Conversion From Regular To Auto Transformer MCQ Quiz in தமிழ் - Objective Question with Answer for Conversion From Regular To Auto Transformer - இலவச PDF ஐப் பதிவிறக்கவும்

Case I:

\({I_1} = \frac{{40 \times {{10}^3}}}{{300}} = 133.33\;A\)

\({I_2} = \frac{{40 \times {{10}^3}}}{{600}} = 66.67\;A\)

\({\rm{Efficiency\;\eta }} = \frac{{{\rm{x}}.\left( {{\rm{VA}}} \right)\cos \phi }}{{{\rm{x}}\left( {{\rm{V}}.{\rm{A}}} \right)\cos \phi + {{\rm{P}}_{\rm{i}}} + {{\rm{x}}^2}{{\rm{P}}_{{\rm{cu}}\left( {{\rm{full}}} \right)}}}}\)

\(0.95 = \frac{{40 \times {{10}^3} \times 1}}{{40 \times {{10}^3} \times 1 + {{\rm{P}}_{\rm{i}}} + {{\rm{P}}_{{\rm{cu}}\left( {{\rm{full}}} \right)}}}}\)

P_i + P_cu(full) = 2105.26 W

Case II:

V.I = 200 × 600 = 120000 = 120 kVA

\({\eta _{auto}} = \frac{{120 \times {{10}^3}}}{{120 \times {{10}^3} + 2105.26}}\times 100 = 98.28\%\)

kVA rating of transformer = 50 kVA

Auto transformer rating = 2300 / 2530

\(K = \frac{{2300}}{{2530}} = 0.909\)

kVA rating of auto transformer \(= \frac{1}{{\left( {1 - k} \right)}} \times 50\)

\(= \frac{1}{{1 - 0.909}} \times 50 = 550\;kVA\)

Alternate Method :

kVA rating of auto transformer\(= \frac{a'}{{\left( {a'-1} \right)}} \times 50\)

where a' = 2530 / 2300 = 1.1

kVA rating of auto transformer = 550 kVA

\(\begin{array}{l} {V_{1\left( {2wind} \right)}} = 480\;V,\;{V_{2\left( {2wind} \right)}} = 8\;kV,\;{S_{2wind}} = 480\;kVA\\ {I_{1\left( {2wind} \right)}} = \frac{{480 \times {{10}^3}}}{{480}} = 1000\;A\\ {I_{2\left( {2wind} \right)}} = \frac{{480 \times {{10}^3}}}{{8 \times {{10}^3}}} = 60\;A \end{array}\)

Two winding transformer is connected as an auto transformer

\({V_{1\left( {auto} \right)}} = 8000\;V,\;{V_{2\left( {auto} \right)}} = 8480\;V,\;{I_{2\left( {auto} \right)}} = {I_{1\left( {2wind} \right)}} = 1000\;A\)

Rating of auto transformer, \({S_{auto}} = {V_{2\left( {auto} \right)}}{I_{2\left( {auto} \right)}} = 8.48\;MVA\)

Efficiency of two winding transformer,

\(\begin{array}{l} \eta = \frac{{480}}{{480 + Full\;load\;losses}} = 0.978\\ \Rightarrow Full\;load\;losses = 10.8\;kW \end{array}\)

In auto transformer connection full load losses remains same. Thus, efficiency of auto transformer is

\(\eta = \frac{8.48}{{8.48 + 10.8 \times {{10}^{ - 3}}}} \times 100 = 99.8\;\% \)

For two winding transformer,

\(\begin{array}{l} {I_1} = \frac{{200 \times {{10}^3}}}{{1500}} = 133.33A\\ {I_2} = \frac{{200 \times {{10}^3}}}{{800}} = 250A \end{array}\)

When connected as autotransformer

Power transferred inductively

= Power transferred in common winding ≈ 200 kVA