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Consider a coil of N turns wound around a magnetic core and is connected to voltage source (see the figure).

By applying KVL, we get,

$$\mathrm{V = e+iR\:\:\:\:\:\:...(1)}$$

Where,

e is the induced EMF in the coil,

R is the resistance of the coil circuit.

The instantaneous power input is given by,

$$\mathrm{p = Vi = e+i^{2}R\:\:\:\:\:\:...(2)}$$

Hence, the energy input to the system is,

$$\mathrm{W_{i} =\int_{0}^{T}=p\:dt=\int_{0}^{T}ei\:dt+\int_{0}^{T}i^{2}Rdt\:\:\:\:\:\:...(3)}$$

The eq. (3) shows that the total input energy consists of two parts. The first part is energy stored in magnetic field and the second part is the energy dissipated in the circuit resistance in the form of heat. Therefore, the energy stored in the magnetic field is given by

$$\mathrm{W_{f}=\int_{0}^{T}ei\:dt}$$

Also, according to Faraday’s law of electromagnetic induction, the induced emf is given by,

$$\mathrm{e=N\frac{dψ}{dt}=\frac{d(Nψ)}{dt}=\frac{dψ}{dt}}$$

Where, ψ = $N_{ψ}$ is the magnetic flux linkage.

$$\mathrm{∴W_{f}=\int_{0}^{T}\frac{dψ}{dt}i dt=\int_{0}^{ψ}idψ\:\:\:\:\:\:...(4)}$$

The eq.(4) shows that the energy stored in the magnetic field is equal to the area between the ψ-i curve for the system and the flux linkage (ψ) axis.

In a magnetically linear system, the field energy is given by,

$$\mathrm{W_{f}=\int_{0}^{ψ}idψ}$$

$$\mathrm{\because i=\frac{ψ}{L}orψ=Li}$$

$$\mathrm{∴W_{f}=\int_{0}^{ψ}\frac{ψ}{L}dψ=\frac{ψ^{2}}{2L}}$$

$$\mathrm{\Longrightarrow W_{f}=\frac{(Li)^{2}}{2L}=\frac{1}{2}Li^{2}}$$

$$\mathrm{\Longrightarrow W_{f}=\frac{ψ^{2}}{2L}=\frac{1}{2}Li^{2}\:\:\:\:\:\:...(5)}$$

The expression in the eq.(5) gives the field energy stored in a magnetically linear system.

The *coenergy* is a non-physical quantity measured in the units of energy, which is used to derive expressions for force or torque developed in an electromagnetic system. The concept of coenergy cannot be applied to non-linear systems.

For a magnetically linear system, the coenergy is given by,

$$\mathrm{W_{f}^{'}=\int_{0}^{i}ψdi=\int_{0}^{i}Lidi=\frac{1}{2}Li^{2}\:\:\:\:\:\:...(6)}$$

The eq. (6) shows that the coenergy in the magnetically linear system is equal to the area between the ψ-i curve for the system and the current (i) axis.

It is clear from the eq. (5) and (6) that for a magnetically linear system, the field energy and coenergy are equal.

$$\mathrm{W_{f}=W_{f}^{'}=\frac{ψ^{2}}{2L}=\frac{1}{2}Li^{2}\:\:\:\:\:\:...(5)}$$

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