For a conceptual single-phase, phase-shifting transformer, the primary voltage leads the secondary voltage by $30^{\circ}$. A load connected to the secondary winding absorbs $50 \mathrm{kVA}$ at 0.9 power factor leading and at a voltage $E_{2}=277 \angle 0^{\circ}$ volts. Determine (a) the primary voltage, (b) primary and secondary currents, (c) load impedance referred to the primary winding, and (d) complex power supplied to the primary winding.

Question

Quizplus · Accepted Answer

The Answer is $(a) \bar{E}_{2}=277 \angle 0^{\circ} \mathrm{V} \quad \bar{E}_{1}=e^{j 30^{\circ}} \bar{E}_{2}=\underline{\underline{277 \angle 30^{\circ}}} \mathrm{V} (b) \bar{I}_{2}=\frac{S_{2}}{E_{2}} \angle+\cos ^{-1}(P . F)=.\frac{50 imes 10^{3}}{277} \angle \cos ^{-1}(0.9)=180.5 \angle+25.84^{\circ} \mathrm{A} \bar{I}_{1}=\frac{\bar{I}_{2}}{\left(e^{j 30^{\circ}} ight)^{*}}=\bar{I}_{2} e^{j 30^{\circ}}=\underline{\underline{180.5 \angle 55.84^{\circ}} \mathrm{A}} (c) \bar{Ƶ}_{2}=\frac{\bar{E}_{2}}{\bar{I}_{2}}=\frac{277 \angle 0^{\circ}}{180.5 \angle 25.84^{\circ}}=1.5346 \angle-25.84^{\circ} \Omega \bar{Ƶ}_{2}^{1}=\bar{Ƶ}_{2}=1.5346 \angle-25.84^{\circ} \Omega (d) \bar{S}_{1}=\bar{S}_{2}=50 \angle-\cos ^{-1}(0.9)=\underline{\underline{50 \angle-25.84^{\circ}}} \mathrm{kVA} \bar{S}_{1}=45 \mathrm{~kW}-j 21.79 \mathrm{kVARS} delivered to primary$
(a)

\bar{E}_{2}=277 \angle 0^{\circ} \mathrm{V} \quad \bar{E}_{1}=e^{j 30^{\circ}} \bar{E}_{2}=\underline{\underline{277 \angle 30^{\circ}}} \mathrm{V}

(b)

\bar{I}_{2}=\frac{S_{2}}{E_{2}} \angle+\cos ^{-1}(P . F)=.\frac{50 imes 10^{3}}{277} \angle \cos ^{-1}(0.9)=180.5 \angle+25.84^{\circ} \mathrm{A}

\bar{I}_{1}=\frac{\bar{I}_{2}}{\left(e^{j 30^{\circ}} ight)^{*}}=\bar{I}_{2} e^{j 30^{\circ}}=\underline{\underline{180.5 \angle 55.84^{\circ}} \mathrm{A}}

(c)

\bar{Ƶ}_{2}=\frac{\bar{E}_{2}}{\bar{I}_{2}}=\frac{277 \angle 0^{\circ}}{180.5 \angle 25.84^{\circ}}=1.5346 \angle-25.84^{\circ} \Omega

\bar{Ƶ}_{2}^{1}=\bar{Ƶ}_{2}=1.5346 \angle-25.84^{\circ} \Omega

(d)

\bar{S}_{1}=\bar{S}_{2}=50 \angle-\cos ^{-1}(0.9)=\underline{\underline{50 \angle-25.84^{\circ}}} \mathrm{kVA}

\bar{S}_{1}=45 \mathrm{~kW}-j 21.79 \mathrm{kVARS}

delivered to primary

For a Conceptual Single-Phase, Phase-Shifting Transformer, the Primary Voltage Leads 30∘30^{\circ}30∘

For a Conceptual Single-Phase, Phase-Shifting Transformer, the Primary Voltage Leads $30^{\circ}$