Question

The currents in a $∆$load are localid="1655466938675" ${\mathbf{l}}_{\mathbf{ab}}\mathbf{=}\mathbf{10}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{,}{\mathbf{l}}_{\mathbf{bc}}\mathbf{=}\mathbf{12}\mathbf{\angle }\mathbf{-}{\mathbf{90}}^{\mathbf{°}}$and localid="1655466944329" ${\mathbf{l}}_{\mathbf{ca}}\mathbf{=}\mathbf{15}\mathbf{\angle }{\mathbf{90}}^{\mathbf{°}}\mathbf{A}$ . Calculate (a) the sequence components of thelocalid="1655466952375" $\mathbf{∆}$-load currents, denoted localid="1655466957040" ${\mathbf{l}}_{\mathbf{∆}\mathbf{0}}\mathbf{}{\mathbf{l}}_{\mathbf{∆}\mathbf{1}}$and localid="1655466961516" ${\mathbf{l}}_{\mathbf{∆}\mathbf{2}}$(b) the line currents localid="1655466965605" ${\mathbf{l}}_{\mathbf{a}}\mathbf{,}{\mathbf{l}}_{\mathbf{b}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{c}}$which feed thelocalid="1655466969204" $\mathbf{∆}$load and (c) sequence components of the line currentslocalid="1655466973046" ${\mathbf{l}}_{\mathbf{Lo}}\mathbf{,}{\mathbf{l}}_{\mathbf{L}\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{L}\mathbf{2}}$. Also, verify the following general relation: localid="1655466976846" ${\mathbf{l}}_{\mathbf{Lo}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{l}}_{\mathbf{L}\mathbf{1}}\mathbf{=}{\sqrt{\mathbf{31}}}_{\mathbf{∆}\mathbf{1}}\mathbf{\angle }\mathbf{-}{\mathbf{30}}^{\mathbf{°}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{L}\mathbf{2}}\mathbf{=}{\sqrt{\mathbf{31}}}_{\mathbf{∆}\mathbf{2}}\mathbf{\angle }{\mathbf{30}}^{\mathbf{°}}$.

Short answer

$$\begin{gathered} \left(a\right)Thesequencecomponentofthe∆loadcurrents{l}_{∆0},l_{∆1}and{l}_{∆2}are \\ 3.48\angle 16.{72}^{°}A,11.14\angle 357.{43}^{°}Aand4.49\angle {40}^{°}Arespectively. \\ \left(b\right)Thelinecurrent{l}_a,l_{b}and{l}_{c}are18.03\angle 303.69A,15.62\angle 230.{19}^{°}Aand27\angle {90}^{°}A. \\ \left(c\right)Thesequencecomponentofthelinecurrents{l}_{L0},l_{L1}and{l}_{L2} \\ are0A,19.30\angle 327.{42}^{°}Aand7.77\angle 216.{37}^{°}A.Thegeneralrelation \\ {l}_{L0}=0A,l_{L1}=\sqrt{3}{l}_{∆1}\angle -{30}^{°}and{l}_{L2}=\sqrt{3}{l}_{∆2}\angle -{30}^{°}areverified. \end{gathered}$$

Step-by-step solution

Write the given data from the question.

$$\begin{gathered} Thecurrents, \\ {l}_{ab}=10\angle 0^{°}\mathrm{A} \\ {l}_{bc}=12\angle -{90}^{°}\mathrm{A} \\ {l}_{ca}=15\angle {90}^{°}\mathrm{A} \end{gathered}$$

determine the formulas to calculate the sequence component of the∆ -load current, line current and the sequence component of line current.

$$\begin{gathered} Theequationtocalculatethezerosequence∆-loadcurrentisgivenasfollows. \\ {\mathbf{l}}_{\mathbf{∆}\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{ab}}\mathbf{+}{\mathbf{l}}_{\mathbf{bc}}\mathbf{+}{\mathbf{l}}_{\mathbf{ca}}\mathbf{)}\dots \dots \left(1\right) \\ Theequationtocalculatethepositivesequence∆-loadcurrentisgivenasfollows. \\ {\mathbf{l}}_{\mathbf{∆}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{ab}}\mathbf{+}{\mathbf{al}}_{\mathbf{bc}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{ca}}\mathbf{)}.................\left(2\right) \\ Theequationtocalculatethenegativesequence∆-loadcurrentisgivenasfollows. \\ {\mathbf{l}}_{\mathbf{∆}\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{ab}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{bc}}\mathbf{+}{\mathbf{al}}_{\mathbf{ca}}\mathbf{)}..............\left(3\right) \\ Theequationtocalculatetheline{l}_{a}isgivenasfollows. \\ {\mathrm{l}}_{\mathrm{a}}={\mathrm{l}}_{\mathrm{ab}}-{\mathrm{l}}_{\mathrm{ca}}......................\left(4\right) \\ Theequationtocalculatetheline{l}_{b}isgivenasfollows. \\ {\mathrm{l}}_{\mathrm{a}}={\mathrm{l}}_{\mathrm{bc}}-{\mathrm{l}}_{\mathrm{ab}}.............\left(5\right) \\ Theequationtocalculatetheline{l}_{c}isgivenasfollows. \\ {\mathrm{l}}_{\mathrm{cc}}={\mathrm{l}}_{\mathrm{ca}}-{\mathrm{l}}_{\mathrm{bc}}....................\left(6\right) \end{gathered}$$

$$\begin{gathered} Theequationtocalculatethelinecurrent{l}_{L0}isgivenasfollows. \\ {\mathbf{l}}_{\mathbf{L}\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{a}}\mathbf{+}{\mathbf{l}}_{\mathbf{b}}\mathbf{+}{\mathbf{l}}_{\mathbf{c}}\mathbf{)}................\left(7\right) \\ Theequationtocalculatethelinecurrent{l}_{L1}isgivenasfollows. \\ {\mathbf{l}}_{\mathbf{L}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{a}}\mathbf{+}{\mathbf{al}}_{\mathbf{b}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{c}}\mathbf{)}.............\left(8\right) \\ Theequationtocalculatethelinecurrent{l}_{L2}isgivenasfollows. \\ {\mathbf{l}}_{\mathbf{L}\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{a}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{b}}\mathbf{+}{\mathbf{al}}_{\mathbf{c}}\mathbf{)}....................\left(9\right) \end{gathered}$$

Calculate the sequence components of the ∆-load currents

Consider the diagram for the$∆$-load shown below.

$$\begin{gathered} \mathrm{Calculate}\mathrm{the}\mathrm{zero}\mathrm{sequence}∆-\mathrm{load}\mathrm{current}. \\ \mathrm{Substitute}10\angle 0^{°}A\mathrm{for}{l}_{ab},12\angle -{90}^{°}A\mathrm{for}{l}_{bc}\mathrm{and}15\angle {90}^{°}A \\ \mathrm{for}{l}_{ca}\mathrm{into}\mathrm{equation}\left(1\right). \\ {l}_{∆0}=\frac{1}{3}\left(10\angle 0^{°}+12\angle -{90}^{°}+15\angle {90}^{°}\right) \\ {l}_{∆0}=\frac{1}{3}\left(10+j12+j15\right) \\ {l}_{∆0}=\frac{1}{3}(10+j3) \\ {l}_{∆0}=\frac{1}{3}3.33+j1A \\ \mathrm{Convert}\mathrm{the}\mathrm{current}\mathrm{into}\mathrm{polar}\mathrm{form}. \\ {l}_{∆0}=3.48\angle 16.{72}^{°}A..................\left(10\right) \end{gathered}$$

$$\begin{gathered} Calculatethepositivesequence∆-loadcurrent. \\ Substitute10\angle 0^{°}Afor{l}_{ab},12\angle -{90}^{°}Afor{l}_{bc}and15\angle {90}^{°}Afor{l}_{ca} \\ intoequation\left(1\right). \\ {l}_{∆0}=\frac{1}{3}\left(10\angle 0^{°}+12\angle -{90}^{°}+15\angle {90}^{°}\right) \\ {l}_{∆0}=\frac{1}{3}\left(10+j12+j15\right) \\ {l}_{∆0}=\frac{1}{3}(10+j3) \\ {l}_{∆0}=\frac{1}{3}3.33+j1A \\ \mathrm{Convert}\mathrm{the}\mathrm{current}\mathrm{into}\mathrm{polar}\mathrm{form}. \\ {l}_{∆0}=3.48\angle 16.{72}^{°}A..................\left(10\right) \end{gathered}$$

$$\begin{gathered} Calculatethepositivesequence∆-loadcurrent. \\ Substitute10\angle 0^{°}Afor{l}_{ab},12\angle -{90}^{°}Afor{l}_{bc}and15\angle {90}^{°}Afor{l}_{ca} \\ intoequation\left(2\right). \\ {l}_{∆1}=\frac{1}{3}\left(10\angle 0^{°}+12\angle {(-{90}^{°}+120)}^{°}+15\angle ({90}^{°}+{240}^{°}\right)) \\ {l}_{∆1}=\frac{1}{3}\left(10+12\angle {30}^{°}+15\angle {330}^{°}\right) \\ {l}_{∆1}=\frac{1}{3}(10+10.39+j6-12.99-j7.5) \\ {l}_{∆1}=11.13+j0.5A \\ \mathrm{Convert}\mathrm{the}\mathrm{current}\mathrm{into}\mathrm{polar}\mathrm{form}. \\ {l}_{∆1}=11.14\angle 357.{43}^{°}A..................(11) \\ Calculatethepositivesequence∆-loadcurrent. \\ Substitute10\angle 0^{°}Afor{l}_{ab},12\angle -{90}^{°}Afor{l}_{bc}and15\angle {90}^{°}Afor{l}_{ca} \\ intoequation(3). \\ {l}_{∆2}=\frac{1}{3}\left(10\angle 0^{°}+12\angle {(-{90}^{°}+240)}^{°}+15\angle ({90}^{°}+{120}^{°}\right)) \\ {l}_{∆2}=\frac{1}{3}\left(10+12\angle {150}^{°}+15\angle {210}^{°}\right) \\ {l}_{∆2}=\frac{1}{3}(10-10.39+j6-12.99-j7.5) \\ {l}_{∆2}=4.46+j0.5A \\ \mathrm{Convert}\mathrm{the}\mathrm{current}\mathrm{into}\mathrm{polar}\mathrm{form}. \\ {l}_{∆2}=4.49\angle 186.{40}^{°}A..................\left(12\right) \\ Hence,thesequencecomponentofthe∆loadcurrents{l}_{∆0},l_{∆1}and{l}_{∆2}are3.48\angle 16.{72}^{°}A \\ ,11.14\angle 357.{43}^{°}Aand4.49\angle 186.{40}^{°}Arespectively. \end{gathered}$$localid="1655467204980" $$\begin{gathered} Calculatethepositivesequence∆-loadcurrent. \\ Substitute10\angle 0^{°}Afor{l}_{ab},12\angle -{90}^{°}Afor{l}_{bc}and15\angle {90}^{°}Afor{l}_{ca} \\ intoequation\left(2\right). \\ {l}_{∆1}=\frac{1}{3}\left(10\angle 0^{°}+12\angle {(-{90}^{°}+120)}^{°}+15\angle ({90}^{°}+{240}^{°}\right)) \\ {l}_{∆1}=\frac{1}{3}\left(10+12\angle {30}^{°}+15\angle {330}^{°}\right) \\ {l}_{∆1}=\frac{1}{3}(10+10.39+j6-12.99-j7.5) \\ {l}_{∆1}=11.13+j0.5A \\ \mathrm{Convert}\mathrm{the}\mathrm{current}\mathrm{into}\mathrm{polar}\mathrm{form}. \\ {l}_{∆1}=11.14\angle 357.{43}^{°}A..................(11) \end{gathered}$$

localid="1655467214345" $$\begin{gathered} Calculatethepositivesequence∆-loadcurrent. \\ Substitute10\angle 0^{°}Afor{l}_{ab},12\angle -{90}^{°}Afor{l}_{bc}and15\angle {90}^{°}Afor{l}_{ca} \\ intoequation\left(3\right). \\ {l}_{∆2}=\frac{1}{3}\left(10\angle 0^{°}+12\angle {(-{90}^{°}+240)}^{°}+15\angle ({90}^{°}+{120}^{°}\right)) \\ {l}_{∆2}=\frac{1}{3}\left(10+12\angle {150}^{°}+15\angle {210}^{°}\right) \\ {l}_{∆2}=\frac{1}{3}(10-10.39+j6-12.99-j7.5) \\ {l}_{∆2}=4.46+j0.5A \\ \mathrm{Convert}\mathrm{the}\mathrm{current}\mathrm{into}\mathrm{polar}\mathrm{form}. \\ {l}_{∆2}=4.49\angle 186.{40}^{°}A..................(12) \\ Hence,thesequencecomponentofthe∆loadcurrents{l}_{∆0},l_{∆1}and{l}_{∆2}are \\ 3.48\angle 16.{72}^{°}A,11.14\angle 357.{43}^{°}Aand4.49\angle 186.{40}^{°}Arespectively. \end{gathered}$$

Calculate line currents.

Calculate the line $l_a$.

$$\begin{gathered} \mathrm{Substitute}10\angle 0^{°}A\mathrm{for}{l}_{ab}\mathrm{and}15\angle {90}^{°}A\mathrm{for}{l}_{ca}\mathrm{into}\mathrm{equation}\left(4\right). \\ {l}_a=10\angle 0^{°}-15\angle {90}^{°} \\ {l}_a=18.03\angle 303.69A \\ \mathrm{Calculate}\mathrm{the}\mathrm{line}{l}_a. \\ \mathrm{Substitute}10\angle 0^{°}A\mathrm{for}{l}_{ab}\mathrm{and}12\angle {90}^{°}A\text{for}{l}_{bc}\mathrm{into}\mathrm{equation}\left(5\right). \\ {l}_b=12\angle -{90}^{°}-10\angle 0^{°} \\ {l}_b=15.62\angle 230.{19}^{°}A \\ \mathrm{Calculate}\mathrm{the}\mathrm{line}{l}_a. \\ \mathrm{Substitute}12\angle -{90}^{°}A\mathrm{for}{l}_{bc}\mathrm{and}15\angle {90}^{°}A\mathrm{for}{l}_{ca}\mathrm{into}\mathrm{equation}\left(6\right). \\ {l}_a=15\angle {90}^{°}-12\angle -{90}^{°} \\ {l}_a=27\angle {90}^{°}A \\ \mathrm{Hence}\mathrm{the}\mathrm{line}\mathrm{current}{l}_a,l_b\mathrm{and}{l}_c\mathrm{are}18.03\angle 303.69A,15.62\angle 230.{19}^{°}A\mathrm{and}27\angle {90}^{°}A. \end{gathered}$$

Calculate the sequence component of the line current.

$$\begin{gathered} \mathrm{Calculate}\mathrm{the}\mathrm{line}\mathrm{current}{l}_{L0}. \\ \mathrm{Substitute}18.03\angle 303.69A\mathrm{for}{l}_a,15.62\angle 230.{19}^{°}A\mathrm{for}{l}_{b}and27\angle {90}^{°}A\mathrm{for}{l}_c \\ intoequation\left(7\right). \\ {l}_{L0}=\frac{1}{3}\left(18.03\angle 303.69+15.62\angle 230.{19}^{°}+27\angle {90}^{°}\right) \\ {l}_{L0}=\frac{1}{3}\left(10-j15-10-j12+j27\right) \\ {l}_{L0}=0A \\ \mathrm{Calculate}\mathrm{the}\mathrm{line}\mathrm{current}{l}_{L0}. \\ \mathrm{Substitute}18.03\angle 303.69A\mathrm{for}{l}_a,15.62\angle 230.{19}^{°}A\mathrm{for}{l}_b\mathrm{and}27\angle {90}^{°}A\mathrm{for}l \\ \mathrm{into}\mathrm{equation}\left(8\right). \\ {l}_{L1}=\frac{1}{3}\left(18.03\angle 303.{69}^{°}+15.62\angle {(230.19+120)}^{°}+27\angle {(90+240)}^{°}\right) \\ {l}_{L1}=\frac{1}{3}\left(18.03\angle 303.{69}^{°}+15.62\angle 350.{19}^{°}+27\angle {90+330}^{°}\right) \\ {l}_{L1}=\frac{1}{3}(10-j15+15.39-j2.66+23.38-j13.5) \\ {l}_{L1}=\frac{1}{3}\left(48.77-j31.16\right) \end{gathered}$$

$$\begin{gathered} Solvefurtheras, \\ {l}_{L1}=16.26-j10.39A \\ {l}_{L1}=19.30\angle 327.{42}^{°}A.................\left(14\right) \\ Calculatethelinecurrent{l}_{L2}. \\ \mathrm{Substitute}18.03\angle 303.69A\mathrm{for}{l}_a,15.62\angle 230.{19}^{°}A\mathrm{for}{l}_b\mathrm{and}27\angle {90}^{°}A\mathrm{for}l \\ \mathrm{into}\mathrm{equation}\left(8\right). \\ {l}_{L2}=\frac{1}{3}\left(18.03\angle 303.{69}^{°}+15.62\angle {(230.19+120)}^{°}+27\angle {(90+240)}^{°}\right) \\ {l}_{L2}=\frac{1}{3}\left(18.03\angle 303.{69}^{°}+15.62\angle 350.{19}^{°}+27\angle {210}^{°}\right) \\ {l}_{L2}=\frac{1}{3}(10-j15+5.39+j14.66-23.38-j13.5) \\ {l}_{L2}=\frac{1}{3}\left(-18.77-j13.84\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Solve}\mathrm{further}\mathrm{as}, \\ {l}_{L1}=-6.26-j4.61A \\ {l}_{L1}=7.77\angle 216.{37}^{°}A............\left(15\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{sequence}\mathrm{component}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{currents}{l}_{L0},l_{L1}\mathrm{and}{l}_{L2}\mathrm{are} \\ 0\mathrm{A},19.30\angle 327.{42}^{°}\mathrm{A}\mathrm{and}7.77\angle 216.{37}^{°}\mathrm{A}. \\ \mathrm{Divide}\mathrm{the}\mathrm{equation}\left(13\right)\mathrm{from}\mathrm{equation}\left(10\right). \\ \frac{l_{L0}}{l_{∆0}}=\frac{0}{0} \\ {l}_{L0}=0A \\ Dividetheequation\left(14\right)fromequation\left(11\right). \\ \frac{l_{L1}}{l_{∆1}}=\frac{19.30\angle 327.{42}^{°}}{11.14\angle 357.{43}^{°}} \\ \frac{l_{L1}}{l_{∆1}}=1.732\angle -{30}^{°} \\ {l}_{L1}=\sqrt{3}{l}_{∆1}\angle -{30}^{°} \\ Dividetheequation\left(15\right)fromequation\left(12\right). \\ \frac{l_{L2}}{l_{∆2}}=\frac{7.77\angle 216.{37}^{°}}{4.49\angle 186.{40}^{°}} \\ \frac{l_{L2}}{l_{∆2}}=1.732\angle -{30}^{°} \\ {l}_{L2}=\sqrt{3}{l}_{∆2}\angle -{30}^{°}A \\ Hencethegeneralrelation{l}_{L0}=0A,l_{L1}=\sqrt{3}{l}_{∆1}\angle -{30}^{°}and \\ {l}_{L2}=\sqrt{3}{l}_{∆2}\angle -{30}^{°}areverified. \end{gathered}$$