The discharging a capacitor formula calculates the voltage remaining across a capacitor at any given time after it begins discharging through a resistor. The direct answer for the voltage at time t is:

V(t) = V0 × e-t/RC

This equation models the exponential decay of electrical potential in a simple Resistor-Capacitor (RC) circuit. Whether you are designing a power supply bleed network, sizing a camera flash dump resistor, or debugging an RC snubber, understanding how to manipulate and apply this formula is mandatory for predictable circuit behavior.

The Core Formula and Symbol Definitions

To use the formula correctly, every variable must be defined and converted to base SI units before calculation. The natural logarithm base (e) is a mathematical constant, while the remaining variables represent physical circuit properties.

Symbol Definition Standard SI Unit Common Component Markings
V(t) Voltage across the capacitor at time t Volts (V) Measured via multimeter/oscilloscope
V0 Initial voltage across the capacitor at t = 0 Volts (V) Source voltage or pre-charge level
e Euler's number (base of natural logarithm) Dimensionless (~2.71828) N/A
t Time elapsed since discharge began Seconds (s) ms, μs in fast circuits
R Resistance of the discharge path Ohms (Ω) kΩ, MΩ for bleed resistors
C Capacitance of the component Farads (F) μF, nF, pF
τ (tau) Time constant (where τ = R × C) Seconds (s) Calculated value

The product of R × C is known as the time constant (τ). After one time constant (1τ), the capacitor voltage drops to approximately 36.8% of its initial value. According to HyperPhysics, this exponential decay means the voltage theoretically never reaches absolute zero, but for practical engineering, it is considered fully discharged after 5τ.

Rearranged Forms: Solving for Time, Resistance, and Capacitance

In bench work and design, you rarely just solve for V(t). You usually know the target voltage and need to size the resistor or calculate the delay. Here are the algebraic rearrangements of the discharging a capacitor formula, solving for each variable:

  • Solve for time (t):
    t = -R × C × ln(V(t) / V0)
    Use case: Calculating how long a power supply remains lethal after unplugging.
  • Solve for resistance (R):
    R = -t / (C × ln(V(t) / V0))
    Use case: Sizing a bleed resistor to meet a specific safety discharge timeframe.
  • Solve for capacitance (C):
    C = -t / (R × ln(V(t) / V0))
    Use case: Selecting a capacitor to hold a logic line high for a specific dropout duration.
  • Solve for initial voltage (V0):
    V0 = V(t) / e-t/RC
    Use case: Forensic analysis to determine the original charge state based on a later measurement.

Worked Examples with Strict Unit Tracking

The most common reason calculations fail on the workbench is unit mismatch. The formula requires base SI units: Farads, Ohms, and Seconds. Below are two real-world scenarios with explicit dimensional tracking.

Example 1: High-Voltage Power Supply Bleed Resistor

Scenario: A 400V DC bus uses a 470 μF electrolytic capacitor. A 100 kΩ bleed resistor is placed in parallel to discharge the cap when the unit is turned off. What is the voltage across the capacitor 15 seconds after power-off?

  1. Identify and convert variables to SI units:
    • V0 = 400 V
    • C = 470 μF = 470 × 10-6 F = 0.00047 F
    • R = 100 kΩ = 100 × 103 Ω = 100,000 Ω
    • t = 15 s
  2. Calculate the time constant (τ):
    • τ = R × C
    • τ = (100,000 Ω) × (0.00047 F)
    • Unit check: [Ω] × [F] = [s]
    • τ = 47 seconds
  3. Apply the discharging a capacitor formula:
    • V(t) = 400 × e-15 / 47
    • V(t) = 400 × e-0.3191
    • V(t) = 400 × 0.7269
    • V(t) = 290.7 V

Result: After 15 seconds, the capacitor still holds nearly 291V. This highlights why 15 seconds is insufficient for safety; at 5τ (235 seconds), it drops below 3V.

Example 2: Sizing a Strobe Light Dump Resistor

Scenario: A camera strobe circuit charges a 200 μF capacitor to 300V. The flash tube triggers, acting as a dynamic load. We want the voltage to drop to 50V in exactly 2 milliseconds (0.002 s) to freeze motion. What effective resistance must the flash tube path present?

  1. Identify and convert variables:
    • V0 = 300 V
    • V(t) = 50 V
    • C = 200 μF = 0.0002 F
    • t = 2 ms = 0.002 s
  2. Select the rearranged formula for R:
    • R = -t / (C × ln(V(t) / V0))
  3. Calculate the natural log term:
    • ln(50 / 300) = ln(0.1667) = -1.7917
  4. Solve for R:
    • R = -0.002 / (0.0002 × -1.7917)
    • R = -0.002 / -0.0003583
    • R = 5.58 Ω

Result: The flash tube and series wiring must present an effective resistance of 5.58 Ω during the discharge event. For deeper transient analysis of RC networks, All About Circuits provides excellent foundational breakdowns of time constant behavior.

Boundary Conditions: Assumptions, Unit Traps, and Realistic Magnitudes

The discharging a capacitor formula is an idealized mathematical model. Applying it blindly to complex physical circuits without understanding its boundaries leads to design failures.

When the Formula Applies (and its Assumptions)

  • Constant Resistance: The formula assumes R is static. In reality, components like incandescent bulbs, thermistors, and gas-discharge tubes (like the strobe in Example 2) have dynamic resistance that changes with temperature or voltage. The formula provides an effective average R, not an exact instantaneous model.
  • Ideal Capacitor: It assumes zero Equivalent Series Resistance (ESR) and zero Equivalent Series Inductance (ESL). In high-speed snubber circuits (where t is in nanoseconds), ESL dominates the initial discharge spike, rendering the pure RC formula inaccurate for the first few microseconds.
  • No Parasitic Leakage: It assumes the only discharge path is through R. In high-impedance sample-and-hold circuits, the capacitor's internal dielectric leakage and the PCB surface contamination will alter the decay curve.

Unit Mistakes That Break the Math

The most frequent errors occur when engineers plug component markings directly into the equation without converting to base SI units:

  • The Microfarad Trap: Plugging "470" into C instead of "0.00047". This inflates the time constant by a factor of one million.
  • The Millisecond Trap: Using "5" for time instead of "0.005" when working with oscilloscope timebases.
  • The Reciprocal Error: Forgetting the negative sign in the exponent. If you calculate e+t/RC, you model an impossible infinite voltage explosion rather than a decay.

Realistic Answer Magnitudes

Knowing what a "normal" time constant looks like prevents sanity-check failures. If your calculation yields a τ of 400 seconds for an audio crossover, you have made a math error.

Application Typical R Range Typical C Range Expected τ Magnitude
High-Frequency Snubber 10 Ω - 100 Ω 1 nF - 10 nF 10 ns - 1 μs
Audio Crossover / Filter 1 kΩ - 10 kΩ 10 nF - 1 μF 10 μs - 10 ms
Microcontroller Reset Delay 10 kΩ - 100 kΩ 1 μF - 10 μF 10 ms - 1 s
HV Power Supply Bleed 100 kΩ - 2 MΩ 100 μF - 1000 μF 10 s - 5 minutes

Frequently Asked Questions

How long does it take for a capacitor to fully discharge?

Mathematically, the exponential decay curve of the discharging a capacitor formula never reaches exactly 0V. Practically, electrical engineering standards define "fully discharged" as 5 time constants (5τ). At 5τ, the remaining voltage is less than 1% (specifically 0.67%) of the initial voltage V0. For safety-critical applications like high-voltage DC bus bleeds, some standards require calculating the time to drop below an absolute threshold (e.g., 50V) rather than relying solely on the 5τ rule.

Does the discharging a capacitor formula work for AC circuits?

No. The formula V(t) = V0e-t/RC specifically models the transient DC decay of stored energy. In continuous AC circuits, capacitors do not "discharge" to zero and stay there; they continuously charge and reverse-charge. For AC analysis, you must abandon the time-domain decay formula and instead use complex impedance (Zc = 1 / jωC) to calculate RMS voltage drops and phase shifts.

Why is my measured discharge time slower than the formula predicts?

If your oscilloscope shows a slower decay than your math dictates, you are likely encountering dielectric absorption (also known as battery action or soakage). In electrolytic and certain film capacitors, the dielectric material physically absorbs some charge during the initial charging phase and slowly releases it back into the circuit after the primary discharge path is activated. Additionally, if you are measuring a very high-impedance discharge with a standard multimeter, the meter's internal input impedance (typically 10 MΩ) becomes a parallel resistor, altering the effective R and skewing your measurement.

What happens if I short-circuit the capacitor instead of using a resistor?

If R approaches zero (a dead short), the formula predicts an instantaneous discharge (t = 0). In reality, the discharge rate is limited by the capacitor's internal Equivalent Series Resistance (ESR) and the parasitic inductance (ESL) of the shorting wire. This results in a massive, near-instantaneous current spike (I = V / ESR) that can vaporize the shorting tool, weld the capacitor terminals, or cause the electrolyte to vent explosively. Never use the ideal formula to justify shorting high-energy capacitors.