Power Equation for Capacitors: DC/AC Derivation, Examples, and Sizing

The Core Power Equations for Capacitors (DC & AC)

When bench-testing or designing power supplies, you quickly learn that a capacitor does not dissipate real power like a resistor. Instead, it stores and releases energy. Because of this, the power equation for capacitor circuits splits into two distinct domains: instantaneous transient power (DC/switching) and reactive power (AC steady-state). Using the wrong equation for your domain is the fastest way to undersize a component and watch it vent electrolyte across your workbench.

In DC or transient circuits, we calculate instantaneous power $p(t)$, which is the rate of energy transfer at any exact microsecond. In AC circuits, we calculate reactive power $Q$, which represents the energy sloshing back and forth between the source and the capacitor's electric field.

The fundamental instantaneous power equation is $p(t) = v(t) \cdot i(t)$. Since capacitor current is defined as $i(t) = C \frac{dv(t)}{dt}$, we substitute to get the primary DC/transient power equation:

$$p(t) = C \cdot v(t) \cdot \frac{dv(t)}{dt}$$

For AC sinusoidal steady-state, the reactive power equation is derived from $Q = \frac{V_{RMS}^2}{X_C}$, where capacitive reactance $X_C = \frac{1}{2\pi f C}$:

$$Q = V_{RMS}^2 \cdot 2\pi f C$$

Rearranged Forms and Unit Traps

On the bench, you rarely solve for power directly. You usually know your power budget or reactive power requirement and need to back-solve for capacitance or voltage slew rate. Here are the rearranged forms:

• Solve for Capacitance (DC): $C = \frac{p(t)}{v(t) \cdot \frac{dv}{dt}}$
• Solve for Slew Rate (DC): $\frac{dv}{dt} = \frac{p(t)}{C \cdot v(t)}$
• Solve for Capacitance (AC): $C = \frac{Q}{V_{RMS}^2 \cdot 2\pi f}$
• Solve for Voltage (AC): $V_{RMS} = \sqrt{\frac{Q}{2\pi f C}}$

Worked Problem 1: DC Transient Instantaneous Power

Scenario: You are designing a soft-start circuit for a 48V DC motor drive. The DC bus uses a $1000\mu F$ bulk capacitor. The soft-start ramps the voltage linearly from 0V to 48V over 10 milliseconds. What is the peak instantaneous power demanded from the supply during this ramp?

1. Convert units to base SI: $C = 1000\mu F = 1000 \times 10^{-6} F = 0.001 F$. Time $t = 10ms = 0.01s$.
2. Calculate the voltage slew rate ($dv/dt$): Because the ramp is linear, $\frac{dv}{dt} = \frac{\Delta V}{\Delta t} = \frac{48V - 0V}{0.01s} = 4800 \text{ V/s}$.
3. Identify the condition for peak power: Looking at $p(t) = C \cdot v(t) \cdot \frac{dv}{dt}$, $C$ and $\frac{dv}{dt}$ are constant. Therefore, $p(t)$ is maximum when $v(t)$ is at its maximum. Peak voltage is $48V$ at $t = 10ms$.
4. Substitute and solve: $p_{peak} = 0.001 F \cdot 48V \cdot 4800 \text{ V/s}$.
5. Track the units: $F \cdot V \cdot \frac{V}{s} = \left(\frac{C}{V}\right) \cdot V \cdot \frac{V}{s} = \frac{C \cdot V}{s} = \text{Joules/s} = \text{Watts}$.
6. Final Calculation: $p_{peak} = 0.001 \cdot 48 \cdot 4800 = 230.4 \text{ W}$.

Bench Reality Check: Your power supply must be able to source a momentary 230W surge at the very end of the ramp, even though the average power over the 10ms is exactly half of that (115.2W). If your supply folds back at 150W, the soft-start will stall.

Worked Problem 2: AC Reactive Power Sizing

Scenario: You are replacing a failed motor run capacitor on a 240VAC, 60Hz single-phase compressor. The nameplate specifies a $45\mu F$ capacitor. You need to verify the reactive power rating to ensure the replacement can handle the continuous AC stress.

1. Convert units: $C = 45\mu F = 0.000045 F$. $V_{RMS} = 240V$. $f = 60Hz$.
2. Calculate angular frequency: $2\pi f = 2 \cdot 3.14159 \cdot 60 = 376.99 \text{ rad/s}$.
3. Apply the AC reactive power equation: $Q = V_{RMS}^2 \cdot 2\pi f \cdot C$.
4. Substitute values: $Q = (240)^2 \cdot 376.99 \cdot 0.000045$.
5. Calculate intermediate steps: $240^2 = 57,600$. Then, $57,600 \cdot 376.99 = 21,714,624$.
6. Final Calculation: $21,714,624 \cdot 0.000045 = 977.16 \text{ VAR}$.

The capacitor must continuously exchange 977 VAR with the motor windings. This does not mean it dissipates 977 Watts of heat (which would instantly vaporize it), but it dictates the physical size and dielectric thickness required to handle the internal displacement current without breaking down. For a deep dive on how reactive power impacts grid efficiency, refer to the All About Circuits guide on AC Reactive Power.

Real-World Decision Path: Sizing the Right Capacitor

Knowing the math is only half the battle; picking the right physical component is where designs succeed or fail. Use this decision tree to terminate your calculation into a concrete part selection.

Realistic Magnitudes and ESR Reality Checks

The equations above assume an ideal capacitor. In reality, every capacitor has Equivalent Series Resistance (ESR). While the reactive power $Q$ might be 977 VAR, the real power dissipated as heat is governed by $P_{real} = I_{RMS}^2 \cdot R_{ESR}$. Understanding this magnitude is critical for thermal management.

Let us look at realistic magnitudes for a $1000\mu F$ aluminum electrolytic capacitor in a switching power supply output:

• Ripple Current ($I_{RMS}$): 4.5 A
• Datasheet ESR ($R_{ESR}$): 0.08 $\Omega$ (80 m$\Omega$) at 100kHz
• Real Dissipated Power: $P = (4.5)^2 \cdot 0.08 = 20.25 \cdot 0.08 = 1.62 \text{ W}$.

A realistic answer magnitude for real power dissipation in a through-hole capacitor is between 0.5W and 3W. If your ESR calculation yields 15W of dissipation in a standard 10x16mm can, your math is wrong, or your capacitor is about to vent. The Georgia State University HyperPhysics capacitor energy module provides excellent baseline theory on how this stored energy translates to physical thermal limits. Always cross-reference your calculated $I_{RMS}$ against the manufacturer's ripple current rating, and apply the temperature derating curves found in the datasheet.