This calculator is used to calculate the discharge of a capacitor through a fixed-value resistor. This calculator solves for time or resistance, measuring the resulting initial power dissipation in the resistance and the total energy discharged to zero volts, given a capacitance value and beginning and end voltages.
Capacitor Safety Discharge Calculator

Capacitor Safety Discharge Calculator
Use this calculator to analyze the ideal exponential discharge of a capacitor through a resistor. Depending on the available inputs, it can calculate capacitor voltage after a specified time, the time required to reach a target voltage, or the resistance or capacitance needed for a specified transition.
The calculator is an engineering estimate, not proof that equipment is safe to touch. Capacitor voltage must be verified with an appropriately rated instrument and the applicable equipment, workplace, and lockout procedures must be followed. Real circuits may contain additional energy sources, isolated capacitor banks, dielectric absorption, leakage paths, switching delays, and component tolerances that are not represented by an ideal RC model.
How to Use the Capacitor Discharge Calculator
Enter the initial capacitor voltage before discharge begins.
Enter the desired final or target voltage.
Enter the capacitance and discharge resistance using the displayed units.
Enter the discharge time, or leave the one variable being solved for blank if the calculator supports reverse calculations.
Calculate the result.
Check the initial resistor current, initial power, stored energy, component ratings, and worst-case tolerances before using the result in a real design.
For a passive discharge to zero, the final equilibrium or supply voltage is 0 V. If a source remains connected, the capacitor approaches that source voltage rather than zero, and the general transition formula must be used.
Capacitor Discharge Formula
For an ideal capacitor initially charged to V0 and discharged through a resistor toward 0 V:
VC(t) = V0e−t/(RC)
where:
VC(t) is the capacitor voltage after time t.
V0 is the initial capacitor voltage.
R is the discharge resistance.
C is the capacitance.
t is the elapsed discharge time.
e is the base of the natural logarithm.
General RC Transition Formula
If the capacitor approaches a nonzero final equilibrium voltage VS, use:
VC(t) = VS + (V0 − VS)e−t/(RC)
This equation describes both charging and discharging through a linear resistance after a step change. The target voltage must lie along the exponential path from V0 toward VS. In an ideal first-order RC circuit, the capacitor approaches VS asymptotically and does not reach it in a finite time.
Solve for Discharge Time
For discharge toward 0 V from V0 to a positive target voltage VT:
t = RC ln(V0/VT)
For a nonzero equilibrium voltage VS:
t = −RC ln[(VT − VS)/(V0 − VS)]
The logarithm ratio must be greater than zero and no greater than one for a forward-time transition toward VS.
Solve for Resistance or Capacitance
For a pure discharge toward zero:
R = t ÷ [C ln(V0/VT)]
C = t ÷ [R ln(V0/VT)]
These equations provide nominal values. A real discharge design must account for component tolerances, voltage coefficients, temperature, aging, parasitic paths, and the resistor's allowable voltage, pulse energy, and thermal load.
RC Time Constant
The time constant is:
τ = RC
After one time constant, a capacitor discharging toward zero retains e−1, or about 36.8%, of its initial voltage. Each additional time constant multiplies the remaining voltage by the same factor.
| Elapsed Time | Remaining Voltage | Voltage Discharged |
|---|---|---|
| 1τ | 36.7879% of V0 | 63.2121% |
| 2τ | 13.5335% of V0 | 86.4665% |
| 3τ | 4.97871% of V0 | 95.0213% |
| 4τ | 1.83156% of V0 | 98.1684% |
| 5τ | 0.673795% of V0 | 99.3262% |
| 6τ | 0.247875% of V0 | 99.7521% |
| 7τ | 0.0911882% of V0 | 99.9088% |
“Five time constants” is a useful approximation in some low-energy calculations, but it is not a universal safe-voltage criterion. The remaining voltage depends on V0, and the acceptable target must come from the applicable standard, equipment documentation, and safety procedure.
Time Required to Reach a Percentage of Initial Voltage
| Target Voltage | Required Time |
|---|---|
| 50% of V0 | 0.693147τ |
| 10% of V0 | 2.30259τ |
| 5% of V0 | 2.99573τ |
| 1% of V0 | 4.60517τ |
| 0.1% of V0 | 6.90776τ |
Current, Power, and Stored Energy
For discharge toward zero, the resistor current magnitude is:
I(t) = (V0/R)e−t/(RC)
Initial current is highest at t = 0:
I0 = V0/R
Resistor power also begins at its maximum value:
P(t) = VC(t)2/R
P0 = V02/R
Energy stored in the capacitor at voltage V is:
E = ½CV2
For an isolated capacitor discharging through a resistor from V0 to VT, the reduction in stored energy is:
ΔE = ½C(V02 − VT2)
For a complete ideal discharge to zero, the resistor must absorb the capacitor's initial stored energy. Its suitability cannot be determined from steady-state wattage alone; pulse-energy capability, overload duration, surface temperature, voltage rating, mounting, and repetition rate may also control the design.
Unit Conversions
| Quantity | Conversion |
|---|---|
| Resistance | 1 kΩ = 1,000 Ω; 1 MΩ = 1,000,000 Ω |
| Capacitance | 1 µF = 10−6 F; 1 nF = 10−9 F |
| Time constant | Ω × F = s |
| Useful combination | kΩ × µF = ms |
| Useful combination | MΩ × µF = s |
| Useful combination | kΩ × nF = µs |
Capacitor Discharge Example
Consider an isolated 48 V capacitor with C = 2,200 µF discharging through R = 10 kΩ. Find the time required to reach 5 V.
Convert the values: C = 0.0022 F and R = 10,000 Ω.
τ = RC = 10,000 Ω × 0.0022 F = 22 s
t = 22 s × ln(48/5) = 49.7588 s
Initial current:
I0 = 48 V/10,000 Ω = 4.8 mA
Initial resistor power:
P0 = 482/10,000 = 0.2304 W
Initial capacitor energy:
E0 = ½ × 0.0022 × 482 = 2.5344 J
These calculations are nominal. A practical design must still verify resistor pulse capability and voltage rating, capacitor tolerance, remaining connected circuitry, and the actual voltage before contact.
Example: Solve for the Discharge Resistor
A 470 µF capacitor initially at 400 V must reach 60 V in 90 s in the ideal model.
R = 90 ÷ [0.000470 × ln(400/60)] = 100,937 Ω
The nominal resistance is approximately 100.9 kΩ. At the initial voltage:
I0 = 400/100,937 = 3.96287 mA
P0 = 4002/100,937 = 1.58515 W
E0 = ½ × 0.000470 × 4002 = 37.6 J
This is an analytical example, not a component recommendation. High-voltage capacitor work can be lethal and must be performed only by qualified personnel using approved equipment-specific procedures.
Choosing and Checking a Discharge Resistor
| Check | Why It Matters |
|---|---|
| Resistance tolerance | The largest R generally gives the slowest discharge; the smallest R gives the highest initial current and power. |
| Capacitance tolerance | The largest C generally gives the slowest discharge and the greatest stored energy at a given voltage. |
| Working voltage | The resistor or resistor string must withstand the initial voltage and any transient voltage distribution. |
| Initial power | Power is highest when discharge begins and then decays exponentially. |
| Pulse energy | The resistor must tolerate the energy and pulse duration without damage. |
| Repetition rate | Repeated discharges can cause cumulative heating even when one pulse is acceptable. |
| Continuous connection | A permanently connected bleeder dissipates V2/R while the supply is energized. |
| Environment and layout | Temperature, enclosure, spacing, contamination, and airflow affect safety and resistor temperature. |
Safety Limitations
Disconnect and isolate every energy source before relying on passive discharge.
Follow the equipment manufacturer's service procedure and applicable lockout or tagout requirements.
Use a discharge resistor and leads rated for the available voltage and energy; do not improvise a direct short across a charged high-energy capacitor.
Verify the voltage with an appropriately rated, functioning instrument before contact.
Continue to control or monitor stored energy if voltage can reaccumulate.
Account for series or parallel capacitor banks, hidden supplies, isolated nodes, and switching devices that may prevent the assumed discharge path.
Allow for dielectric absorption, which can cause voltage to rebound after an apparent discharge.
Treat a failed, damaged, swollen, leaking, or unknown capacitor according to the manufacturer's safety guidance.
Common Calculation Mistakes
Using µF directly where the formula requires farads.
Using kΩ directly where the formula requires ohms without keeping units consistent.
Assuming the capacitor reaches exactly 0 V after five time constants.
Ignoring the initial resistor power and total discharge energy.
Using nominal R and C values without checking worst-case tolerances.
Assuming the supply is disconnected when it still sets a nonzero equilibrium voltage.
Using an exponential resistor-load model for a constant-current or constant-power load.
Treating a calculated voltage as a substitute for measurement.
Capacitor Discharge FAQ
What determines capacitor discharge time?
In an ideal resistor-capacitor circuit, the rate is set by the time constant RC. A larger resistance or capacitance produces a slower voltage decay.
Does a capacitor fully discharge after five time constants?
No. After 5τ, an ideal capacitor discharging toward zero retains about 0.6738% of its initial voltage. Whether that voltage is acceptable depends on the initial voltage and applicable safety requirements.
Why is the discharge resistor hottest at the beginning?
The initial voltage and current are highest at t = 0, so P0 = V02/R is the maximum resistor power for a simple passive discharge toward zero.
Can I calculate discharge to exactly zero volts?
The ideal exponential reaches zero only as time approaches infinity. Use a positive target voltage for finite-time calculations and verify the actual voltage in the equipment.
Why might voltage return after a capacitor was discharged?
Dielectric absorption can cause some capacitors to recover part of their terminal voltage after the discharge path is removed. Connected circuitry or another energy source can also recharge the capacitor.
Does this formula work for every load?
No. The exponential equation applies to a linear resistive discharge path. Constant-current, constant-power, switched, or nonlinear loads produce different voltage-versus-time behavior.
Capacitor Discharge Video
Watch the original capacitor discharge video on YouTube.
References
How do you calculate the discharge of a capacitor?
Discharge of a capacitor through a resistor Capacitor discharge (voltage decay): V = Voe-(t/RC) Capacitor discharge (charge decay): Q = Qoe-(t/RC) V = Voe-(t/RC) and also I = Ioe-(t/RC) Q = Qoe-(t/RC) Capacitor charging (potential difference): V = Vo[1-e-(t/RC)]
How do you safely discharge a capacitor?
Another way to discharge a capacitor would be to source an incandescent light bulb that can tolerate the voltage held in the capacitor. Hook this up and once the bulb is no longer lit, the capacitor is discharged. Again, you always want to measure the voltage after it's supposedly discharged just to be safe.
How long will a capacitor take to discharge?
After 5 time constants, the capacitor will discharge to almost 0% of all its voltage. After 5 time constants, for all extensive purposes, the capacitor will be discharged of nearly all its voltage. A capacitor never discharges fully to zero volts, but does get very close.
Does discharging a capacitor ruin it?
To discharge a capacitor there must be a circuit, a loop, that passes through both terminals of the capacitor. In regard to the "health" of the capacitor, high discharge currents can damage it or reduce its lifespan, so it is favorable to discharge through a resistor.
How do you calculate the initial discharge of a capacitor?
A capacitor with capacitance 0.1F in an RC circuit is initially charged up to an initial voltage of Vo = 10V and is then discharged through an R=10Ω resistor as shown. The switch is closed at time t=0. Immediately after the switch is closed, the initial current is Io =Vo /R=10V/10Ω.
Is it safe to discharge a capacitor with a screwdriver?
It's often safe to discharge a capacitor using a common insulated screwdriver; however, it is usually a good idea to put together a capacitor discharge tool and use that for electronics with larger capacitors such as household appliances.
What is discharging of capacitor?
Discharging means, the capacitor giving up the stored charge in it. Let us assume, the voltage of the capacitor at fully charged condition is V volt. As soon as the capacitor is short circuited, the discharging current of the circuit would be, – V / R ampere.
What happens when a capacitor is discharging?
If a capacitor is discharging, current exits the more positive terminal rather than entering. That's really all there is to it. When current enters the more positive terminal, power is delivered to the capacitor and, thus, the stored energy increases.
Why do capacitors need to be discharged?
A capacitor can be slowly charged to the necessary voltage and then discharged quickly to provide the energy needed. It is even possible to charge several capacitors to a certain voltage and then discharge them in such a way as to get more voltage (but not more energy) out of the system than was put in.
How long does a capacitor take to discharge?
A fully charged capacitor discharges to 63% of its voltage after one time period. After 5 time periods, a capacitor discharges up to near 0% of all the voltage that it once had. Therefore, it is safe to say that the time it takes for a capacitor to discharge is 5 time constants.
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