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Main >> Applications >> Sample problems >> Cylindrical capacitor simulation

Cylindrical capacitor dissipation factor

This is an example of the cylindrical capacitor simulation, performed with QuickField software. Capacitor consists of ceramic tube with silver electrodes mounted at the surface.

Problem Type:
Axisymmetric problem of AC conduction.

Geometry:
Cylindrical ceramic capacitor dissipation factor Capacitor consists of ceramic tube with silver electrodes mounted at the surface. Ø2 mm 3 mm ceramic silver silver

Given:
Relative permittivity of air ε = 1.
Relative permittivity of ceramic ε = 6.
Conductivity of ceramic g = 10 nS/m;
Voltage U = 10 V.
Frequency f = 1 kHz.

Task:
Determine capacitance and dissipation factor of the capacitor.

Solution:
The value of dissipation factor can be calculated as tg(δ) = PA/PR. Capacitance can be calculated as C = q/U, where U is potential difference between electrodes and q - is a charge on electrodes.

Results:
Charge q = 27.7 pC.
Capacitance C = 2.77 pF.
Active power PA = 24.5 nW.
Reactive power PR = 817 nW.
Dissipation factor tg(δ) = 0.03.

Field distribution in cylindrical capacitor:

Field distribution in cylindrical capacitor

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