GATE Electrical Engineering (EE) 2018 Solved Paper

Question : 34 of 65

Marks: +1, -0

A single-phase

100 {kVA}, 1000 {V} {/} 100 {V}, 50 {Hz}

transformer has a voltage drop of

5 \%

across its series impedance at full load. Of this,

3 \%

is due to resistance. The percentage regulation of the transformer at full load with

0.8

lagging power factor is

Solution:

Percent voltage regulation

=(\;{I_{2} R_{02}}/{E_{2}} × 100) \cos ϕ_{2} ±(\;{I_{2} X_{02}}/{E_{2}} × 100) \sin ϕ_{2}

\% \;\; \V ⋅ \R.=V_{r} \cos φ_{2} ± V_{x} \sin ϕ_{2}

(where, '+' lag p.f. and '-' lead p.f.)
At full load:
Given,

V_{r}=3 \%

Impedance drop,

\;\; v_{z}=5 \%

Reactance drop,

\;\; V_{x}=√{5^{2}-3^{2}}=4 \%

Voltage regulation at full load at

0.8

p.f. lagging

\;\text " V.R. "\; \;\; =3(0.8)+4(0.6)

\;\; =2.4+2.4=4.8 \%

Go to Question:

Prev Question

Next Question