Question

The identical impedances ${\mathbf{Z}}_{\mathbf{∆}}\mathbf{=}\mathbf{30}\mathbf{\angle }{\mathbf{30}}^{\mathbf{°}}\mathbf{\Omega }$are connected in $\mathbf{∆}$to a balanced three-phase$\mathbf{208}\mathbf{}\mathbf{V}$ source by three identical line conductors with impedance ${\mathit{Z}}_{\mathbf{L}}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{+}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{)}\mathit{\Omega }$per line. (a) Calculate the line-to-line voltage at the load terminal. (b) Repeat part (a) when a $\mathbf{∆}$-connected capacitor bank with reactance localid="1655119206863" $\mathbf{(}\mathbf{-}\mathit{j}\mathbf{60}\mathbf{)}\mathit{\Omega }$ per phase is connected in parallel with the load.

Short answer

(a) The load voltage is$189.3V.$

(b) The load voltage is$194.3\mathrm{V}$.

Step-by-step solution

Determine the formulas of line current

Write the formula of phase voltage

${\mathrm{V}}_{\mathrm{ph}}=\frac{{\mathrm{V}}_{\mathrm{line}}}{\sqrt{3}}$ ……. (1)

Write the formula of per-phase impedance in star connection

$\mathbf{z}\mathbf{=}\frac{{\mathbf{z}}_{\mathbf{∆}}}{\mathbf{3}}$ ……. (2)

Write the formula for load voltage using voltage divider rule

$\mathbf{V}\mathbf{=}{\mathbf{V}}_{\mathbf{p}\mathbf{h}}\left(\frac{Z_y}{Z_Y+Z_L}\right)$ ……. (3)

Determine the line-to-line voltage at load terminal.

(a)

Determine the phase voltage.

$$\begin{gathered} \mathrm{Substitute}208\mathrm{V}\mathrm{for}{V}_{\mathrm{line}}\mathit{}\mathrm{in}\mathrm{equation}\left(1\right). \\ {V}_{ph}\mathit{=}\frac{208}{\sqrt{3}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=120.09\mathrm{V} \\ \mathrm{Determine}\mathrm{the}\mathrm{per}-\mathrm{phase}\mathrm{impedance}\mathrm{in}\mathrm{star}-\mathrm{connection}. \\ \mathrm{Substitute}30\angle {30}^{°}\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{∆}\mathrm{equation}\left(2\right). \\ {Z}_y\mathit{=}\frac{Z_{\mathit{∆}}}{3} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=\frac{30\angle {30}^{°}}{3} \\ =10\angle {30}^{°}\mathrm{\Omega } \\ \mathrm{Determine}\mathrm{the}\mathrm{load}\mathrm{voltage} \\ \mathrm{Substitute}10\angle {30}^{°}\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{y}}\mathrm{and}(0.8+j0.6)\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{L}}\mathrm{in}\mathrm{equation}\left(3\right). \\ \mathrm{V}=120.09\times \frac{10\angle {30}^{°}}{10\angle {30}^{°}+(0.8+j0.6)} \\ =109.23\angle -0.{623}^{°}\mathrm{V} \\ \mathrm{The}\mathrm{line}-\mathrm{to}-\mathrm{line}\mathrm{load}\mathrm{voltage}\mathrm{using}\mathrm{equation}\left(1\right)\mathrm{is} \\ {V}_s\mathit{=}\sqrt{3}\times 109.23\angle -0.{623}^{°} \\ =189.3\angle -0.{623}^{°}\mathrm{V} \end{gathered}$$

Determine the line-to-line voltage at load when capacitor is connected.

Now, a capacitor bank is connected in parallel to the load.

The per phase load capacitance using equation $\left(2\right)\mathrm{is}{\mathrm{Z}}_{\mathrm{c}}=-j20\mathrm{\Omega }$

The equivalent load resistance is,

$$\begin{gathered} Z_{eq}=Z_y\left|\right|Z_c \\ =10\angle {30}^{°}\left|\right|(-j20) \\ =11.547\angle 0^{°}\mathrm{\Omega } \\ \mathrm{Determine}\mathrm{the}\mathrm{load}\mathrm{voltage} \\ \mathrm{substitute}11.547\angle -0^{°}\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{eq}}\mathrm{and}(0.8+j0.6)\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{L}}\mathrm{in}\mathrm{equation}\left(3\right). \\ \mathrm{v}=120.09\times \frac{11.547\angle -0^{°}\mathrm{\Omega }}{11.547\angle -0^{°}\mathrm{\Omega }+(0.8+j0.6)} \\ =112.2\angle -2.{78}^{°}\mathrm{V} \\ \mathrm{The}\mathrm{line}-\mathrm{to}-\mathrm{line}\mathrm{load}\mathrm{voltage}\mathrm{using}\mathrm{equation}\left(1\right)\mathrm{is} \\ {\mathrm{V}}_{\mathrm{s}}=\sqrt{3}\times 112.2\angle -2.{78}^{°} \\ =194.3\angle -2.{78}^{°}\mathrm{V} \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{to}\mathrm{line}\mathrm{voltage}\mathrm{is}194.3\angle -2.{78}^{°} \end{gathered}$$