To understand a single excited magnetic system, we consider an attracted armature type electromagnet relay shown in the figure below.
The above system itself is a singly excited magnetic system. As shown in the figure, a coil having N number of turns is wound on the magnetic core. And this magnetic core is connected to a supply voltage source Vs.
Now assume that the armature is fixed. And the current that passes through the circuit is increased from zero (0) to .
Due to the current passes through the coil, the magnetic flux will be set up in the magnetic system. The magnitude of magnetic flux depends on the reluctance of the magnetic path and MMF.
By applying KVL to the electrical circuit, we can find the equation of instantaneous voltage and it is;
Where, = Flux linkage = N (Wb-Turns)
Now, multiply the above with
Here we assume that the flux links with all turns of the coil. Hence;
Where, F = instantaneous MMF
dWe = incremental electrical energy
From the above equation;
Here, we kept the armature fixed. Hence, the mechanical energy output or work done will be zero.
dWf = Energy supplied to the magnetic field
From equation-1, when the movable part of any magnetic system is stationary, the total electrical energy input is stored in the magnetic field.
When the flux increases from zero (0) to , the energy stored in the field is;
Where B = magnetic flux density
l = length of flux path
A = area of core
H = magnetic field intensity
Now, put these values in the above equation and we get;
For a linear magnetic circuit, equation-2 is written as;
As we know, self-inductance (L) is defined as the magnetic flux linkage per unit current.
For angular movement of an armature, the electromagnetic torque for a linear system is defined as;
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