ACElec2: Cylindrical capacitor

AC current in the cylindrical capacitor

Problem Type:
An axisymmetric problem of AC conduction.

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

Given:
Relative permittivity of air e = 1.

Relative permittivity of ceramic e = 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) = 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:

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

See the ACElec2.pbm problem in the Examples folder.