An ideal capacitor does not dissipate any power. However there is no such thing as an ideal cap. In reality, there’s an associated series resistance called the Equivalent Series Resistance (use this calculator to find its value).

Enter the

- ESR
- Current

to find the dissipated power

**Formula**

**P _{d} = I_{RMS}^{2} * ESR**

**Example Calculation**

If the capacitor has an ESR of **5 mΩ** and the RMS current is **7 Amperes**, this results in **245 mW** of dissipated power.

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