Capacitor Calculator

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Capacitor Energy & Charge Calculator

Calculate the stored electrical charge and energy in a capacitor based on capacitance and voltage.

Farads (F) Millifarads (mF) Microfarads (µF) Nanofarads (nF) Picofarads (pF)
2 Places 4 Places 6 Places 9 Places
Total Charge (Q): 0
Stored Energy (E): 0
Capacitance in Farads: 0

How to Use the Capacitor Calculator

A capacitor is a passive two-terminal electronic component that stores electrical energy in an electric field. This calculator helps you determine the amount of charge (measured in Coulombs) and the total energy (measured in Joules) stored within a capacitor for a given voltage level.

To use this tool, follow these steps:

  • Capacitance Value: Enter the numerical value printed on your capacitor (e.g., 220, 0.1, 4700).
  • Unit: Select the correct multiplier. Most commercial capacitors are labeled in microfarads (µF), nanofarads (nF), or picofarads (pF).
  • Voltage: Enter the DC voltage applied across the capacitor terminals.

Fundamental Formulas

The calculations are based on two primary laws of physics governing electrostatics:

1. Charge Formula (Q = CV)

The charge stored in a capacitor is directly proportional to the capacitance and the voltage applied.
Q = C × V
Where Q is Charge (Coulombs), C is Capacitance (Farads), and V is Voltage (Volts).

2. Stored Energy Formula (E = ½CV²)

Energy storage increases linearly with capacitance but quadratically with voltage. This means doubling the voltage results in four times the energy storage.
E = 0.5 × C × V²
Where E is Energy (Joules), C is Capacitance (Farads), and V is Voltage (Volts).

Unit Conversion Reference

Unit Name Symbol Multiplier (Farads)
Millifarad mF 10-3
Microfarad µF 10-6
Nanofarad nF 10-9
Picofarad pF 10-12

Practical Example

Imagine you have a 470µF electrolytic capacitor connected to a 25V power supply. How much energy is it holding?

  1. Convert 470µF to Farads: 470 × 10⁻⁶ = 0.00047 F.
  2. Calculate Charge: 0.00047 F × 25V = 0.01175 Coulombs.
  3. Calculate Energy: 0.5 × 0.00047 F × (25V)² = 0.5 × 0.00047 × 625 = 0.146875 Joules.

Why Knowing Stored Energy Matters

In power supply design, the energy stored in capacitors helps smooth out voltage ripples. In high-voltage applications, knowing the stored energy is critical for safety, as even a small capacitor charged to a high voltage can hold enough energy (Joules) to cause a painful or dangerous electrical shock.

function calculateCapacitanceLogic() { var val = parseFloat(document.getElementById('capValue').value); var unitMult = parseFloat(document.getElementById('capUnit').value); var voltage = parseFloat(document.getElementById('capVoltage').value); var precision = parseInt(document.getElementById('capPrecision').value); if (isNaN(val) || isNaN(voltage)) { alert("Please enter valid numerical values for Capacitance and Voltage."); return; } if (val < 0 || voltage < 0) { alert("Values must be positive."); return; } // Convert input to base Farads var farads = val * unitMult; // Calculate Charge (Q = CV) var charge = farads * voltage; // Calculate Energy (E = 0.5 * C * V^2) var energy = 0.5 * farads * Math.pow(voltage, 2); // Update Display document.getElementById('resCharge').innerHTML = charge.toFixed(precision) + " C (Coulombs)"; document.getElementById('resEnergy').innerHTML = energy.toFixed(precision) + " J (Joules)"; document.getElementById('resFarads').innerHTML = farads.toExponential(4) + " F"; // Show Results document.getElementById('resultsArea').style.display = "block"; }

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