The Core Discharging Equation and Symbol Definitions
When a charged capacitor is connected across a resistor, Kirchhoff’s Voltage Law dictates that the sum of the voltages around the loop must be zero: Vc + Vr = 0. Substituting the component definitions (q/C + R(dq/dt) = 0) and integrating with respect to time yields the standard exponential decay function. For a deeper physics-based derivation of this differential equation, refer to the HyperPhysics RC circuit documentation. The governing discharging equation of a capacitor is:V(t) = V₀ × e-t / RC
| Symbol | Definition | Standard SI Unit |
|---|---|---|
| V(t) | Voltage across the capacitor at time t | Volts (V) |
| V₀ | Initial voltage across the capacitor at t = 0 | Volts (V) |
| e | Euler's number (mathematical constant ≈ 2.71828) | Dimensionless |
| t | Time elapsed since the discharge began | Seconds (s) |
| R | Resistance of the discharge path | Ohms (Ω) |
| C | Capacitance of the component | Farads (F) |
| τ (tau) | RC Time Constant (where τ = R × C) | Seconds (s) |
Rearranged Forms: Solving for Any Variable
On the bench, you rarely need to find V(t). Usually, you know your target voltage threshold and need to size the resistor or calculate the delay time. By applying the natural logarithm (ln) to both sides, we can isolate any variable in the discharging equation of a capacitor.- 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/RC
Worked Examples with Strict Unit Tracking
The most common point of failure in RC calculations is unit mismatch. The formula demands base SI units: Ohms, Farads, and Seconds. Let's walk through two real-world scenarios.Problem 1: Timing Circuit Voltage Decay
Scenario: A 555 timer delay circuit uses a 10µF capacitor and a 100kΩ resistor. The capacitor is initially charged to 12V. What is the voltage across the capacitor exactly 2 seconds after the discharge pin is triggered?
- Convert to Base SI Units:
C = 10µF = 10 × 10-6 F = 0.00001 F
R = 100kΩ = 100,000 Ω - Calculate the Time Constant (τ):
τ = R × C = 100,000 Ω × 0.00001 F = 1 second. - Apply the Discharging Equation:
V(2) = 12V × e-(2 / 1)
V(2) = 12V × e-2
V(2) = 12V × 0.13534 - Final Answer:
V(2) ≈ 1.62 V
Problem 2: Sizing a High-Voltage Bleed Resistor
Scenario: You are designing a motor drive inverter with a 4700µF DC bus capacitor charged to 400V. Safety standards require the bus to discharge to below 50V within 60 seconds of power-off. What resistance value is required, and what is the initial power dissipation?
- Convert to Base SI Units:
C = 4700µF = 0.0047 F
t = 60 s
V₀ = 400 V, V(t) = 50 V - Rearrange for R and Substitute:
R = -t / (C × ln(V(t) / V₀))
R = -60 / (0.0047 × ln(50 / 400))
R = -60 / (0.0047 × ln(0.125))
R = -60 / (0.0047 × -2.0794) - Solve for R:
R = -60 / -0.009773 = 6,139 Ω (Select a standard 6.2kΩ resistor). - Calculate Initial Power Dissipation (Crucial Step):
P = V₀² / R = 400² / 6139 = 160,000 / 6139 = 26.06 Watts.
Assumptions, Unit Traps, and Realistic Magnitudes
When the Formula Applies (and When It Doesn't)
The discharging equation of a capacitor applies strictly to ideal, lumped-element RC circuits where a capacitor discharges through a purely resistive, linear load. It assumes constant capacitance and constant resistance. In reality, if you are using Class II MLCC ceramics (like X7R or Y5V), capacitance drops drastically as voltage increases, meaning your effective C is much lower at V₀ than at V(t). Furthermore, it ignores Equivalent Series Resistance (ESR) and parasitic inductance, which dominate in high-speed snubber circuits.
The Unit Mistakes That Break Your Math
The 'Microfarad Trap' is responsible for 90% of RC calculation errors. Plugging '4700' into C instead of '0.0047' will yield a time constant that is off by a factor of one million. Similarly, mixing milliseconds (ms) with seconds (s) in the time variable will result in drastically undersized resistors. Always convert to Farads, Ohms, and Seconds before touching your calculator.
What a Realistic Answer Magnitude Looks Like
A realistic magnitude for a bench-top RC timing circuit τ is between 1ms and 10s. If your calculation yields a τ of 0.000004s or 45,000s for a standard 555 timer delay, you have likely fallen victim to a unit trap. Additionally, the industry standard for 'fully discharged' is 5τ (five time constants). At 5τ, the voltage has decayed to 0.7% of V₀ (e-5 ≈ 0.0067). If your 5τ calculation exceeds your safety window, you must decrease R or C.
Frequently Asked Questions
How does the discharging equation of a capacitor change with a constant current load?
If the capacitor discharges into an active constant-current sink (like a linear regulator or an active electronic load) rather than a passive resistor, the exponential decay becomes a linear ramp. The equation changes to V(t) = V₀ - (I × t / C). This is highly preferred in applications like defibrillators or camera flashes where a predictable, linear voltage drop is required to maintain constant power delivery.
Why doesn't my physical capacitor match the discharging equation of a capacitor perfectly?
Physical components exhibit non-ideal behaviors. The primary culprit in electrolytic and film capacitors is dielectric absorption (also known as 'soakage' or 'battery action'). After you short a capacitor to 0V and remove the short, the voltage will spontaneously 'rebound' to a few millivolts or even volts as trapped charges in the dielectric relax. Furthermore, electrolytic capacitors have significant internal leakage current, which acts as a parallel, unintended bleed resistor, slightly altering the effective R of your circuit. For precision timing, use polypropylene film or C0G/NP0 ceramic capacitors, which exhibit virtually zero dielectric absorption.
At what point in the discharging equation of a capacitor is the energy half depleted?
This is a common trap: voltage is not energy. The stored energy in a capacitor is defined by E = 0.5 × C × V². Because energy scales with the square of the voltage, half of the total energy is depleted when the voltage drops to V₀ / √2 (approximately 0.707 × V₀). By plugging 0.707V₀ into the rearranged time equation, we find that 50% of the energy is dissipated in just t ≈ 0.346τ. This means the first third of a time constant does the heavy lifting in terms of thermal dissipation in your bleed resistor.
For further reading on non-ideal capacitor behaviors and dielectric absorption, review the Electronics Tutorials RC Discharge Guide.
