SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

${\mathrm{V}}_1=2000\mathrm{V},{\mathrm{V}}_2=250\mathrm{V},$ ${\mathrm{R}}_1=1.15\mathrm{\Omega },{\mathrm{R}}_2=0.0155\mathrm{\Omega }$
From equation(1)
$\mathrm{K}=\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$
$\mathrm{K}=\frac{250}{2000}$
$\mathrm{K}=\frac{1}{8}$
Effective resistance of the transformer as referred to the secondary of the transformer.
From equations (2) & (3)
${\mathrm{R}}_{02}=0.0155+\frac{1.15}{8^2}$
${\mathrm{R}}_{02}=0.0335\mathrm{\Omega }$
Power output $\mathrm{P}=40\mathrm{kVA}$
${\mathrm{V}}_2{\mathrm{I}}_2=40\times {10}^3$
${\mathrm{I}}_2=\frac{40000}{250}$
${\mathrm{I}}_2=160\mathrm{A}={\mathrm{I}}_{\mathrm{fl}}$
${\mathrm{V}}_2{\mathrm{I}}_2=40\times {10}^3$
${\mathrm{I}}_2=\frac{40000}{250}$
${\mathrm{I}}_2=160\mathrm{A}={\mathrm{I}}_{\mathrm{fl}}$
${\mathrm{I}}_{\mathrm{fl}}$ is the full load current flowing through the secondary.
We required to find power loss at half full load condition
So, current at half full load is given as,
${\mathrm{I}}_{\mathrm{hfl}}=\frac{160}{2}=80\mathrm{A}$
∴ Total power loss at half full load condition is given as,
${\mathrm{P}}_{\mathrm{hfl}}={\mathrm{I}}_{\mathrm{hfl}}^2\times {\mathrm{R}}_{02}$
${\mathrm{P}}_{\mathrm{hfl}}={80}^2\times 0.0335$
${\mathrm{P}}_{\mathrm{hfl}}=214.2\mathrm{W}$