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

Cylindrical capacitor

cylindrical capacitor, ceramic capacitor dissipation factor, ceramic capacitor loss angle tangent, cylindric capacitor dissipation factor

This is an example of the cylindrical capacitor simulation, performed with QuickField software.

Problem Type:
Axisymmetric problem of AC conduction.

Geometry:

Capacitor consists of ceramic tube with silver electrodes mounted at the surface.

Given:

Relative permittivity of air ε = 1.

Relative permittivity of ceramic ε = 6.
Conductivity of ceramic g = 10^-8 S/m;
Voltage U = 10 V.
Frequency f = 1000 Hz.

Problem:

Determine capacitance and dissipation factor of the capacitor.

Solution:

The value of dissipation factor can be calculated as

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

Results:
Field distribution in cylindrical capacitor:

	QuickField
q, C	2.77·10-11
C, F	2.77·10-12
PA, W	2.45·10-8
PR, W	8.17·10-7
tg(δ)	0.03

Watch on YouTube.

View simulation report in PDF.

Download simulation files (files may be viewed using any QuickField Edition).