For the device shown in figure 1 draw the lumped parameter magnetic equivalent c
ID: 2265758 • Letter: F
Question
For the device shown in figure 1 draw the lumped parameter magnetic equivalent circuit. Neglect fringing flux and iron reluctance. From the magnetic equivalent circuit draw the electrical equivalent circuit. 1. 20% a. Calculate the flux in each limb when la-22Amps and Ib-12Amps; b. Determine the mutual inductance between the windings; c. If the core material saturates abruptly at a flux density of 1.5T 40% 20% 20% ge that may be applied to winding "a". 12Amps 22Anps 40Turns Depth R S0nn Dinensions n nn figure 1Explanation / Answer
Data is insufficient because window length(top to bottom) and breadth are not given(which will be equal to parallel beam). But idea to solve problem is to calculate the total reluctance by formula l/(mu0*mur*A), where l is length and A is area of the beam or air gap.We need to make calculations separately for air gap reluctance and other material(generally silicon steel) reluctance, where area is changed make separate calculation and add to total reluctance.
Now to draw magnetic circuit eqivalent consider net mmf =(22*40)-(150*12) { mmf =Number of turns*Current in winding)
=-920 Ampere Turns(AT)
We can draw a winding with -920 Ampere turns (eg 920 turns and -1 A current, product must be -920AT)
then this winding is across a material(silicon steel) of above calculated reluctance and also air gap reluctance.Choose l and A appropriately.[I is current and A is area]
To draw electrical equivalent voltage=net mmf(calculated above -920 AT)
=-920 V
Two resistances one value equal to silicon steel reluctance and other air gap reluctance.
(a) To determine flux in each limb divide net flux by respective limb reluctance(above calculated)
(b)Mutual flux M= k(L1*L2)^(0.5) {L1 and L2 are self inductances of a and b winding respectively)
k<=1
so, M<=(L1*L2)^0.5
L1=(40*22)/(total reluctance) {numerator is mmf=coil a turns*its current)
Similarly
L2=(150*12)/(total reluctance)
(c) We know that
V(rms)=4.44*frequency*max flux*Number of turns
=4.44f(fi)(Tph) (Symbolically)
max flux=B(1.5 T given)* Area
Area is area of limb at which winding is turned=100*50 mm sq=5000mm sq
so, max flux=1.5*5000*(10^(-6)) [1mm=10^(-3) m]
=7.5*10^(-3) weber
So, Vrms=4.44*50*7.5*10^(-3) *40
=66.6 V
So, Vmax=66.6*1.414
=94.18 V