Home / Examples / Electric Analysis [Coulomb] / Example 18: Electric Resistance Set on Boundary Face

Example 18: Electric Resistance Set on Boundary Face

General

 

 

Analysis Space

Item

Settings

Analysis Space

3D

Model Unit

mm

 

Show Results

Item

Settings

Solver

Electric Analysis [Coulomb]

Analysis Type

Static Analysis (Resistance)

Options

None

Model

The model consists of two rectangular solid bodies made of aluminum. They share a boundary face.

The electric wall boundary condition is set on the bottom face and the top face. The electric resistance is set as a boundary condition on the boundary face of the two bodies.

 

 

Body Attributes and Materials

Body Number/Type

Body Attribute Name

Material Name

0/Solid

conductor

001_Al *

1/Solid

conductor

001_Al *

* Available from the material DB

Boundary Condition

Boundary Condition Name/Topology

Tab

Boundary Condition Type

Setting

100volt/Face

Electric

Electric Wall

100 [V]

0volt/Face

Electric

Electric Wall

0 [V]

Electric Resistance/Face

Electric

Electric Resistance

1.0 X 10-6 [ohm]

 

Results

The electric potential distribution is shown below.

The electric potential changes discontinuously on the face, on which the electric resistance is set.

 

 

The current density vectors are shown below.

 

 

 

 

The integral of the heat flux is calculated by Integration on Results Window. Calculation result of the current density on the top face of the body is shown below.

 

 

Theoretical value is calculated as shown below. Analysis results almost match with the theoretical value.

 

Electric resistance of the body [ohm] = Thickness in current density direction [m] / (cross sectional area [mm^2] x conductivity [1/ohm/m] ),

Body dimensions: 10 [mm] x 10 [mm] x 2 [mm], Electric Conductivity of Aluminum: 3.767×10^7 [1/ohm/m]

With these items above, calculation is done as below.

Total electric resistance = electric resistance of upper body + electric resistance at boundary + electric resistance of lower body

= 2 x 10-3 / ( 3.767 x 107 x 100 x 10-6 ) + 1.0 x 10-6 + 2 x 10-3 / ( 3.767 x 107 x 100 x 10-6 )

=2.0619 x 10-6 [ohm]

 

Current [A] = Electric Potential [V] / Total Electric Resistance [ohm]

= (100 - 0 ) / ( 2.0619 x 10-6 )

= 4.85 x 107 [A]