Capacitor Discharging Voltage

Describes the exponential decay of voltage across a discharging capacitor.

Written by the Vorzep Team

Circuit AnalysisTransient Response

Capacitor Discharging Calculator RC Circuit · V(t) = V₀·e⁻ᵗ/ᴿᶜ

V(t) = V₀ · e^−t/(R·C)

V(t) = voltage at time t (V) · V₀ = initial voltage (V) · t = time (s) · R = resistance (Ω) · C = capacitance (F)

⟹ Solve ⟷ V(t), V₀, t, R, C

Voltage at time (V(t))

Initial Voltage (V₀)

Time (t)

Resistance (R)

Capacitance (C)

Please fix the errors above.

Solve for: V(t) V₀ t R C

Presets:

Voltage at time t —

—

V(t): — V₀: — t: — R: — C: —

✓ Copied!

Voltage Remaining —

High (> 50%) Medium (10–50%) Low (< 10%)

V(t) = V₀ · e^−t/(R·C) · Time constant τ = R·C · After one τ, voltage drops to ~36.8% of V₀.

V(t) = V₀·e^(−t/RC)

Capacitor Discharging Voltage

Variables

Symbol	Quantity	Unit
V(t)	Voltage at time t	V
V₀	Initial capacitor voltage	V
t	Time elapsed	s
R	Resistance	Ω
C	Capacitance	F
τ	Time constant (R×C)	s

What it means

The voltage across a discharging capacitor in an RC circuit as a function of time is V(t) = V₀ * e^(−t/τ), where V₀ is the initial voltage and τ = RC is the time constant. The voltage decays exponentially from V₀ towards 0. This equation is derived from the same differential equation as the charging case, with the opposite sign. Discharge circuits are used in timing, pulse generation, and in flash cameras. The discharge time to reach a certain fraction of V₀ is analogous to charging: after one τ, the voltage drops to 36.8% of its initial value. Example: V₀=10V, τ=1s. At t=1s, V = 10 * e^(−1) = 3.68V. At t=3s, V = 10 * e^(−3) = 0.50V. The discharge curve is essential for designing blinking LEDs, relaxation oscillators, and for timing applications.

Worked example

Capacitor Discharging – Practical Example

Real‑World

Scenario: The same capacitor charged to 10 V is now discharged through the resistor. What is the voltage after 100 ms (one time constant)?

Parameter	Value
V₀	10 V
τ	0.1 s
t	0.1 s
Formula	`V(t) = V₀ · e^−t/τ`

1At t = τ: V(τ) = 10 · e⁻¹

2e⁻¹ ≈ 0.368, so:V = 10 × 0.368 = 3.68 V

Final Design V(100ms) ≈ 3.68 V ✓ 36.8% of V₀

Why: During discharge, the voltage decays exponentially; after one τ, it drops to 36.8% of the initial value.

Common mistakes

V₀: The initial voltage across the capacitor (at t=0).
Decay: The voltage decays exponentially to zero.
Time constant: Same as charging: τ = RC.
Energy: The energy is dissipated in the resistor.
Discharge path: Ensure there is a closed path for the current to flow.

Applications

The capacitor discharging voltage decays exponentially as V(t) = V₀·e^(−t/RC), falling to 36.8% of the initial value after one time constant. This formula is used to design discharge circuits, to ensure safe residual voltage removal, and to analyse the decay of stored energy. Engineers apply it in timing circuits, in reset circuits, and in the design of capacitance touch sensors. It is also used to calculate the time for a capacitor to discharge to a safe level for servicing. By understanding the discharge process, professionals can predict when the voltage will drop below a threshold, enabling reliable logic transitions and safe operation.

Discharge path design for safety (bleeder resistors)
Reset and power‑on‑reset circuits
Capacitive sensing and touch detection
Timer and monostable multivibrator circuits
Educational understanding of exponential decay