# BusBar

In this example, we will model a busbar simply modeled as rectangular parallelepiped.A differential electrical potential is applied to the entry/exit of the busbar. We note respectively $V_0$ the electrical potential on the entry and and $V_1$ on the exit.

## 1. Running the case

The command line to run this case in 2D is

``mpirun -np 4 feelpp_toolbox_electric --case "github:{path:toolboxes/electric/busbar}" --case.config-file 2d.cfg``

The command line to run this case in 3D is

``mpirun -np 4 feelpp_toolbox_thermoelectric --case "github:{path:toolboxes/electric/busbar}" --case.config-file 3d.cfg``

## 2. Geometry

The busbar conductor consists in a rectangular cross section extruded along the x axis.+ In 2D, the geometry is as follow:
In 3D, this is the same geometry, but extruded along the z axis.

## 3. Input parameters

Name Description Value Unit

$Lx$

1

$m$

$Ly$

2

$m$

$Lz$

angle

$\pi/2$

$rad$

$V_D$

electrical potential

9

$V$

### 3.1. Model & Toolbox

• This problem is fully described by the Electric model, namely a poisson equation for the electrical potential $V$ with Dirichlet boundary conditions on entry /exit.

• toolbox: electric

### 3.2. Materials

Name Description Marker Value Unit

$\sigma$

electric conductivity

omega

$4.8e7$

$S.m^{-1}$

### 3.3. Boundary conditions

The boundary conditions for the electrical probleme are introduced as simple Dirichlet boundary conditions for the electric potential on the entry/exit of the conductor. For the remaining faces, as no current is flowing througth these faces, we add Homogeneous Neumann conditions.

Marker Type Value

V0

Dirichlet

0

V1

Dirichlet

$V_D$

Lside, Rside, top*, bottom*

Neumann

0

*: only in 3D

## 4. Outputs

The main fields of concern are the electric potential $V$, the current density $\mathbf{j}$ and the electric field $\mathbf{E}$. // presented in the following figure.

## 5. Verification Benchmark

The analytical solution is given by:

$V(x) = V_0 + (\frac{x}{L}-1) (V_1-V_0)$

Note that we may get an expression for the resistance $R$ of the busbar from this equation. We recall that $R$ is defined as $R = V_D/I$ where $I$ stands for the total current flowing in the busbar ($V_D$ corresponds to the differential applied voltage).

By definition:

$I = \int_{V0} \mathbf{j} \cdot \mathbf{n} \,d\Gamma$

From Gauss law we have: $\mathbf{j} = \sigma\,\mathbf{E} = -\sigma \nabla V$. It follows:

$R = \frac{1}{\sigma} \frac{Lx}{S}$

with $S=Ly*Lz$.

We will check if the approximations converge at the appropriate rate:

• k+1 for the $L_2$ norm for $V$

• k for the $H_1$ norm for $V$

• k for the $L_2$ norm for $\mathbf{E}$ and $\mathbf{j}$

• k-1 for the $H_1$ norm for $\mathbf{E}$ and $\mathbf{j}$