The Core Capacitor Discharging Equation and Symbol Definitions

When you discharge a charged capacitor through a fixed resistor, the voltage does not drop linearly. It decays exponentially. The capacitor discharging equation defines the exact voltage remaining across the capacitor plates at any specific moment in time. For DC transient analysis, this formula is non-negotiable for sizing bleeder resistors, calculating timing circuits, and ensuring high-voltage bus safety.

The standard formula for the voltage across a discharging capacitor is:

V(t) = V0 · e-t / (R · C)

Every variable in this equation carries strict unit requirements. If you mix milliamps with microfarads without converting to base SI units, your exponent will be wrong by orders of magnitude. Below is the definitive symbol table for the capacitor discharging equation.

Symbol Parameter Name Required SI Unit Typical Bench Range
V(t) Instantaneous voltage at time t Volts (V) 0 V to 400 V
V0 Initial voltage (at t = 0) Volts (V) 3.3 V to 800 V
e Euler's number (mathematical constant) Dimensionless ≈ 2.71828
t Elapsed discharge time Seconds (s) 1 ms to 300 s
R Discharge resistance (load + ESR) Ohms (Ω) 10 Ω to 1 MΩ
C Capacitance Farads (F) 100 pF to 10,000 μF
τ (tau) RC Time Constant (R · C) Seconds (s) 1 μs to 10 s

Rearranged Forms: Solving for Time, Resistance, and Capacitance

On the workbench, you rarely need to find V(t). Usually, you know your target safe voltage and need to calculate the required bleed time (t), or you have a fixed timing requirement and need to select the correct resistor (R). By applying natural logarithms (ln) to both sides of the capacitor discharging equation, we can isolate any variable.

Solving for Time (t):
t = -R · C · ln(V(t) / V0)

Solving for Resistance (R):
R = -t / [C · ln(V(t) / V0)]

Solving for Capacitance (C):
C = -t / [R · ln(V(t) / V0)]

Solving for Initial Voltage (V0):
V0 = V(t) / e-t / (R · C)

Assumptions, Unit Traps, and Realistic Magnitudes

The capacitor discharging equation is an idealized model. Before plugging numbers into your calculator, you must understand its physical assumptions and where real-world components deviate from the math.

When the Formula Applies (and Its Assumptions)

  • Constant Resistance: The equation assumes R is perfectly linear and constant. If your discharge path includes a semiconductor (like a diode or transistor) whose resistance changes with voltage, this simple exponential model fails.
  • Ideal Capacitor: It assumes zero Equivalent Series Inductance (ESL) and ignores Equivalent Series Resistance (ESR). In high-speed pulse discharge (like a camera flash), ESL dominates the first few microseconds, causing ringing that the equation does not predict.
  • No Dielectric Absorption: Real capacitors, especially electrolytics and polypropylene film caps, exhibit dielectric absorption. This is the tendency of a capacitor to become internally charged after being rapidly discharged, causing a 'rebound' voltage. The equation predicts 0 V at infinity; a real 400V electrolytic might rebound to 15V after the bleeder resistor is disconnected.

Unit Mistakes That Break the Math

The most common reason engineers get wildly incorrect answers is failing to convert to base SI units before calculating the exponent. The product of R and C must yield seconds.

  • The Microfarad Trap: Capacitance is almost always printed in μF (microfarads). You must multiply by 10-6. A 470 μF cap is 0.00047 F.
  • The Kilo-ohm Trap: Resistors are usually in kΩ or MΩ. A 100 kΩ resistor must be entered as 100,000 Ω.
  • The Missing Negative Sign: The exponent is negative. If you forget the minus sign, your calculator will output a voltage higher than your initial supply, which is physically impossible for a passive discharge.

What a Realistic Answer Magnitude Looks Like

The RC time constant (τ = R · C) dictates the speed of decay. At t = 1τ, the voltage drops to 36.8% of V0. At t = 5τ, it drops to 0.67% of V0, which is universally accepted in electrical engineering as 'fully discharged'. If you calculate a bleed time of 5 seconds for a 10,000 μF capacitor, your resistor should be around 100 Ω. If your math yields 0.002 seconds or 4,000 seconds for a standard bench circuit, you have likely missed a unit conversion.

Worked Examples with Step-by-Step Unit Tracking

Let us apply the capacitor discharging equation to two real-world scenarios, tracking every unit conversion and intermediate calculation.

Problem 1: Finding Voltage at a Specific Time

Scenario: You are testing a 24V DC motor control board. The smoothing capacitor is 2,200 μF, and the bleeder resistor is 4.7 kΩ. You pull the power plug. What is the voltage across the capacitor exactly 15 seconds later?

Step 1: Identify and convert variables to base SI units.

  • V0 = 24 V
  • C = 2,200 μF = 2,200 × 10-6 F = 0.0022 F
  • R = 4.7 kΩ = 4,700 Ω
  • t = 15 s

Step 2: Calculate the time constant (τ).

  • τ = R · C
  • τ = 4,700 [Ω] × 0.0022 [F] = 10.34 seconds

Step 3: Apply the capacitor discharging equation.

  • V(t) = V0 · e-t / τ
  • V(15) = 24 · e-15 / 10.34
  • V(15) = 24 · e-1.4506
  • V(15) = 24 · 0.2344
  • V(15) = 5.62 V

Result: After 15 seconds, the capacitor has discharged from 24V down to a relatively safe 5.62V.

Problem 2: Sizing a Bleeder Resistor for Safety Compliance

Scenario: You are designing the power supply for a tube amplifier. The main DC bus runs at 400V. The filter capacitor is 470 μF. Safety standards require the bus to discharge below 50V within 10 seconds of unplugging. What maximum resistance (R) must you install across the capacitor?

Step 1: Identify and convert variables.

  • V0 = 400 V
  • V(t) = 50 V (target threshold)
  • t = 10 s
  • C = 470 μF = 0.00047 F

Step 2: Select the rearranged formula for Resistance.

  • R = -t / [C · ln(V(t) / V0)]

Step 3: Calculate the natural log term.

  • ln(V(t) / V0) = ln(50 / 400) = ln(0.125)
  • ln(0.125) ≈ -2.0794

Step 4: Solve for R.

  • R = -10 / [0.00047 × (-2.0794)]
  • R = -10 / [-0.0009773]
  • R = 10,232 Ω

Result: You must use a resistor of 10 kΩ or lower. In practice, you would select a standard 9.1 kΩ or 8.2 kΩ resistor rated for at least 2W (since continuous power dissipation at 400V is V2/R = 160,000 / 8200 = 19.5W — wait, a single resistor will burn up. You would use a series string of high-wattage resistors to handle the continuous 20W heat load while maintaining the < 10.2 kΩ total resistance). For deep component selection guidance, refer to the All About Circuits guide on RC time constants.

Frequently Asked Questions

How does the capacitor discharging equation change for an AC circuit?

The standard capacitor discharging equation applies strictly to DC transients (a capacitor discharging through a resistor after a DC source is removed). In an AC circuit, the capacitor is continuously charging and discharging in alternating half-cycles. Instead of a transient exponential decay, you must calculate capacitive reactance (Xc = 1 / (2πfC)) to determine the steady-state AC current and voltage drops using complex impedance.

Why does the capacitor discharging equation use the natural logarithm base e?

The base e arises directly from the calculus derivation of the circuit. The current flowing out of the capacitor is proportional to the rate of change of the voltage across it (I = -C(dV/dt)). By Kirchhoff's Voltage Law, this current also equals V/R. Setting -C(dV/dt) = V/R creates a first-order linear differential equation. The only mathematical function whose derivative is proportional to itself is the exponential function with base e. For a deeper mathematical proof, see the LibreTexts OpenStax Physics module on RC Circuits.

What happens to the capacitor discharging equation if the resistor is removed (open circuit)?

If you remove the external resistor, R approaches infinity, making the exponent zero, and the equation predicts the voltage will remain at V0 forever. In reality, an open-circuit capacitor discharges through its own internal leakage resistance and the dielectric material. This leakage is modeled as a very high-value resistor (often 10 MΩ to 10 GΩ) in parallel with the ideal capacitor. The equation still applies, but you must substitute the external R with the capacitor's datasheet-specified leakage resistance.

How do I calculate the total energy dissipated during capacitor discharge?

The capacitor discharging equation tells you voltage over time, but not energy. The total energy (E) stored in the capacitor and eventually dissipated as heat in the resistor is calculated using E = ½ · C · V02. Interestingly, the value of the resistor (R) does not change the total energy dissipated; it only changes the rate at which that energy is released. A smaller resistor dissipates the same total joules, but in a much shorter, more violent burst of heat.