coenergy

Consider a system with a single coil and a non-moving armature (i.e. no mechanical work is done). Hence, all of the electric energy supplied to the device is stored in the magnetic field.

$$dW\_\backslash text\{input\}\; =\; dW\_\backslash text\{stored\}\; \backslash qquad\; (dW\_\backslash text\{mechanical\}\; =\; 0)$$

where (e is the voltage, i is the current, and $\backslash lambda$ is the flux linkage):

$$dW\_\backslash text\{input\}\; =\; e\; i\; \backslash ,\; dt$$

$$dW\_\backslash text\{stored\}\; =\; i\; \backslash ,\; d\backslash lambda$$

therefore $$dW\_\backslash text\{stored\}\; =\; e\; i\; \backslash ,\; dt\; =\; i\; \backslash ,\; d\backslash lambda$$

For a general problem the relationship $i-\backslash lambda$ is non-linear (see also magnetic hysteresis).

If there is a finite change in flux linkage from one value to another (e.g. from $\backslash lambda\_1$ to $\backslash lambda\_2$), it can be calculated as:

$$\backslash Delta\; W\_\backslash text\{stored\}\; =\; \backslash int\_\{\backslash lambda\_1\}^\{\backslash lambda\_2\}\; i(\backslash lambda)\; \backslash ,\; d\backslash lambda$$

(If the changes are cyclic there will be losses for hysteresis and eddy currents. The additional energy for this would be taken from the input energy, so that the flux linkage to the coil is not affected by the losses and the coil can be treated as an ideal lossless coil. Such system is therefore conservative.)

For calculations either the flux linkage $\backslash lambda$ or the current i can be used as the independent variable.

The total energy stored in the system is equal to the area OABO, which is in turn equal to OACO, therefore:

$$\backslash mathrm\{Energy\}\; =\; \backslash operatorname\{Area\}(OABO)\; =\; W\_\backslash text\{stored\}\; =\; \backslash int\_\{0\}^\{\backslash lambda\}\; i(\backslash lambda)\; \backslash ,\; d\backslash lambda$$

$$\backslash mathrm\{Coenergy\}\; =\; \backslash operatorname\{Area\}(OACO)\; =\; W\text{'}\_\backslash text\{stored\}\; =\; \backslash int\_\{0\}^\{i\}\; \backslash lambda(i)\; \backslash ,\; di$$

For linear lossless systems the coenergy is equal in value to the stored energy. The coenergy has no real physical meaning, but it is useful in calculating mechanical forces in electromagnetic systems. To distinguish it from the "real" energy in calculations it is usually marked with an apostrophe.

The total area of the rectangle OCABO is equal to the sum of the two triangles (energy + coenergy), so:

$$\backslash operatorname\{Area\}(OCABO)\; =\; \backslash operatorname\{Area\}(OABO)\; +\; \backslash operatorname\{Area\}(OACO)$$

Hence for at a given operating point with current i and flux linkage $\backslash lambda$:

$$\backslash mathrm\{Energy\}\; (W)\; +\; \backslash mathrm\{Coenergy\}\; (W\text{'})\; =\; i\; \backslash lambda$$

The self inductance is defined as flux linkage over current:

$$L\; =\; \backslash frac\{\backslash lambda\}\{i\}$$

and the energy stored in a coil is:

$$W\_\backslash text\{energy\}\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{\backslash lambda^2\}\{L\}\; =\; \backslash frac\{1\}\{2\}\; L\; i^2$$

In a magnetic circuit with a movable-armature the inductance {{math|L(x)}} will be a function of position {{mvar|x}}.

Therefore the field energy can be written as a function of two mathematically independent variables $\backslash lambda$ and {{mvar|x}}:

$$W(\backslash lambda,x)\_\backslash text\{energy\}\; =\; \backslash frac\{1\}\{2\}\; \backslash ,\; \backslash frac\{\backslash lambda^2\}\{L(x)\}$$

And the coenergy is a function of two independent variables {{mvar|i}} and {{mvar|x}}:

$$W\text{'}(i,x)\_\backslash text\{coenergy\}\; =\; \backslash frac\{1\}\{2\}\; L(x)\; i^2$$

The last two expressions are general equations for energy and coenergy in magnetostatic system.

The concept of coenergy is practically used for instance in finite element analysis for calculations of mechanical forces between magnetized parts.

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Category:Electromagnetism