# Constant Flux Linkage Theorem of Synchronous Machines

Electronics & ElectricalElectronDigital Electronics

A synchronous machine may be subjected to various disturbances. Any cause of disturbance will produce electrical and mechanical transients. These transients may result from switching, from sudden changes of load, from sudden short-circuits between line and ground or between double lines or between all the three lines. These disturbances produce large mechanical stresses which may damage the machine. The synchronous machine may also lose synchronism.

## Constant Flux Linkage Theorem: Statement

The constant flux linkage theorem is used in studying alternator transients. This theorem is stated as follows −

“The flux linkage after sudden disturbance in a closed circuit having zero resistance and zero capacitance remain constant at their pre-disturbed value. In other words, the magnetic flux linkage in an inductor cannot change abruptly.”

The armature and field windings of an alternator have no capacitance. Also, their resistances are negligibly small as compared to their inductances. Therefore, the armature and field windings may be assumed to be purely inductive and the flux linkages in the armature and field windings cannot be changed abruptly by the application of short-circuit to the armature winding. Hence, any sudden change of the current in one winding must be accompanied by a change of current in the other to keep the flux linkages constant.

## Constant Flux Linkage Theorem: Proof

For any closed circuit, the mesh voltage equations may be written as follows −

$$\mathrm{\sum 𝑒 =\sum 𝑉_{𝑅} + \sum 𝑉_{𝐿} + \sum 𝑉_{𝐶}}$$

$$\mathrm{\Rightarrow\:\sum 𝑒 =\sum 𝑖𝑅 +\sum 𝑁\frac{𝑑𝜑}{𝑑𝑡}+\sum \frac{𝑞}{𝑐}… (1)}$$

Also,

$$\mathrm{Flux\:linkage,\:\psi = 𝑁_{𝜑} … (2)}$$

From Eqns. (1) & (2),

$$\mathrm{\sum 𝑒-\sum i𝑅-\sum \frac{𝑞}{ 𝑐}=\sum \frac{𝑑\psi}{𝑑𝑡}= \frac{𝑑}{𝑑𝑡}\sum \psi}$$

$$\mathrm{\Rightarrow\:\frac{𝑑}{𝑑𝑡}\sum \psi=\sum 𝑒_{1}… (3)}$$

Where, e1 is the resultant voltage, which will be some function of time.

On integrating Eqn. (3), the change in linkage flux will be,

$$\mathrm{\Delta \left(\sum \psi \right)=\int_{0}^{\Delta t} e_{1}\:𝑑𝑡 … (4)}$$

Here, $\Delta t$ is a small interval of time. When tends to zero, so will be the integral, thus,

$$\mathrm{\sum \psi= 0}$$

That is the instantaneous change of flux linkage is zero.

Published on 14-Oct-2021 11:31:19