Three Identical Impedances$Z_{\Delta}=20 / 60^{\circ} \Omega$ Are Connected In$\Delta$ To a Balanced Threephase

Question

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The Answer is (a)
$(a) Using voltage division: \bar{V}_{A N}=\bar{V}_{a n} \frac{\bar{Z}_{\Delta} / 3}{\left(Z_{\Delta} / 3 ight)+\bar{Z}_{L I N E}} \bar{V}_{A N}=\frac{208}{\sqrt{3}} \frac{6.667 \angle 60^{\circ}}{6.667 \angle 60^{\circ}+(0.8+j 0.6)}=105.4 \angle 2.96^{\circ} \mathrm{V} Load voltage V_{A B}=\sqrt{3} 105.4=182.6 \mathrm{~V}(\mathrm{~L}-\mathrm{L}) (b) \begin{aligned} & \bar{V}_{A N}=\bar{V}_{a n} \frac{\bar{Ƶ}_{e q}}{\bar{Ƶ}_{e q}+\bar{Ƶ}_{L I N E}} ; \bar{Ƶ}_{e q}=\left(6.667 \angle 60^{\circ} ight) \|(-j 6.667) \ & =12.88 \angle-15^{\circ} \Omega \ & \bar{V}_{A N}=\left(\frac{208}{\sqrt{3}} \angle 0^{\circ} ight)\left[\frac{12.88 \angle-15^{\circ}}{\left(12.88 \angle-15^{\circ} ight)+0.8+j 0.6} ight]=114.4 \angle-3.35^{\circ} \mathrm{V} \end{aligned} The load line to line voltage is V_{A B}=\sqrt{3} 114.4=198.1 \mathrm{~V}$
Using voltage division:

\bar{V}_{A N}=\bar{V}_{a n} \frac{\bar{Z}_{\Delta} / 3}{\left(Z_{\Delta} / 3 ight)+\bar{Z}_{L I N E}}

\bar{V}_{A N}=\frac{208}{\sqrt{3}} \frac{6.667 \angle 60^{\circ}}{6.667 \angle 60^{\circ}+(0.8+j 0.6)}=105.4 \angle 2.96^{\circ} \mathrm{V}

Load voltage

V_{A B}=\sqrt{3} 105.4=182.6 \mathrm{~V}(\mathrm{~L}-\mathrm{L})

(b)
$(a) Using voltage division: \bar{V}_{A N}=\bar{V}_{a n} \frac{\bar{Z}_{\Delta} / 3}{\left(Z_{\Delta} / 3 ight)+\bar{Z}_{L I N E}} \bar{V}_{A N}=\frac{208}{\sqrt{3}} \frac{6.667 \angle 60^{\circ}}{6.667 \angle 60^{\circ}+(0.8+j 0.6)}=105.4 \angle 2.96^{\circ} \mathrm{V} Load voltage V_{A B}=\sqrt{3} 105.4=182.6 \mathrm{~V}(\mathrm{~L}-\mathrm{L}) (b) \begin{aligned} & \bar{V}_{A N}=\bar{V}_{a n} \frac{\bar{Ƶ}_{e q}}{\bar{Ƶ}_{e q}+\bar{Ƶ}_{L I N E}} ; \bar{Ƶ}_{e q}=\left(6.667 \angle 60^{\circ} ight) \|(-j 6.667) \ & =12.88 \angle-15^{\circ} \Omega \ & \bar{V}_{A N}=\left(\frac{208}{\sqrt{3}} \angle 0^{\circ} ight)\left[\frac{12.88 \angle-15^{\circ}}{\left(12.88 \angle-15^{\circ} ight)+0.8+j 0.6} ight]=114.4 \angle-3.35^{\circ} \mathrm{V} \end{aligned} The load line to line voltage is V_{A B}=\sqrt{3} 114.4=198.1 \mathrm{~V}$

\begin{aligned}& \bar{V}_{A N}=\bar{V}_{a n} \frac{\bar{Ƶ}_{e q}}{\bar{Ƶ}_{e q}+\bar{Ƶ}_{L I N E}} ; \bar{Ƶ}_{e q}=\left(6.667 \angle 60^{\circ} ight) \|(-j 6.667) \& =12.88 \angle-15^{\circ} \Omega \& \bar{V}_{A N}=\left(\frac{208}{\sqrt{3}} \angle 0^{\circ} ight)\left[\frac{12.88 \angle-15^{\circ}}{\left(12.88 \angle-15^{\circ} ight)+0.8+j 0.6} ight]=114.4 \angle-3.35^{\circ} \mathrm{V}\end{aligned}

The load line to line voltage is

V_{A B}=\sqrt{3} 114.4=198.1 \mathrm{~V}

Three Identical Impedances $Z_{\Delta}=20 / 60^{\circ} \Omega$ Are Connected In $\Delta$ To a Balanced Threephase

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Three Identical Impedances ZΔ=20/60∘ΩZ_{\Delta}=20 / 60^{\circ} \OmegaZΔ​=20/60∘Ω Are Connected In Δ\DeltaΔ To a Balanced Threephase

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Three Identical Impedances $Z_{\Delta}=20 / 60^{\circ} \Omega$ Are Connected In $\Delta$ To a Balanced Threephase

Correct Answer:
Verified