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The energy of an ideal capacitor is the work required to build voltage in its electric field.
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Formula
W in J; C in F; V in V
| Symbol | Meaning | Unit |
|---|---|---|
| Stored electrical energy | J | |
| Capacitance | F | |
| Voltage across the terminals | V |
W_C=\frac{1}{2}CV^2Equivalent forms are W=½CV2=½QV=Q2/(2C). They apply to a linear capacitor with constant capacitance; energy resides in the electric field and remains stored in an ideally isolated component. Leakage, ESR, voltage rating and discharge paths limit usable energy in real circuits.
Equivalent forms are W=½CV2=½QV=Q2/(2C). They apply to a linear capacitor with constant capacitance; energy resides in the electric field and remains stored in an ideally isolated component. Leakage, ESR, voltage rating and discharge paths limit usable energy in real circuits.
The energy of an ideal capacitor is the work required to build voltage in its electric field.
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Work through the numerical example for “Energy stored in a capacitor” and check its unit and magnitude.
A 100 μF capacitor at 12 V stores W=0.5·100·10−6·122=7.2 mJ.
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: W in J; C in F; V in V.
Formula cluster
The energy of an ideal capacitor is the work required to build voltage in its electric field.
A 100 μF capacitor at 12 V stores W=0.5·100·10−6·122=7.2 mJ.
Estimate energy storage in pulse circuits, filters and capacitor banks.
Here is how to work through a typical Energy stored in a capacitor (WC = ½CV²) task step by step:
Task
Work through the numerical example for “Energy stored in a capacitor” and check its unit and magnitude.
Solution path
A 100 μF capacitor at 12 V stores W=0.5·100·10−6·122=7.2 mJ.
During charging, capacitor voltage rises from zero to V. Average voltage over the transferred charge is V/2, so work is QV/2. Substituting Q=CV gives W=½CV2. Omitting the half incorrectly assumes final voltage existed throughout charging. Assume a linear capacitor and use terminal voltage; real discharge circuits must account for ratings and stored-energy hazards. Application: Estimate energy storage in pulse circuits, filters and capacitor banks. The numerical example A 100 μF capacitor at 12 V stores W=0.5·100·10−6·122=7.2 mJ anchors the idea in a concrete measurement case.
If charge is known, W=Q2/(2C) may be convenient; with charge and voltage use W=½QV. All three forms give the same energy for a linear capacitor. Check units: farad times volt squared is joules. Assume a linear capacitor and use terminal voltage; real discharge circuits must account for ratings and stored-energy hazards. The corresponding units are W in J; C in F; V in V. A 100 μF capacitor at 12 V stores W=0.5·100·10−6·122=7.2 mJ. This checks that the reference quantities are compatible.
Not necessarily. Switches, wiring resistance, ESR and controls dissipate energy, and a load may use only part of the voltage range. High-voltage capacitors can remain dangerously charged after shutdown. Real discharge requires suitable bleeders, component limits and safe procedures. Assume a linear capacitor and use terminal voltage; real discharge circuits must account for ratings and stored-energy hazards. Scope: Equivalent forms are W=½CV2=½QV=Q2/(2C). They apply to a linear capacitor with constant capacitance; energy resides in the electric field and remains stored in an ideally isolated component. Leakage, ESR, voltage rating and discharge paths limit usable energy in real circuits. A practical application is Estimate energy storage in pulse circuits, filters and capacitor banks.
Write the quantity and unit from the variable list next to every value: W in J; C in F; V in V. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example A 100 μF capacitor at 12 V stores W=0.5·100·10−6·122=7.2 mJ. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. Assume a linear capacitor and use terminal voltage; real discharge circuits must account for ratings and stored-energy hazards.
The background states this starting condition: Equivalent forms are W=½CV2=½QV=Q2/(2C). They apply to a linear capacitor with constant capacitance; energy resides in the electric. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. Assume a linear capacitor and use terminal voltage; real discharge circuits must account for ratings and stored-energy hazards.
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Which formula describes Energy stored in a capacitor?
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How do you rearrange Energy stored in a capacitor for Base equation for the target quantity?
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