Understanding Field Energy and Co-energy in Linear Electromagnetic Circuits

In electromagnetic systems, energy is stored in the magnetic field. This stored energy is often referred to as field energy. Related to field energy is a concept called co-energy, which is useful in calculating forces in magnetic systems, especially when the relationship between flux linkage and current is non-linear.

Let's define these terms mathematically for a simple magnetic system where flux linkage ($\lambda$) is a function of current ($i$).

• Field Energy ($\boldsymbol{W_f}$): This is the energy stored in the magnetic field. It is given by the area under the $\lambda$ vs $i$ curve, integrated with respect to flux linkage:

  \( W_f = \int_{0}^{\lambda} i(\lambda') d\lambda' \)

• Co-energy ($\boldsymbol{W_c}$): This is given by the area under the $i$ vs $\lambda$ curve, integrated with respect to current:

  \( W_c = \int_{0}^{i} \lambda(i') di' \)

The total area of the rectangle formed by the point $(\lambda, i)$ on the characteristic curve is given by $\lambda i$. For any electromagnetic system (linear or non-linear), the sum of field energy and co-energy is equal to this product:

\( W_f + W_c = \lambda i \)

Analyzing Linear Electromagnetic Circuits

A linear electromagnetic circuit is one where the flux linkage ($\lambda$) is directly proportional to the current ($i$). This relationship is defined by the inductance ($L$), which is a constant value in a linear system:

\( \lambda = L i \)

Let's calculate the field energy ($W_f$) for a linear system using the formula \( W_f = \int_{0}^{\lambda} i(\lambda') d\lambda' \). From $\lambda = Li$, we have $i = \lambda/L$. Substituting this into the integral:

\( W_f = \int_{0}^{\lambda} \frac{\lambda'}{L} d\lambda' = \frac{1}{L} \int_{0}^{\lambda} \lambda' d\lambda' \)

Integrating $\lambda'$ with respect to $\lambda'$ gives $\frac{(\lambda')^2}{2}$. Evaluating the definite integral from 0 to $\lambda$:

\( W_f = \frac{1}{L} \left[ \frac{(\lambda')^2}{2} \right]_{0}^{\lambda} = \frac{1}{L} \left( \frac{\lambda^2}{2} - \frac{0^2}{2} \right) = \frac{\lambda^2}{2L} \)

Now, let's calculate the co-energy ($W_c$) for the linear system using the formula \( W_c = \int_{0}^{i} \lambda(i') di' \). From $\lambda = Li$, we directly substitute $\lambda(i') = L i'$ into the integral:

\( W_c = \int_{0}^{i} L i' di' = L \int_{0}^{i} i' di' \)

Integrating $i'$ with respect to $i'$ gives $\frac{(i')^2}{2}$. Evaluating the definite integral from 0 to $i$:

\( W_c = L \left[ \frac{(i')^2}{2} \right]_{0}^{i} = L \left( \frac{i^2}{2} - \frac{0^2}{2} \right) = \frac{1}{2} L i^2 \)

Since $\lambda = Li$, we can substitute this back into the expression for $W_f$ or substitute $i = \lambda/L$ into the expression for $W_c$ to see they are equivalent:

• Using $W_f = \frac{\lambda^2}{2L}$ and $\lambda = Li$, we get \( W_f = \frac{(Li)^2}{2L} = \frac{L^2 i^2}{2L} = \frac{1}{2} L i^2 \).
• Using $W_c = \frac{1}{2} L i^2$ and $i = \lambda/L$, we get \( W_c = \frac{1}{2} L \left(\frac{\lambda}{L}\right)^2 = \frac{1}{2} L \frac{\lambda^2}{L^2} = \frac{\lambda^2}{2L} \).

Alternatively, we can also express both in terms of both $\lambda$ and $i$ for the linear case:

\( W_f = \frac{1}{2} \lambda i \)

\( W_c = \frac{1}{2} \lambda i \)

From the derivations, it is clear that for a linear electromagnetic circuit, the expressions for field energy ($W_f$) and co-energy ($W_c$) are identical.

\( W_f = W_c = \frac{1}{2} L i^2 = \frac{\lambda^2}{2L} = \frac{1}{2} \lambda i \)

Comparing Field Energy and Co-energy

Based on our analysis for a linear electromagnetic circuit:

• Field energy is calculated as \( W_f = \frac{1}{2} L i^2 \).
• Co-energy is calculated as \( W_c = \frac{1}{2} L i^2 \).

Therefore, for a linear system, the field energy is equal to the co-energy.

This is generally not true for non-linear electromagnetic circuits, where the inductance $L$ is not a constant and $\lambda$ is a non-linear function of $i$. In non-linear systems, $W_f$ and $W_c$ are typically different, but their sum still equals $\lambda i$.

The statement that holds good for a linear electromagnetic circuit is that the field energy is equal to the co-energy.