### Capacitor impedance Calculation Formula:

Capacitor impedance is nothing but an amount of reactance that can be offered by a pure capacitor at the given frequency. It is also called as capacitive reactance. Hence, the capacitor impedance Xc _{(ohm)} in ohm is equal to the inverse product of 6.28 times of frequency f_{(Hz)} in hertz and capacitance C_{(F)} in farad.

Hence the calculation formula can be written as,

**Xc _{(ohm)} = 1/ (6.28 x f_{(Hz)} x C_{(F)})**

In some time, the value of the capacitance is denoted by nF (Nano farad), pF (Pico farad), µF (micro farad) mF (milli Farad). Also, the capacitor value will be kHz or MHz or GHz.

Here Capacitor

Units |
Farad |

pF | 1 x 10^{-12} |

nF | 1 x 10^{-9} |

µF |
1 x 10^{-6} |

mF | 1 x 10^{-3} |

Here Frequency,

Units |
Frequency (Hz) |

Hz | 1 |

kHz | 1 x 10^{3} |

MHz | 1 x 10^{6} |

GHz |
1 x 10^{9} |

Or

**Capacitive impedance = 1 / (6.28 x frequency x capacitance).**

__Case-1:__

You may do not know the capacitance value, hence you can use the below formula for calculating the capacitance.

**Capacitance (F) = charge (q) / potential difference (volts).**

**C=q/V**

**Farad = Coulomb / volts**

Therefore, the capacitance impedance will be,

**Xc _{(ohm)} = 1/ (6.28 x f_{(Hz)} x q_{(c)} x V_{(V)})**

__Case 2:__

Capacitance of the parallel plate will be permittivity times of the surface area A_{(mm}^{2}_{) } divided by the distance between the plates.

**C = ε x A / d**

Hence the formula can be written as,

**Xc = 1 / (6.28 x f _{(Hz)} x ε x A / d)**

**= d / (6.28 x f _{(Hz)} x ε x A)**

__Case 3:__

Capacitance impedance also is equal to inverse of capacitance (F) times of the angular frequency (rad/s).

**Xc = 1 / ωC**

__Examples:__

Calculate the capacitance impedance of the 2 micro farad two parallel plates working in 50Hz?

Solution :

Xc = 1 / (6.28 x 2 x 10^{-6} x 50)

= 0.000628 ohms