Current flow (v-formulation)

What you’ll learn

• Current flow interactions — constraint, current density source, Lump V/I, and Lump V/I Cut.
• Simply connected domains — when and why cut surfaces are required for a unique potential solution.
• Periodicity — reducing domain size for symmetric conducting problems.
• Weak formulation — Laplace’s equation for steady-state current flow using potential $v$.

This page describes the available interactions and couplings for steady-state current flow using the electric scalar potential $v$ in Allsolve. The weak formulation is stated up front; the full derivation is at the end of this page.

Weak formulation

The weak form solved by Allsolve is

$$\begin{align} \int_{\Omega}\; (-\sigma \nabla v \cdot \nabla v^{\prime})\; d \Omega + \int_{\Gamma}\; (-\boldsymbol{J} \cdot \boldsymbol{n})\; v^{\prime}\; d \Gamma = 0. \end{align}$$

where $v$ is the electric scalar potential, $\boldsymbol{J} = \sigma \boldsymbol{E}$ the current density, and $\sigma$ the conductivity.

See Formulation derivation for the full strong-to-weak derivation.

Interactions

Constraint

Applies a fixed value to the electric scalar potential $v$. Use this when you need to fix the electric potential at a node or within a region. This can be used to drive a potential difference between two capacitor plates.

How to use	Provide an electric potential value in point or region in Volts
Example	$v = 100$ applies an electric scalar potential of $100\:V$ to the specified region.
Unit	Electric potential in Volts (V)

Current density source

Applies a current density vector $\boldsymbol{J}$ as a source term to a region. Used to model externally driven current distributions, such as in coils, where the current is prescribed rather than solved for. Can be specified as a constant vector or a spatially varying field.

How to use	You can use either the matrix editor or the expression editor. With the matrix editor, provide a 3×1 matrix for a 3D problem (2×1 for 2D, etc.). In the expression editor, write the direct expression.
Example	[1000; 5000; 0] drives a current density of $J_x = 1000$ A/m^regdim, $J_y = 5000$ A/m^regdim and $J_z = 0$ A/m^regdim to the specified region.
Unit	Ampere/m^regdim. The regdim is the dimension of the target region (e.g., A/m^2 for surface targets).

Lump V/I

Applies a lumped voltage $V$ or current $I$ to a specific node or region. Unlike the Lump V/I Cut, this interaction does not require a cut surface and is used for simpler geometries where the domain is already simply connected. Commonly used to define electrode potentials or inject current into a conducting region.

How to use	Specify the target curve. From Actuation mode, select either voltage, current or circuit coupling. Fill in the value.
Example	$I = 100$ applies 100 Amperes to the curve.
Unit	Voltage in Volts (V) or current in Amperes (A)

Lump V/I Cut

Applies a lumped voltage $V$ or current $I$ through a cut surface in the domain. In the electric potential formulation, cuts are required to make the domain simply connected, ensuring a unique solution for v. This interaction is used to drive current or voltage through conducting regions such as coils, without explicitly resolving the detailed field distribution inside the conductor.

How to use	Specify the target curve (must be a closed boundary of some surface). From Actuation mode, select either voltage, current or circuit coupling. Fill in the value.
Example	$I = 100$ applies 100 Amperes through the cut surface. See also Lump I/V Cut in Induction motor.
Unit	Current in Amperes (A) or voltage in Volts (V)

Periodicity

Imposes periodic boundary conditions on the electric potential $v$ between two boundaries. Reduces the computational domain size for geometrically symmetric or antisymmetric problems, avoiding the need to model the full geometry.

Example	Periodicity follows the same principles, regardless of which physics module it belongs to. See how periodicity is used in magnetism as a reference: Periodicity in electric motors

Compatibilities with analysis types

Current flow supports all analysis types: Static, Harmonic, Multiharmonic, Transient, and Eigenmode.

See the full compatibility table in Simulations overview.

Use cases

• Heat solid - Current flow coupling in AC system

Formulation derivation

Maxwell’s equations

$$\text{Gauss's law}$$$$\nabla \cdot \boldsymbol{D} = \rho$$$$\text{Gauss's law for magnetism}$$$$\nabla \cdot \boldsymbol{B} = 0$$$$\text{Faraday's law}$$$$\nabla \times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t}$$$$\text{Ampere-Maxwell law}$$$$\nabla \times \boldsymbol{H} = \boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}$$

Strong formulation

In the case of a non-zero constant current flow Ampere-Maxwell law states, that

$$\begin{align*} \nabla \times \boldsymbol{H} = \boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}. \tag{1} \end{align*}$$

Taking the divergence of both sides we get

$$\begin{align*} \nabla \cdot (\nabla \times \boldsymbol{H}) &= \nabla \cdot (\boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}) \tag{2} \end{align*}$$

and

$$\begin{align*} 0 &= \nabla \cdot (\boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}), \tag{3} \end{align*}$$

since the divergence of a curl is zero.

Under the assumption of constant current flow (or approximating a slowly varying current), the time derivative of the electric displacement field $\frac{\partial \boldsymbol{D}}{\partial t}$ equals zero.

$$\begin{align*} \nabla \cdot \boldsymbol{J} = 0 \tag{4} \end{align*}$$

This indicates that the current flow is non-divergent. We can write a current density using Ohm’s law in differential form as

$$\begin{align*} \boldsymbol{J} &= \sigma \boldsymbol{E}, \tag{5} \end{align*}$$

where $\sigma$ is the conductivity of the material.

Thus we have

$$\begin{align*} \nabla \cdot (\sigma \boldsymbol{E}) &= 0 \tag{6} \\[5pt] \nabla \cdot (- \sigma \nabla v) &= 0 \tag{7} \\[5pt] \nabla \cdot (\sigma \nabla v) &= 0 \tag{8} \\[5pt] \end{align*}$$

resulting in Laplace’s equation.

Deriving the weak form

To obtain the weak form, multiply by the test function $v^{\prime}$ and integrate over the domain $\Omega$:

$$\begin{align*} \int_{\Omega}\; (\nabla \cdot (\sigma \nabla v))\; v^{\prime}\; d \Omega = 0 \tag{9} \end{align*}$$

Applying the Leibniz rule for nabla a operator we get

$$\begin{align*} \int_{\Omega}\; \nabla \cdot (v^{\prime}\; \sigma \nabla v)\; d \Omega + \int_{\Omega}\; -(\nabla v^{\prime} \cdot \sigma \nabla v)\; d \Omega = 0 \tag{10} \end{align*}$$

For the first term we can use the Divergence theorem

$$\begin{align*} \int_{\Gamma}\; (v^{\prime}\; \sigma \nabla v) \cdot \boldsymbol{n}\; d \Gamma + \int_{\Omega}\; -(\nabla v^{\prime} \cdot \sigma \nabla v)\; d \Omega = 0 \tag{11} \end{align*}$$

Rearranging the terms and using relation $\boldsymbol{E} = - \nabla v$ we obtain

$$\begin{align*} \int_{\Omega}\; (-\sigma \nabla v \cdot \nabla v^{\prime})\; d \Omega + \int_{\Gamma}\; -\sigma \boldsymbol{E} \cdot \boldsymbol{n}\; v^{\prime}\; d \Gamma = 0 \tag{12} \end{align*}$$

Using relation $\boldsymbol{J} = \sigma \boldsymbol{E}$ on the Neumann term results in the final weak formulation stated at the top of this page.