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## Capacitor Basics

Any two electrical conductors separated by an insulating medium possess the characteristic called capacitance: the ability to store energy in the form of an electric field. Capacitance is symbolized by the capital letter C and is measured in the unit of the Farad (F). The relationship between capacitance, stored electric charge (Q), and voltage (V ) is as follows:

**Q = CV**

For example, a capacitance having a value of 33 microfarads charged to a voltage of 5 volts would store an electric charge of 165 microcoulombs.

Capacitance is a non-dissipative quantity. Unlike resistance, a pure capacitance does not dissipate energy in the form of heat; rather, it stores and releases energy from and to the rest of the circuit.

Capacitors are devices expressly designed and manufactured to possess capacitance. They are constructed of a “sandwich” of conductive plates separated by an insulating dielectric. Capacitors have voltage ratings as well as capacitance ratings. Here are some schematic symbols for capacitors:

A capacitor’s capacitance is related to the electric permittivity of the dielectric material (symbolized by the Greek letter “epsilon,” ε), the cross-sectional area of the overlapping plates (A), and the distance separating the plates (d):

**C = εA / d**

Capacitance adds when capacitors are connected in parallel. It diminishes when capacitors are connected in series:

**C _{parallel} = C_{1} + C_{2} + ・ ・ ・C_{n}**

**C _{series} = 1 / ((1 / C_{1}) + (1 / C_{2}) + ・ ・ ・(1 / C_{n}))**

The relationship between voltage and current for a capacitor is as follows:

**I = C ( dV / dt)**

As such, capacitors oppose changes in voltage over time by creating a current. This behavior makes capacitors useful for stabilizing voltage in DC circuits. One way to think of a capacitor in a DC circuit is as a temporary voltage source, always “wanting” to maintain voltage across its terminals at the same value.

The amount of potential energy (Ep, in units of joules) stored by a capacitor may be determined by altering the voltage/current/capacitance equation to express power (P = IV ) and then applying some calculus (recall that power is defined as the time-derivative of work or energy, P = *d*W / *d*t = *d*E / *d*t ):

**I = C ( dV / dt )**

**P = IV = CV ( dV / dt )**

*d*Ep / dt = CV ( *d*V / *d*t )

**( dEp / dt) / dt = CV dV**

**∫ ( dEp / dt) / dt = ∫ CV dV**

∫

∫

*d*Ep = C ∫*d*V**Ep = (1/2 ) CV ^{2}**

In an AC circuit, the amount of capacitive reactance (XC) offered by a capacitor is inversely proportional to both capacitance and frequency:

**X _{c} = 1 / (2πfC)**

This means an AC signal finds it “easier” to pass through a capacitor (i.e. less ohms of reactance) at higher frequencies than at lower frequencies.

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