The standard equation for discharging a capacitor through a fixed resistor is V(t) = V₀ × e(-t / RC). This formula calculates the remaining voltage across the capacitor at any specific time t after the discharge begins. Whether you are sizing a bleeder resistor for a 400V DC bus in a motor drive or calculating the dump time for a camera flash circuit, this exponential decay model is the foundational math you need.

The Core Equation and Symbol Definitions

Before plugging in numbers, you must understand what each variable represents and the physical assumptions baked into the math. The formula models a first-order RC (Resistor-Capacitor) circuit where a charged capacitor is connected across a resistive load.

Symbol Parameter Standard SI Unit Practical Description
V(t) Instantaneous Voltage Volts (V) The voltage remaining across the capacitor at time t.
V₀ Initial Voltage Volts (V) The voltage across the capacitor at exactly t = 0 (the moment discharge starts).
e Euler's Number Dimensionless Mathematical constant ≈ 2.71828. The base of the natural logarithm.
t Time Seconds (s) The elapsed time since the discharge path was completed.
R Resistance Ohms (Ω) The total series resistance in the discharge path (includes external resistor + ESR).
C Capacitance Farads (F) The nominal capacitance value of the component.
RC Time Constant (τ) Seconds (s) The time required for the voltage to decay to ~36.8% of V₀.

When the Formula Applies (and Its Assumptions)

This equation is not a universal law for all electrical phenomena; it relies on specific boundary conditions. According to HyperPhysics, the derivation assumes:

  • Constant Resistance: The resistor does not change value as it heats up (no thermistor effects).
  • Ideal Capacitor: The model assumes zero dielectric leakage and ignores Equivalent Series Inductance (ESL). For high-frequency AC or ultra-fast pulse dumps, parasitic inductance will cause ringing that this DC transient equation cannot predict.
  • Lumped Parameter Model: The physical dimensions of the circuit are small enough that electromagnetic wave propagation delays are negligible.

Deriving the Formula from Kirchhoff's Laws

To understand why the equation takes this shape, we apply Kirchhoff’s Voltage Law (KVL) to the closed discharge loop. The sum of voltages must be zero:

V_c + V_r = 0

Substituting the component definitions (V_c = q/C and V_r = iR):

q/C + iR = 0

Since current is the rate of change of charge (i = dq/dt), we get a first-order differential equation:

dq/dt = -q / RC

Separating variables and integrating both sides yields the natural exponential function, resulting in q(t) = Q₀ × e(-t/RC). Because voltage is directly proportional to charge (V = q/C), the voltage equation follows the exact same exponential decay profile.

Rearranged Forms: Solving for Time, Resistance, and Capacitance

On the bench, you rarely just solve for V(t). More often, you have a safety threshold (like 50V) and need to size a bleeder resistor or calculate the wait time before touching a PCB. Here are the algebraically rearranged forms of the equation for discharging a capacitor:

  • Solving for Time (t):
    t = -RC × ln(V(t) / V₀)
    Use case: Calculating how long to wait before servicing a high-voltage DC bus.
  • Solving for Resistance (R):
    R = -t / [C × ln(V(t) / V₀)]
    Use case: Sizing a bleeder resistor to ensure a power supply drops to a safe voltage within a mandated time limit.
  • Solving for Capacitance (C):
    C = -t / [R × ln(V(t) / V₀)]
    Use case: Determining the maximum allowable bus capacitance for a specific discharge resistor and time limit.
  • Solving for Initial Voltage (V₀):
    V₀ = V(t) / e(-t/RC)
    Use case: Forensic analysis to determine the original charge voltage based on a measurement taken at a known time delay.

Worked Examples with Unit Tracking

Abstract math is useless if you drop a zero. Let's run two realistic bench scenarios, tracking every unit conversion to ensure the final magnitude makes physical sense.

Problem 1: Finding Voltage at a Specific Time (Bleeder Resistor Check)

Scenario: You are testing a motor controller with a 400V DC bus. The bus uses a 470µF electrolytic capacitor (e.g., a Cornell Dubilier 381LX series). You install a 100kΩ bleeder resistor across the terminals. What is the voltage across the capacitor exactly 5 seconds after power is removed?

  1. Convert to Base SI Units:
    C = 470 µF = 470 × 10⁻⁶ F = 0.00047 F
    R = 100 kΩ = 100,000 Ω
    V₀ = 400 V
    t = 5 s
  2. Calculate the Time Constant (τ = RC):
    τ = 100,000 Ω × 0.00047 F = 47 seconds
  3. Calculate the Exponent (-t / RC):
    -5 / 47 = -0.10638
  4. Apply Euler's Number:
    e-0.10638 ≈ 0.89907
  5. Multiply by Initial Voltage:
    V(5) = 400 V × 0.89907 = 359.6 V

Realistic Magnitude Check: 5 seconds is roughly 0.1 time constants. In an RC decay curve, the voltage drops rapidly at first, but 0.1τ is very early in the curve. Retaining ~90% of the initial charge (359.6V) aligns perfectly with the physics of the exponential decay graph.

Problem 2: Finding Time to Reach a Safe Touch Threshold

Scenario: Using the same 400V, 470µF, and 100kΩ circuit from Problem 1, how long must you wait for the voltage to drop to 50V (the standard safe-to-touch DC threshold)?

  1. Identify Knowns and Select Formula:
    V₀ = 400 V, V(t) = 50 V, RC = 47 s.
    We need t, so we use: t = -RC × ln(V(t) / V₀)
  2. Calculate the Voltage Ratio:
    50 / 400 = 0.125
  3. Take the Natural Logarithm (ln):
    ln(0.125) ≈ -2.0794
  4. Multiply by Negative Time Constant:
    t = -47 × (-2.0794) = 97.73 seconds

Realistic Magnitude Check: A common rule of thumb is that a capacitor is 'fully' discharged at 5τ (which would be 235 seconds here, dropping to ~2.7V). However, reaching 12.5% of the initial voltage (50V) takes roughly 2.1 time constants. 2.1 × 47s = 98.7s. Our precise answer of 97.73 seconds matches the rule-of-thumb estimate perfectly.

Common Unit Mistakes That Break the Math

When the math spits out an impossible answer (like a negative time or a voltage higher than the supply), you almost certainly made a unit conversion error. Watch out for these three traps:

1. The Microfarad Trap
Datasheets list capacitance in µF, mF, or nF. If you plug '470' directly into the C variable instead of '0.00047', your time constant will be off by a factor of one million. Always convert to base Farads before calculating.
2. The Millisecond Mismatch
Oscilloscopes often measure time in milliseconds (ms) or microseconds (µs). If your t is in milliseconds but your RC is in seconds, the exponent will be skewed by 1,000. Standardize everything to seconds.
3. The Inverted Log Ratio (Domain Error)
When solving for time, you calculate ln(V(t) / V₀). Because the capacitor is discharging, V(t) must be smaller than V₀, making the ratio less than 1. The natural log of a fraction is negative, which cancels out the negative sign in front of the RC to give a positive time. If you accidentally flip the ratio to ln(V₀ / V(t)), the log will be positive, resulting in a mathematically impossible negative time.

Frequently Asked Questions

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

Mathematically, the equation for discharging a capacitor is asymptotic; it approaches zero but never truly reaches it. In practical engineering, we use the 5τ (five time-constant) rule. At t = 5RC, the voltage has decayed to e⁻⁵, which is roughly 0.67% of the initial voltage. For most low-voltage logic circuits, this is considered 0V. For high-voltage DC buses (e.g., 800V in EV inverters), 0.67% is still 5.3V, which may require a longer wait or an active discharge circuit to meet strict safety regulations.

Does the equation for discharging a capacitor apply to AC circuits?

No, not directly. The formula V(t) = V₀ × e(-t/RC) models a DC transient event (a step response). In a continuous AC circuit, the capacitor is constantly charging and reversing polarity. To analyze AC circuits, you must abandon the time-domain exponential equation and use the frequency-domain concept of capacitive reactance (Xc = 1 / (2πfC)) and complex impedance. For a deep dive into AC behavior, refer to the RC Circuit tutorials on Electronics Tutorials.

What happens to the discharge equation if the capacitor has high ESR?

Real-world capacitors have Equivalent Series Resistance (ESR). The R in the equation represents the total resistance in the discharge loop. Therefore, R_total = R_external + ESR. For a standard bleeder resistor (e.g., 100kΩ), an ESR of 0.05Ω is mathematically irrelevant. However, in low-resistance, high-current dump circuits (like a spot welder or a railgun dump using a 0.1Ω shunt), an ESR of 0.02Ω represents a 20% error in your total resistance. In those cases, you must add the ESR to your external R value to get an accurate time constant.

Why do we use the natural logarithm (ln) instead of log base 10 in the rearranged formula?

The natural logarithm (ln, base e) is used because the original calculus derivation of the circuit's differential equation (dq/dt = -q/RC) naturally yields Euler's number (e) when integrated. The rate of change of the voltage is directly proportional to the voltage itself, which is the exact definition of an exponential function with base e. If you prefer using base-10 logarithms (log₁₀), you must apply the change-of-base formula: ln(x) = log₁₀(x) / log₁₀(e), which adds an unnecessary conversion factor of ~2.302 to your calculations.

Can I use this equation to calculate the energy dissipated by the resistor?

The voltage equation tells you the potential difference, but to find energy, you need to integrate the power over time. The total energy stored in the capacitor initially is E = 0.5 × C × V₀². Assuming an ideal circuit with no radiated electromagnetic losses or dielectric absorption, 100% of this initial energy will be dissipated as heat in the resistor over the course of the discharge, regardless of the resistance value. A smaller resistor will dissipate the exact same total energy, just at a much higher instantaneous wattage over a shorter time.