Mastering Charging and Discharging a Capacitor Equations: Formulas & Worked Examples

The fundamental charging and discharging a capacitor equations describe the exponential voltage change across a capacitor in a series resistor-capacitor (RC) circuit. For charging, the voltage at time t is V(t) = V_s × (1 - e^-t/RC). For discharging, it is V(t) = V₀ × e^-t/RC. These formulas are the bedrock of timing circuits, power supply filtering, and signal coupling. Below, we break down every symbol, map out the rearranged forms, and solve real-world bench problems with strict unit tracking.

The Core RC Time Constant Equations

Before plugging numbers into a calculator, you must understand the anatomy of the formulas. The charging equation models a capacitor starting at 0V and asymptotically approaching the source voltage. The discharging equation models a capacitor starting at an initial voltage and asymptotically decaying to 0V.

Symbol Definition and Standard Units

Symbol	Meaning	Standard SI Unit
V(t)	Voltage across the capacitor at time t	Volts (V)
Vs	Source (supply) voltage for charging	Volts (V)
V0	Initial voltage across the capacitor before discharging begins	Volts (V)
e	Euler's number (mathematical constant ≈ 2.71828)	Dimensionless
t	Elapsed time since the charge/discharge cycle began	Seconds (s)
R	Series resistance limiting the current	Ohms (Ω)
C	Capacitance value	Farads (F)
τ (tau)	RC Time Constant (τ = R × C)	Seconds (s)

Rearranged Forms

On the bench, you rarely just solve for V(t). Usually, you know the target voltage and need to find the required time, resistance, or capacitance. Here are the rearranged forms solving for each variable:

• Solving for time (t): t = -R × C × ln(1 - V(t)/V_s) [Charging] | t = -R × C × ln(V(t)/V₀) [Discharging]
• Solving for resistance (R): R = -t / (C × ln(1 - V(t)/V_s)) [Charging] | R = -t / (C × ln(V(t)/V₀)) [Discharging]
• Solving for capacitance (C): C = -t / (R × ln(1 - V(t)/V_s)) [Charging] | C = -t / (R × ln(V(t)/V₀)) [Discharging]

Note: 'ln' represents the natural logarithm (base e). For a deeper theoretical background on the derivation of these exponential curves, refer to the Georgia State University HyperPhysics RC circuit module.

Assumptions, Unit Traps, and Realistic Magnitudes

The charging and discharging a capacitor equations are elegant, but they rely on strict assumptions. If your physical circuit violates these, your calculated numbers will not match your oscilloscope traces.

Unit Mistakes That Break the Math

The most common reason DIYers and students get wildly incorrect answers is failing to convert sub-multiples to base SI units before calculating. The term -t/RC in the exponent must be entirely dimensionless.

• The Microfarad Trap: Entering 100 μF as '100' instead of '0.0001' (100 × 10^-6). This makes your time constant 1,000,000 times too large.
• The Kilo-ohm Trap: Entering 4.7 kΩ as '4.7' instead of '4700'.
• The Millisecond Trap: Mixing milliseconds for t while leaving R in Ohms and C in Farads. Always convert t to seconds first, calculate, and then convert the final answer back to milliseconds if desired.

What a Realistic Answer Magnitude Looks Like

Before hitting 'equals' on your calculator, do a sanity check on the time constant (τ = R × C).

• Signal Filtering (High Frequency): R = 1 kΩ, C = 100 pF. τ = 1000 × (100 × 10^-12) = 0.1 μs. (Realistic for RF or high-speed digital lines).
• Timing Circuits (555 Timers): R = 100 kΩ, C = 10 μF. τ = 100,000 × 0.00001 = 1 second. (Realistic for blinkers or delays).
• Power Bank Ride-Through: R = 0.1 Ω (ESR/trace), C = 50 F (Supercap). τ = 5 seconds. (Realistic for memory backup).

If you are designing a 1-second delay and your math yields τ = 4,500 seconds, you forgot to convert microfarads to farads. For standard practical applications, review the Electronics Tutorials RC Circuit guide to see how these magnitudes apply to real waveforms.

Worked Example 1: Charging a Supercapacitor Bank

Scenario: You are building a memory backup circuit using a Maxwell 2.5F supercapacitor. The supply rail is 5.0V DC. To prevent inrush current from tripping the upstream LDO regulator, you place a 10 Ω series resistor. What is the voltage across the supercapacitor exactly 10 seconds after power is applied?

Bench Reality Check: On a real bench, a supercapacitor has significant Equivalent Series Resistance (ESR), often around 50mΩ to 100mΩ for a 2.5F cell. When the 5V is first applied, there will be an instantaneous voltage step across the ESR (V = I × R_ESR) before the exponential curve begins. The ideal equation ignores this initial step.

Worked Example 2: Discharging a Camera Flash Capacitor

Scenario: You are salvaging a Cornell Dubilier 150 μF photoflash capacitor from an old camera. It is charged to 330V. To safely handle it, you solder a 2 kΩ bleeder resistor across the terminals. How long will it take for the voltage to drop to a safe-to-touch threshold of 50V?

Bench Reality Check: While the math says it takes just over half a second to reach 50V, dielectric absorption in high-voltage electrolytic capacitors causes the voltage to 'rebound' slightly after the resistor is removed. Always keep the bleeder resistor permanently attached or short the terminals with an insulated tool before handling. For more on safe high-voltage discharge practices, consult the All About Circuits RC Time Constant chapter.

Frequently Asked Questions

How do the charging and discharging a capacitor equations change when using AC instead of DC?

The standard time-domain equations (V(t) = V_s(1 - e^-t/RC)) only apply to DC step inputs. When driven by an AC sinusoidal voltage, the capacitor never reaches a steady-state DC voltage. Instead, you must transition to the frequency domain using complex impedance (Z_c = 1 / jωC). The voltage across the capacitor becomes a function of the voltage divider ratio between R and Z_c, resulting in a phase shift and amplitude attenuation rather than a simple exponential charge curve.

Why do the charging and discharging a capacitor equations rely on the mathematical constant e?

The constant e (Euler's number) appears naturally because the rate of voltage change across a capacitor is directly proportional to the current flowing into it (I = C × dV/dt). As the capacitor charges, the voltage difference between the source and the capacitor shrinks, which reduces the current (per Ohm's law). This creates a differential equation where the rate of change is proportional to the remaining value. The only mathematical function that is its own derivative—and thus solves this specific type of continuous decay/growth—is the exponential function with base e.

What happens to the charging and discharging a capacitor equations if the component has high ESR?

High Equivalent Series Resistance (ESR) effectively adds to the external series resistor R. If your external resistor is 100 Ω and the capacitor has an ESR of 5 Ω, the actual time constant becomes τ = (100 + 5) × C. More importantly, high ESR causes an instantaneous voltage step at t=0. During charging, the initial voltage jumps to V_step = V_s × (ESR / (R + ESR)) before the exponential curve begins. The standard ideal equations fail to predict this initial step, which can be critical in high-current pulse applications like motor drives or defibrillators.

At what time is a capacitor considered fully charged according to the equations?

Mathematically, the exponential curve is asymptotic; it never truly reaches 100% of V_s. However, in practical electrical engineering, a capacitor is considered 'fully charged' or 'fully discharged' after 5 time constants (5τ). At t = 5τ, the capacitor has reached 99.3% of the source voltage (charging) or decayed to 0.7% of its initial voltage (discharging). For a circuit with τ = 2 seconds, the capacitor is practically fully charged at t = 10 seconds.