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# Example18Electric Resistance Set on Boundary Face

### General

• This model consists of two bodies. Their interface has electric resistance. The voltage is applied to them.

• The distribution of the potentials and the current density are solved.

• Unless specified in the list below, the default conditions will be applied.

### Analysis Space

 Item Setting Analysis Space 3D Model Unit mm

### Show Results

 Item Setting 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 1.0×10^(-6 )[ohm]

### Results

The potential distribution is shown below.

The voltage 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 Integral 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 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 / ( 3.767×10^7 x 100 x 10^-6) + 1.0×10^-6 + 2 x 10^-3 / ( 3.767×10^7 x 100 x 10^-6)

= 2.0619×10^-6 [ohm]

Current[A] = potential[V] / total electric resistance[ohm]

= (100 – 0 ) / (2.0619×10^-6)

= 4.85×10^7 [A]

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