**How much power does a capacitor dissipate in an electrical circuit?**

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 ESR 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. As the ESR increases, so also does the power.

**What is Power Dissipation?**

Power dissipation refers to the process of converting electrical energy into heat energy in a circuit. When current flows through a resistor, some of the electrical energy is converted into heat due to the resistance offered by the resistor. This conversion of electrical energy into heat is known as power dissipation.

The amount of power dissipated in a circuit can be found using the formula

**P = V _{RMS}^{2}/R = I_{RMS}^{2} * R**

where P is the power dissipated, V_{RMS} is the RMS voltage across the resistor, and R is the resistance of the resistor. It can also be expressed in terms of the RMS current.

From this formula, it is evident that power dissipation is directly proportional to the square of the voltage across or current through the resistor. This means that higher voltages or current will result in greater power dissipation.

Power dissipation is an important consideration in circuit design as it helps in understanding the efficiency and heat generated by the circuit. By managing power dissipation, analog design engineers can ensure that the components in the circuit do not overheat as this can lead to potential damage or long term degradation of performance. The unit of power dissipation is Watt.

**Practical Applications**

An analog design engineer can use the ESR value to determine the power dissipated by the capacitor. From this they can calculate the rise in temperature above ambient conditions and then determine whether the capacitor is within operating limits.

For example, if the ESR is 5 mΩ and the maximum RMS current is 20 A, then the calculator can be used to find the power dissipated P_{d} = 2 Watt.

If the conductivity G is 100 mW/^{o}C, then the rise in temperature ΔT is given by

ΔT = P_{d}/G = 2000/100 = 20^{o}C

*Additional details in the capacitor heat dissipation calculator*

The temperature of the capacitor under these conditions is given by

**T _{C} = T_{amb} + ΔT**

If the ambient temperature **T _{amb}** = 60

^{o}C, then

**T**= 60

_{C}^{o}C + 20

^{o}C = 80

^{o}C

The electrical data in the data sheet should provide the maximum rated voltage at two different temperatures and using a straight line extrapolation we will be able to determine, what the max voltage at 80^{o}C will be. *In general the higher the operating temperature, the lower the max voltage. *

**Related Calculators**

- Capacitor Dissipation Factor. Calculates the Dissipation Factor (DF) – a measure of a capacitor’s dielectric losses

**References**

[1] Calculating and Interpreting Power Dissipation for Polypropylene Film DC-Link Capacitors