Capacitor Charging Discharging Equations: Formulas, Tables, and Worked Examples

The capacitor charging discharging equations describe the exponential voltage change across a capacitor in a resistor-capacitor (RC) circuit over time. Governed by the time constant τ (tau), where τ = R × C, these formulas dictate that a charging capacitor reaches exactly 63.2% of its source voltage after one time constant, while a discharging capacitor drops to 36.8% of its initial voltage in the same timeframe. Below is the complete mathematical framework, real-world data, and strict unit-tracking examples you need for bench work and circuit design.

The Core RC Equations and Symbol Definitions

There are two primary equations depending on the state of the circuit. For a capacitor charging from 0V toward a DC source voltage, the voltage at any time t is:

V(t) = V_s * (1 - e^(-t / (R * C)))

For a capacitor discharging from an initial voltage toward 0V, the equation is:

V(t) = V_0 * e^(-t / (R * C))

Symbol	Parameter	Unit (SI Base)	Definition
V(t)	Voltage at time t	Volts (V)	The instantaneous voltage across the capacitor at time t.
V_s	Source Voltage	Volts (V)	The constant DC supply voltage the capacitor is charging toward.
V_0	Initial Voltage	Volts (V)	The starting voltage across the capacitor before discharging begins.
R	Resistance	Ohms (Ω)	The total series resistance limiting current flow.
C	Capacitance	Farads (F)	The capacitance value of the component.
t	Time	Seconds (s)	Elapsed time since the charge or discharge cycle began.
e	Euler's Number	Dimensionless	Mathematical constant ≈ 2.71828, the base of the natural logarithm.
τ (tau)	Time Constant	Seconds (s)	The product of R and C (τ = R × C). Represents the time to reach 63.2% charge.

Real-World RC Time Constants

To ground these formulas in reality, here is a data-dense table of common RC combinations you will encounter on the bench, showing their time constants and practical applications.

Application	Resistance (R)	Capacitance (C)	Time Constant (τ)	Time to ~99% (5τ)
Microcontroller GPIO Debounce	10 kΩ	100 nF (0.1 µF)	1.0 ms	5.0 ms
Audio High-Pass Filter (Crossover)	10 kΩ	4.7 µF	47.0 ms	235.0 ms
555 Timer Monostable Delay	1 MΩ	10 µF	10.0 s	50.0 s
Power Supply Soft-Start Gate	100 kΩ	100 µF	10.0 s	50.0 s
High-Voltage Snubber Network	47 Ω	10 nF	470 ns	2.35 µs

Rearranged Forms: Solving for Any Variable

On the bench, you rarely just solve for V(t). Usually, you know the target voltage and need to find the required resistor, or you are measuring a delay and need to calculate parasitic capacitance. Here are the algebraically rearranged forms for the charging equation. (Note: For discharging, replace (1 - V(t)/V_s) with (V(t)/V_0)).

• Solving for Time (t): t = -R * C * ln(1 - (V(t) / V_s))
• Solving for Resistance (R): R = -t / (C * ln(1 - (V(t) / V_s)))
• Solving for Capacitance (C): C = -t / (R * ln(1 - (V(t) / V_s)))
• Solving for Source Voltage (V_s): V_s = V(t) / (1 - e^(-t / (R * C)))

Note: 'ln' represents the natural logarithm (base e). Ensure your calculator is not set to base-10 log.

Assumptions, Unit Traps, and Realistic Magnitudes

When the Formula Applies (and When it Fails)

These equations assume an ideal DC step input, constant resistance, and an ideal capacitor. They break down in three common real-world scenarios:

1. Equivalent Series Resistance (ESR): Real capacitors (especially electrolytics) have internal resistance. At t=0, the ideal formula predicts infinite current. In reality, ESR limits the instantaneous current, causing a small immediate voltage step before the exponential curve begins.
2. Dielectric Absorption: In long time-constant circuits (τ > 10s), the dielectric material 'soaks up' charge and releases it slowly, causing the late-stage discharge curve to deviate from the pure exponential model.
3. Non-Linear Loads: If the capacitor is discharging into a constant-current load (like an LED driver or switching regulator) rather than a fixed resistor, the voltage drops linearly, not exponentially. Use I = C(dV/dt) instead.

The Unit Mistake That Breaks 90% of Calculations

If you want to use prefixes without converting to decimals, use these shortcut pairings:

• Megaohms (MΩ) × Microfarads (µF) = Seconds (s)
• Kiloohms (kΩ) × Microfarads (µF) = Milliseconds (ms)
• Kiloohms (kΩ) × Nanofarads (nF) = Microseconds (µs)

What a Realistic Answer Magnitude Looks Like

In 99% of hobbyist and commercial low-voltage circuits, your time constant (τ) will fall between 10 µs and 5 seconds. If you calculate a time constant of 4,500 seconds for a simple 555 timer circuit, you have almost certainly forgotten to convert microfarads to base Farads. Furthermore, a charging capacitor is considered 'fully charged' for all practical digital logic purposes at 5τ (99.3%). Designing a circuit to wait for 100% theoretical charge will result in an infinite wait time.

Worked Examples with Strict Unit Tracking

Let's run two common bench scenarios, explicitly tracking units to prevent the prefix trap. For deeper theoretical derivations of these exponential responses, refer to the All About Circuits RC Time Constant chapter or the Electronics Tutorials RC Waveform guide.

Problem 1: Finding Voltage During Charging

Scenario: You are building a soft-start circuit. A 220 µF capacitor is charged through a 4.7 kΩ resistor from a 12V DC supply. What is the voltage across the capacitor exactly 2.5 seconds after power is applied?

Step 1: Convert to SI Base Units

• V_s = 12 V
• R = 4.7 kΩ = 4,700 Ω
• C = 220 µF = 0.00022 F
• t = 2.5 s

Step 2: Calculate the Time Constant (τ)

• τ = R × C
• τ = 4,700 Ω × 0.00022 F = 1.034 seconds

Step 3: Apply the Charging Equation

• V(t) = V_s * (1 - e^(-t / τ))
• V(2.5) = 12 * (1 - e^(-2.5 / 1.034))
• V(2.5) = 12 * (1 - e^(-2.4178))
• V(2.5) = 12 * (1 - 0.0891) (Intermediate e^x value)
• V(2.5) = 12 * 0.9109
• V(2.5) = 10.93 V

Sanity Check: 2.5 seconds is roughly 2.4 time constants. At 2τ, charge is ~86%; at 3τ, charge is ~95%. An answer of 10.93V (91% of 12V) perfectly aligns with the expected magnitude.

Problem 2: Finding Time During Discharging

Scenario: A sensor node wakes up, reads a 10 µF holding capacitor charged to 5.0V, and goes back to sleep. The sleep current acts as a 100 kΩ discharge resistor. How long will it take for the capacitor voltage to drop to 1.5V (the brownout threshold of the MCU)?

Step 1: Convert to SI Base Units

• V_0 = 5.0 V
• V(t) = 1.5 V
• R = 100 kΩ = 100,000 Ω
• C = 10 µF = 0.00001 F

Step 2: Calculate the Time Constant (τ)

• τ = 100,000 Ω × 0.00001 F = 1.0 second

Step 3: Rearrange Discharging Equation for Time (t)

• Original: V(t) = V_0 * e^(-t / τ)
• Divide by V_0: V(t) / V_0 = e^(-t / τ)
• Take natural log: ln(V(t) / V_0) = -t / τ
• Isolate t: t = -τ * ln(V(t) / V_0)

Step 4: Execute the Math

• t = -1.0 * ln(1.5 / 5.0)
• t = -1.0 * ln(0.3)
• t = -1.0 * (-1.20397) (Intermediate ln value)
• t = 1.204 seconds

Sanity Check: The target voltage (1.5V) is 30% of the initial voltage (5V). We know that at 1τ (1 second), the voltage drops to 36.8%. Since 30% is lower than 36.8%, the time must be slightly greater than 1τ. Our answer of 1.204 seconds is logically sound.