SSE JE Electrical Engineering 24 March 2021 Shift 1 Paper

${\mathrm{W}}_0=$ wattmeter reading = 50 W
${\mathrm{V}}_1=$ voltmeter reading = 230 V
${\mathrm{I}}_0=$ Ammeter reading = 2 A
$\cos {\mathrm{\varphi }}_0=$ No load power factor.
∴ Iron loss of the transformer, ${\mathrm{P}}_{\mathrm{i}}={\mathrm{W}}_0$
${\mathrm{W}}_0={\mathrm{V}}_1{\mathrm{I}}_0\cos {\mathrm{\varphi }}_0$
$\Rightarrow 50=230\times 2\times \cos {\mathrm{\varphi }}_0$
$\Rightarrow \cos {\mathrm{\varphi }}_0=0.108\mathrm{lagging}$
Working component or core loss component,
${\mathrm{I}}_{\mathrm{W}}={\mathrm{I}}_0\cos {\mathrm{\varphi }}_0$
${\mathrm{I}}_{\mathrm{W}}=2\times 0.108=0.216\mathrm{A}$