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Answer: Given Data :-Assume an ideal single Phase Transformer which supplies the rated load at 0.8 \mathrm{Pf} lagging and at rated voltage:-\begin{array}{l}\text { KVA rating }=100 \mathrm{kVA} \\\text { Voltage rating }=8.05 \mathrm{kV} / 230 \mathrm{~V} \\\text { operational frequency }=50 \mathrm{~Hz}\end{array}cross-sectional Area of the core =1.7 \times 10^{-2} \mathrm{~m}^{2}solution:- (a) calculate fluxdensity in the core:since we know thatV_{1}=4.44(B A) N_{1}Where V_{1} \rightarrow Primary VoltageB \rightarrow flux densityA \rightarrow cross-sectional AreaN_{1} \rightarrow Primary turnAlso,\begin{array}{c}\frac{N_{1}}{N_{2}}=\frac{V_{1}}{V_{2}}=\frac{8.05 \times 1000}{230} \\N_{1}=35 N_{2}\end{array}\begin{array}{l}N_{1}=350 \text { turn } \\N_{2}=10 \text { turn }\end{array}Therefore, \quad V_{1}=4.44 \times B \times 1.7 \times 10^{-2} \times 350\begin{array}{l}B=\frac{V_{1}}{4.44 \times 1.7 \times 10^{-2} \times 350} \\B=\frac{8.05 \times 10^{3}}{4.44 \times 1.7 \times 10^{-2} \times 350} \\B=304.71 \mathrm{wb} / \mathrm{m}^{2}\end{array}Flux density (B)=304.71 \mathrm{wb} / \mathrm{m}^{2} Ave.(b) Determine the number of turns in Primary \& 2^{n d} r y.Primary \operatorname{turn}\left(N_{1}\right)=350Secondary furn \left(N_{2}\right)=10(C) Determine the impedence of the load referred to the Primary windings.since we know That\begin{array}{c}\text { KVA rating }(5)=V_{1} \times I_{1} \\I_{1}=\frac{100 \mathrm{kvA}}{8.05 \mathrm{kV}} \\I_{1}=12.422 \mathrm{~A}\end{array}Impedence referred to Primary =\frac{V}{I}=\frac{8.05 \times 10^{3}}{12.422 \angle-36.86^{\circ}}\text { Impedence }=648436.86^{\circ}Ans.(d) sketch a reflective phasor diagram ...