Question 3. The primary and secondary windings of a 30 kVA, 6000/230 V transformer have resistances of 10 n and 0-016 respectively. The total reactance of the transformer referred to the primary is 23 Calculate the percentage voltage regulation of the transformer when supplying full-load current at a p.f. 0-8 lagging [2.55 %)

Q3. Given data:page 1Rating of transformer S=30 \mathrm{KVA}v_{1} / v_{2}=6000 / 230 \mathrm{~V}primary Reristance R_{1}=10 \OmegaSecondary winding Reristance R_{2}=0.016 \OmegaTotal Reactance Reffered to primary X_{01}=23 \Omega% voitage regulation at full load and 0.8 logging \mathrm{Pf}= ?\% \text { voltage Regulation }=\frac{I_{2} R_{02} \cos \phi_{2}+I_{2} \times_{\mathrm{O}_{2}} \sin \phi_{2}}{V_{2}} \times 100 .\Rightarrow full load current I_{2}=\frac{K V A \text { Rating }}{\text { Rated secondary voltage }}\begin{aligned}& =\frac{30 \times 10^{3}}{230} \\I_{2} & =130.4347 \mathrm{~A}\end{aligned}\begin{aligned}\Rightarrow \cos \phi_{2} & =0.8 \log \\\phi_{2} & =\cos ^{-1}(0.8)=36.869^{\circ} \\\sin \phi_{2} & =\sin (36.869)=0.6\end{aligned}\Rightarrow Total Resistance Reffered to Secondary =R_{02}R_{02}=R_{2}+R_{1}^{\prime}R_{1}^{\prime}= primary Resistance reffered to secondary\begin{array}{l}R_{1}^{\prime}=K^{2} R_{1} \\=\left(\frac{230}{6000}\right)^{2} \times 10 \\R_{1}^{\prime}=0.0146 \Omega \\\left\{k=\frac{V_{2}}{v_{1}}\right\} \\R_{02}=R_{2}+R_{1}^{\prime}=0.016+0.014=0.03 \Omega \\\end{array}page 2\left\{\begin{array}{l}\text { we know that } \\ \text { iransformation Ratio } K=\frac{N_{2}}{N_{1}}=\frac{V_{2}}{V_{1}}\end{array}\right\}R_{02}=R_{2}+R_{1}^{\prime}=0.016+0.014=0.03 \Omega\Rightarrow Total Reactance Refferred to Secondary =X_{\mathrm{O}_{2}}\begin{array}{l}x_{02}=k^{2} x_{01} \Rightarrow\left(\frac{230}{6000}\right)^{2} \times 23 \\x_{02}=0.033 \Omega\end{array}230\frac{3.1304+2.5826}{230} \times 100=2.4839 \%\% regulation =2.5 \% ...