elements like Equivalent Series Resistance (ESR) and DC bias derating alter your results. The table below provides real-world data for common capacitor types you will use in RC networks, including their typical ESR and how they behave at the 1-time-constant (τ) mark.

Table 2: Real-World Capacitor Characteristics in RC Circuits (Assuming 5V Source, 10kΩ Resistor)

Capacitor Type	Typical Value	Voltage Rating	Typical ESR	Time Constant (τ)	Vc at 1τ (63.2%)	Real-World Edge Case
Ceramic (X7R MLCC)	100 nF	50V	< 0.05 Ω	1.0 ms	3.16 V	Capacitance drops up to 50% at rated DC bias voltage.
Aluminum Electrolytic	100 μF	25V	0.2 Ω	1.0 s	3.16 V	High leakage current causes final voltage to fall slightly short of Vs.
Film (Polypropylene)	1 μF	400V	0.01 Ω	10 ms	3.16 V	Highly stable, but physically massive for high capacitance values.
Supercapacitor (EDLC)	1 F	2.7V	0.5 Ω	10,000 s	1.70 V	Strict 2.7V max limit; requires active balancing in series strings.

Note: Data reflects typical 2026 component specifications from major distributors. Always check the manufacturer datasheet for DC bias derating curves, especially for Class II ceramics (X5R, X7R), as detailed in SparkFun's capacitor guide.

Rearranged Forms for Circuit Design

In practical design, you rarely solve for voltage. Usually, you know your target voltage threshold (like a microcontroller's logic-high trigger) and need to calculate the required time, resistor, or capacitor. Here are the algebraically rearranged forms for the charging equation:

• Solving for Time (t): t = -R × C × ln(1 - (Vc / Vs))
• Solving for Resistance (R): R = -t / (C × ln(1 - (Vc / Vs)))
• Solving for Capacitance (C): C = -t / (R × ln(1 - (Vc / Vs)))

For the discharging equation, the rearranged form to find time is:

• Solving for Time (t): t = -R × C × ln(Vc / V0)

Worked Examples with Unit Tracking

Skipping unit conversion is the number one reason RC timing circuits fail on the first prototype. Let's walk through two common scenarios with strict unit tracking.

Problem 1: ESP32 GPIO Delay Trigger (Charging)

Scenario: You want an RC network to delay an ESP32 boot sequence. The GPIO pin registers a logic HIGH at 3.3V. Your system rail is 5V. You have a 10 kΩ resistor and a 10 μF capacitor. How long after power-on will the GPIO trigger?

Knowns:

• V_s = 5 V
• V_c = 3.3 V
• R = 10,000 Ω (converted from kΩ)
• C = 0.00001 F (converted from μF)

Equation: t = -R × C × ln(1 - (Vc / Vs))

Step-by-Step Solution:

1. Calculate the RC time constant: 10,000 Ω × 0.00001 F = 0.1 seconds.
2. Calculate the voltage ratio: 3.3 V / 5 V = 0.66.
3. Subtract from 1: 1 - 0.66 = 0.34.
4. Take the natural log: ln(0.34) ≈ -1.0788.
5. Multiply by -RC: -0.1 × -1.0788 = 0.10788 seconds.

Unit Tracking Verification:
Ω × F = (Volts / Amps) × (Amps × seconds / Volts) = seconds.
Result: The GPIO will trigger 107.9 milliseconds after power is applied.

Problem 2: High-Voltage Power Supply Bleeder (Discharging)

Scenario: A 400V DC bus uses a 470 μF smoothing capacitor. For safety, a 220 kΩ bleeder resistor is placed in parallel. How long after unplugging the unit does the voltage drop to a safe 50V?

Knowns:

• V₀ = 400 V
• V_c = 50 V
• R = 220,000 Ω
• C = 0.00047 F

Equation: t = -R × C × ln(Vc / V0)

Step-by-Step Solution:

1. Calculate RC: 220,000 Ω × 0.00047 F = 103.4 seconds.
2. Calculate voltage ratio: 50 V / 400 V = 0.125.
3. Take the natural log: ln(0.125) ≈ -2.0794.
4. Multiply by -RC: -103.4 × -2.0794 = 215.01 seconds.

Result: It takes roughly 3 minutes and 35 seconds for the capacitor to discharge to a safe touch voltage. Always verify with a multimeter before working on the board.

Assumptions, Unit Traps, and Edge Cases

The equations above are mathematically perfect, but physical components are not. Here is what you need to know to bridge the gap between the math and the multimeter.

When the Formula Applies (and Its Assumptions)

These equations assume a lumped parameter model with an ideal voltage source, an ideal resistor, and an ideal capacitor. They apply perfectly to low-frequency DC timing circuits (like 555 timers or MCU reset delays). They begin to break down in high-frequency RF applications where the parasitic inductance (ESL) of the capacitor leads creates a resonant tank, or in high-current pulse circuits where the ESR causes an immediate voltage drop (V = I × ESR) at t=0 before the exponential curve even begins.

The Unit Mistakes That Break Your Math

If your calculated time is wildly off, check these three traps:

1. The Microfarad Trap: Entering '10' instead of '0.00001' for a 10 μF capacitor. Always convert to base SI units (Farads) before calculating.
2. The Millisecond Trap: Forgetting that the output of the equation is in seconds. If you need milliseconds, multiply the final answer by 1,000.
3. The Natural Log vs Log10 Trap: Using the base-10 log button on your calculator instead of the natural log (ln) button. The 'e' in the equation dictates the use of ln.

Sanity Check: What a Realistic Magnitude Looks Like

Develop an intuition for the RC product. A 1 kΩ resistor and a 1 μF capacitor yield a 1 ms time constant. If you are designing a debounce circuit for a mechanical switch and your math spits out 45 seconds, you missed a decimal place. Conversely, if you are designing a high-voltage bleeder network and your math says it discharges in 0.002 seconds, your resistor value is dangerously low and will likely overheat and fail during normal operation. For deeper component selection criteria, refer to the All About Circuits guide on capacitor types.

Edge Case: DC Bias Derating in MLCCs

If you use a 10 μF X7R ceramic capacitor in a 5V circuit, the actual capacitance might only be 6 μF due to DC bias derating. This shrinks your time constant by 40%. If your timing circuit is running faster than the math predicts, swap the X7R for a C0G/NP0 dielectric (which doesn't suffer from derating but is limited to lower capacitance values) or increase the physical package size (e.g., moving from 0805 to 1206) to mitigate the effect.