Question Solved1 Answer b) (i) N = 110 turns; (ii) IL = 16.5 AThese are the finalanswer Of this Question so please make sure everything 100% make sure answers are same i want to learn the exact method so please make sure everything 100% b) Consider the magnetic circuit has a length of 0.6 m an area of 0.0018 m' and a single airgap of length 2.3 mm. The circuit is energised by a coil. If the core permeability is 1000, by making suitable approximations calculate: (i) the number of turns required to achieve an inductance of 12 mH; (ii) the inductor current which will result in a core flux density of 1.0 T; (iii) A particular application uses the same magnetic core but requires the airgap flux density to be increased by a factor of 1.5 while maintaining the inductance below 17mH. Design a magnetic circuit to do this. (iv) For your design, determine the maximum RMS voltage that can be applied to the coil to ensure that the peak flux remains below 1.5 T 12
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SolutionGiven parameters in the questionMean length of the megnetic core quad(l_(c))=0.6mArea of cross -section(A)=0.0018m^(2)length of Airgap a(l_(g))=2.3mm=2.3 xx10^(-3)mRelative permeability of core (H_(r))=1000i) L=12mH.first of all we have to calculate reluctance of core and airgap.Reluctance of core (R_(c))=(l_(c))/(mu_(0)H_(r)A_(c))=(0.6)/(4pi xx10^(-7)xx1000 xx0.0018)=265258. At/wb.Relut Reluetance of airgap (R_(g))=(l xi)/(mu_(0)H_(r)Ag_(g))=(2.3 xx10^(-3))/(4pi xx10^(-7)xx1xx0.0018)=1016823AT//wb.Neglecting core reluctance, R=1016823AT//wb -we know thatL=(N^(2))/(R)=(N^(2))/(1016823AT):.N=sqrt(12 xx10^(-3)xx1016823)=110.46" tums "(ii) core flux density = 1.0Twe know,{:[B=(phi )/(A)],[phi=B*A],[phi=1xx0.0018omega b.]:}and we know. phi=(NI)/(R_("tral ")){:[:.quad0.0018=(110 xx I)/(1016823)],[I ... See the full answer