
AC current in the plane capacitor
Problem Type:
A planeparallel problem of AC conduction.
Geometry:
Due to symmetry only a small part of 1 mm height is used. The length of capacitor in z direction is L = 10 mm.
Given:
Relative permittivity of substrate ε = 10;
Conductivity of substrate g = 10^{8} S/m.
Voltage U = 5 V,
Frequency f = 50 Hz.
Problem:
Find current and dissipation factor tg(δ) of the plane capacitor with nonideal dielectric inside.
Solution:
Capacitor with nonideal dielectric can be replaced by electric scheme with ideal capacitor C and resistivity R connected in parallel.
The capacitance of the plane capacitor is calculated by the equation
C = εε_{0}S/d, where S is the plate area S = h·l.
Resistance of the substrate is calculated by the equation
R = ρ·d/S.
Current I has two components: active I_{A} and reactive I_{R}.
For parallel scheme
I_{A} = U / R, I_{R} = U / X_{C}.
tg(δ) = P_{A} / P_{R} = U·I_{A} / U·I_{R} = I_{A} / I_{R}
Results:
S = 100 mm^{2}.
R = (1/10^{8})·0.1e3/100e6 = 10^{9} Ohm.
C = 10·8.854e12·100e6/0.1e3 = 8.854·10^{11} F.
X_{C} = 1 / 2·3.1416·50·8.854·10^{11} = 0.3595·10^{9} Ohm.
I_{A} = 5 / 10^{9} = 5·10^{9} A
I_{R} = 5 / 0.3595·10^{9} = 13.908·10^{9} A
tg(δ) = 5·10^{9} / 13.908·10^{9} = 0.36
QuickField 
Theoretical result 

I_{A}, A 
5·10^{9} 
5·10^{9} 
I_{R}, A 
13.908·10^{9} 
13.908·10^{9} 
tg(δ) 
0.36 
0.36 
See the ACElec1.pbm problem in the Examples folder.