Capacitor Energy & Charge Calculator
Calculates the energy stored in a capacitor and its remaining energy at any point during discharge through a resistor. Use it when designing RC circuits, timing circuits, or energy storage systems.
About this calculator
A capacitor stores energy in its electric field, given by E = ½CV², where C is capacitance in farads and V is voltage in volts. When discharged through a resistor R, the voltage decays exponentially: V(t) = V₀ × e^(−t/RC), where RC is the time constant τ. The energy remaining at time t is therefore E(t) = ½C × [V₀ × e^(−t/RC)]² × 1000 (converted to millijoules). The time constant τ = RC tells you how quickly the capacitor loses its charge — after one τ, about 63% of the initial charge has drained. Engineers use these relationships to size capacitors for flash circuits, power supplies, and signal filters.
How to use
Suppose you have a 100 μF capacitor charged to 12 V, discharging through a 1000 Ω resistor. After t = 0.05 s: τ = RC = 1000 × (100 / 1,000,000) = 0.1 s. Decayed voltage = 12 × e^(−0.05 / 0.1) = 12 × e^(−0.5) ≈ 12 × 0.6065 ≈ 7.278 V. Remaining energy = 0.5 × (100 / 1,000,000) × 7.278² × 1000 ≈ 2.648 mJ. Compare that to the initial 7.2 mJ to see how much energy has dissipated in the resistor.
Frequently asked questions
What is the time constant of an RC circuit and why does it matter?
The time constant τ = R × C (in seconds) defines how fast a capacitor charges or discharges. After one τ, the voltage drops to about 36.8% of its initial value; after 5τ the capacitor is considered fully discharged (≈ 0.7% remaining). It matters because it directly controls the timing behaviour of filters, oscillators, and debounce circuits. Choosing R and C to set a specific τ is one of the most common tasks in analog circuit design.
How do you calculate the energy stored in a capacitor?
The energy stored is E = ½CV², where C is in farads and V is the voltage across the capacitor. For example, a 470 μF cap at 5 V stores ½ × 470 × 10⁻⁶ × 25 ≈ 5.875 mJ. This energy is held in the electric field between the capacitor's plates. Larger capacitance or higher voltage both increase stored energy, but voltage has the stronger effect because it appears squared.
Why does a capacitor discharge exponentially rather than linearly?
As the capacitor discharges through a resistor, the current driving the discharge is proportional to the remaining voltage (I = V/R). As voltage drops, current drops too, meaning the rate of discharge slows continuously — producing the characteristic exponential curve V(t) = V₀e^(−t/RC). This self-limiting behaviour is a direct consequence of Ohm's Law combined with the capacitor's charge-voltage relationship Q = CV. A linear discharge would require a constant-current load rather than a resistor.