Question

$$\begin{gathered} \mathbf{The}\mathbf{}\mathbf{currents}\mathbf{}\mathbf{in}\mathbf{}\mathbf{a}\mathbf{}\mathbf{∆}\mathbf{}\mathbf{load}\mathbf{}\mathbf{are}\mathbf{}{\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{°}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{ca}}\mathbf{=}\mathbf{15}\mathbf{\angle }{\mathbf{90}}^{\mathbf{°}}\mathbf{}\mathbf{A} \\ \mathbf{.}\mathbf{}\mathbf{Calculate}\mathbf{}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{the}\mathbf{}\mathbf{sequence}\mathbf{}\mathbf{components}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{∆}\mathbf{}\mathbf{-}\mathbf{load}\mathbf{}\mathbf{currents} \\ \mathbf{,}\mathbf{}\mathbf{denoted}\mathbf{}{\mathbf{l}}_{\mathbf{∆}\mathbf{0}}\mathbf{,}{\mathbf{l}}_{\mathbf{∆}\mathbf{01}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{∆}\mathbf{2}}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{the}\mathbf{}\mathbf{line}\mathbf{}\mathbf{currents}\mathbf{}{\mathbf{l}}_{\mathbf{a}}\mathbf{,}{\mathbf{l}}_{\mathbf{b}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{c}}\mathbf{}\mathbf{}\mathbf{which}\mathbf{}\mathbf{feed}\mathbf{}\mathbf{the}\mathbf{}\mathbf{∆}\mathbf{}\mathbf{load}\mathbf{}\mathbf{and}\mathbf{} \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{sequence}\mathbf{}\mathbf{components}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{line}\mathbf{}\mathbf{currents}\mathbf{}{\mathbf{l}}_{\mathbf{L}\mathbf{0}}\mathbf{,}{\mathbf{l}}_{\mathbf{L}\mathbf{1}}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{L}\mathbf{2}}\mathbf{}\mathbf{.}\mathbf{}\mathbf{Also}\mathbf{,}\mathbf{}\mathbf{verify}\mathbf{}\mathbf{the}\mathbf{}\mathbf{following}\mathbf{} \\ \mathbf{general}\mathbf{}\mathbf{relation}\mathbf{:}\mathbf{}{\mathbf{l}}_{\mathbf{L}\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{l}}_{\mathbf{L}\mathbf{1}}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{l}}_{\mathbf{∆}\mathbf{1}}\mathbf{\angle }\mathbf{-}{\mathbf{30}}^{\mathbf{°}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{L}\mathbf{2}}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{l}}_{\mathbf{∆}\mathbf{2}}\mathbf{\angle }{\mathbf{30}}^{\mathbf{°}}\mathbf{}\mathbf{.} \end{gathered}$$

Short answer

(a)The sequence component of the $∆$load currents $l_{∆0},l_{∆1}\mathrm{and}{l}_{∆2}$ are $3.48\angle 16.{72}^{°}\mathrm{A},11.14\angle 357.{43}^{°}\mathrm{A}\mathrm{and}4.49\angle 186.{40}^{°}\mathrm{A}$ respectively.

(b) The line current $l_a,l_b\mathrm{and}{l}_c$are $18.03\angle 303.69\mathrm{A},15.62\angle 230.{19}^{°}\mathrm{A}\mathrm{and}27\angle {90}^{°}\mathrm{A}.$

(c) The sequence component of the line currents $l_{L0},l_{L1}\mathrm{and}{l}_{L2}$are $0\mathrm{A},19.30\angle 327.{42}^{°}\mathrm{A}$and . $7.77\angle 216.{37}^{°}\mathrm{A}$The general relation $l_{L0}=0A,l_{L1}=\sqrt{3}{l}_{∆1}\angle -{30}^{°}\mathrm{and}{l}_{L2}=\sqrt{3}{l}_{∆2}\angle -{30}^{°}\mathrm{A}$are verified.

Step-by-step solution

Write the given data from the question.

The currents,

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

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

The equation to calculate the zero sequence$∆$-load current is given as follows.

role="math" localid="1655882663734" ${\mathbf{l}}_{\mathbf{∆}\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{ab}}\mathbf{+}{\mathbf{l}}_{\mathbf{bc}}\mathbf{+}{\mathbf{l}}_{\mathbf{ca}}\mathbf{)}$ …… (1)

The equation to calculate the positive sequence$∆$-load current is given as follows.

${\mathbf{l}}_{\mathbf{∆}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{ab}}\mathbf{+}\mathit{a}{\mathbf{l}}_{\mathbf{bc}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{ca}}\mathbf{)}$ …… (2)

The equation to calculate the negative sequence $∆$-load current is given as follows.

${\mathbf{l}}_{\mathbf{∆}\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{ab}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{bc}}\mathbf{+}\mathit{a}{\mathbf{l}}_{\mathbf{ca}}\mathbf{)}$ .........................(3)

The equation to calculate the line $l_a$is given as follows.

${\mathbf{l}}_{\mathbf{a}}\mathbf{=}{\mathbf{l}}_{\mathbf{ab}}\mathbf{-}{\mathbf{l}}_{\mathbf{ca}}$ …… (4)

The equation to calculate the line $l_b$is given as follows.

${\mathbf{l}}_{\mathbf{a}}\mathbf{=}{\mathbf{l}}_{\mathbf{bc}}\mathbf{-}{\mathbf{l}}_{\mathbf{ab}}$ …… (5)

The equation to calculate the line $l_c$is given as follows.

${\mathbf{l}}_{\mathbf{c}}\mathbf{=}{\mathbf{l}}_{\mathbf{ca}}\mathbf{-}{\mathbf{l}}_{\mathbf{bc}}$ …… (6)

The equation to calculate the line current $l_{L0}$is given as follows.

${\mathbf{l}}_{\mathbf{L}\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left(l_a+l_b+l_c\right)$ …… (7)

The equation to calculate the line current $l_{L1}$is given as follows.

${\mathbf{l}}_{\mathbf{L}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{a}}\mathbf{+}{\mathbf{al}}_{\mathbf{b}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{c}}\mathbf{)}$ …… (8)

The equation to calculate the line current $l_{L2}$is given as follows.

${\mathbf{l}}_{\mathbf{L}\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{a}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{b}}\mathbf{+}{\mathbf{al}}_{\mathbf{c}}\mathbf{)}$ …… (8)

 Step3: Calculate the sequence components of the ∆ -load currents

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

Calculate the zero sequence $∆$-load current.

localid="1655887040640" $$\begin{gathered} \mathrm{Substitute}10\angle 0^{°}A\mathrm{for}{l}_{ab},12\angle -{90}^{°}A\mathrm{for}{l}_{bc}\mathrm{and}15\angle {90}^{°}A \\ \mathrm{for}{l}_{ca}\mathrm{into}\mathrm{equation}\left(1\right). \\ {l}_{∆0}=\frac{1}{3}\left(10\angle 0^{°}+12\angle -{90}^{°}\right) \\ {l}_{∆0}=\frac{1}{3}(10+j12+j15) \\ {l}_{∆0}=\frac{1}{3}(10+j3) \\ {l}_{∆0}=3.33+j1\mathrm{A} \end{gathered}$$

Convert the current into polar form.

$l_{∆0}=3.48\angle 16.{72}^{°}\mathrm{A}..........\left(10\right)$

Calculate the positive sequence $∆$-load current.

localid="1655887378104" $$\begin{gathered} \mathrm{Substitute}10\angle 0^{°}A\mathrm{for}{l}_{ab},12\angle -{90}^{°}A\mathrm{for}{l}_{bc}\mathrm{and}15\angle {90}^{°}A \\ \mathrm{for}{l}_{ca}\mathrm{into}\mathrm{equation}\left(2\right). \\ {l}_{∆1}=\frac{1}{3}(10\angle 0^{°}+12\angle {(-90+120)}^{°}+15\angle ({90}^{°}+{240}^{°}\left)\right) \\ {l}_{∆1}=\frac{1}{3}(10+12\angle {30}^{°}+15\angle {330}^{°}) \\ {l}_{∆1}=\frac{1}{3}(10+10.39+j6+12.99-j7.5) \\ {l}_{∆1}=11.13+j0.5\mathrm{A} \end{gathered}$$

Convert the current into polar form.

$l_{∆1}=11.14\angle 357.{43}^{°}\mathrm{A}$ ..........(11)

Calculate the negative sequence $∆$-load current.

$$\begin{gathered} \mathrm{Substitute}10\angle 0^{°}A\mathrm{for}{l}_{ab},12\angle -{90}^{°}A\mathrm{for}{l}_{bc}\mathrm{and}15\angle {90}^{°}A \\ \mathrm{for}{l}_{ca}\mathrm{into}\mathrm{equation}\left(3\right). \\ {l}_{∆2}=\frac{1}{3}(10\angle 0^{°}+12\angle {(-90+240)}^{°}+15\angle ({90}^{°}+{120}^{°}\left)\right) \\ {l}_{∆2}=\frac{1}{3}(10+12\angle {150}^{°}+15\angle {210}^{°}) \\ {l}_{∆2}=\frac{1}{3}(10-10.39+j6-12.99-j7.5) \\ {l}_{∆2}=-4.46+j0.5\mathrm{A} \end{gathered}$$

Convert the current into polar form.

$l_{∆2}=-4.46\angle 186.{40}^{°}\mathrm{A}$ ............(12)

$$\begin{gathered} \mathrm{Hence},\mathrm{the}\mathrm{sequence}\mathrm{component}\mathrm{of}\mathrm{the}∆\mathrm{load}\mathrm{currents}{l}_{∆0},l_{∆1}\mathrm{and}{l}_{∆2}\mathrm{are} \\ 3.48\angle 16.{72}^{°}\mathrm{A},11.14\angle 357.{43}^{°}\mathrm{A}\mathrm{and}4.49\angle 186.{40}^{°}\mathrm{A}\mathrm{respectively}. \end{gathered}$$

Calculate line currents.

Calculate the line $l_a$.

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

Calculate the sequence component of the line current.

$$\begin{gathered} \mathrm{Calculate}\mathrm{the}\mathrm{line}\mathrm{current}{l}_{L0}. \\ \mathrm{Substitute}18.03\angle 303.69\mathrm{A}\mathrm{for}{l}_a,15.62\angle 230.{19}^{°}\mathrm{A}\mathrm{for}{l}_b\mathrm{and}27\angle {90}^{°}\mathrm{A}\mathrm{for} \\ {l}_c\mathrm{into}\mathrm{equation}\left(7\right). \\ {l}_{L0}=\frac{1}{3}(18.03\angle 303.69+15.62\angle 230.{19}^{°}+27\angle {90}^{°}) \\ {l}_{L0}=\frac{1}{3}(10-j45-10-j12+j27)..............\left(13\right) \\ {l}_{L0}=0\mathrm{A} \end{gathered}$$

role="math" localid="1655890500698" $$\begin{gathered} \mathrm{Calculate}\mathrm{the}\mathrm{line}\mathrm{current}{l}_{L1}. \\ \mathrm{Substitute}18.03\angle 303.69\mathrm{A}\mathrm{for}{l}_a,15.62\angle 230.{19}^{°}\mathrm{A}\mathrm{for}{l}_b\mathrm{and}27\angle {90}^{°}\mathrm{A}\mathrm{for} \\ {l}_c\mathrm{into}\mathrm{equation}\left(8\right). \\ {l}_{L1}=\frac{1}{3}(18.03\angle 303.{69}^{°}+15.62\angle {(230.19+120)}^{°}+27\angle {(90+240)}^{°}) \\ {l}_{L1}=\frac{1}{3}(18.03\angle 303.{69}^{°}+15.62\angle {350.19}^{°}+27\angle {330}^{°}) \\ {l}_{L1}=\frac{1}{3}(10-j15+15.39-j2.66+23.38-j13.5) \\ {l}_{L1}=\frac{1}{3}(48.77-j31.16) \end{gathered}$$

Solve further as,

role="math" localid="1655890633664" $$\begin{gathered} l_{L1}=16.26-j10.39\mathrm{A} \\ {l}_{L1}=19.30\angle 327.{42}^{°}\mathrm{A}...........\left(14\right) \\ \mathrm{Calculate}\mathrm{the}\mathrm{line}\mathrm{current}{l}_{L2}. \\ \mathrm{Substitute}18.03\angle 303.69\mathrm{A}\mathrm{for}{l}_a,15.62\angle 230.{19}^{°}\mathrm{A}\mathrm{for}{l}_b\mathrm{and} \\ 27\angle {90}^{°}\mathrm{A}\mathrm{for}{l}_c\mathrm{into}\mathrm{equation}\left(8\right). \\ {l}_{L2}=\frac{1}{3}(18.03\angle 303.{69}^{°}+15.62\angle {(230.19+120)}^{°}+27\angle {(90+240)}^{°}) \\ {l}_{L2}=\frac{1}{3}(18.03\angle 303.{69}^{°}+15.62\angle {470.19}^{°}+27\angle {210}^{°}) \\ {l}_{L2}=\frac{1}{3}(10-j15-5.39+j14.66-23.38-j13.5) \\ {l}_{L2}=\frac{1}{3}(18.77-j13.84) \end{gathered}$$

$$\begin{gathered} \mathrm{Solve}\mathrm{further}\mathrm{as}, \\ {l}_{L1}=-6.26-j4.61\mathrm{A} \\ {l}_{L1}=7.77\angle 216.{37}^{°}\mathrm{A}...........\left(15\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{sequence}\mathrm{component}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{currents}{l}_{L0},l_{L1}\mathrm{and}{l}_{L2}\mathrm{are} \\ 0\mathrm{A},19.30\angle 327.{42}^{°}\mathrm{A}\mathrm{and}7.77\angle 216.{37}^{°}\mathrm{A}. \end{gathered}$$

Divide the equation (13) from equation (10).

$$\begin{gathered} \frac{l_{L0}}{l_{∆0}}=\frac{0}{0} \\ {l}_{L0}=0\mathrm{A} \\ \mathrm{Divide}\mathrm{the}\mathrm{equation}\left(14\right)\mathrm{from}\mathrm{equation}\left(11\right). \\ \frac{l_{L1}}{l_{∆1}}=\frac{19.30\angle 327.{42}^{°}}{11.14\angle 357.{43}^{°}} \\ \frac{l_{L1}}{l_{∆1}}=1.732\angle -{30}^{°} \\ {l}_{L1}=\sqrt{3}{l}_{∆1}\angle -{30}^{°} \\ Dividetheequation\left(15\right)fromequation\left(12\right). \\ \frac{l_{L2}}{l_{∆2}}=\frac{7.77\angle 216.{37}^{°}}{4.49\angle 186.{40}^{°}} \\ \frac{l_{L2}}{l_{∆2}}=1.732\angle {30}^{°} \\ {l}_{L2}=\sqrt{3}{l}_{∆2}\angle -{30}^{°}\mathrm{A} \\ \mathrm{Hence}\mathrm{the}\mathrm{general}\mathrm{relation}{l}_{L0}=0\mathrm{A},l_{L1}=\sqrt{3}{l}_{∆1}\angle -{30}^{°}\mathrm{and}{l}_{L2}=\sqrt{3}{\mathrm{l}}_{∆1}\angle -{30}^{°} \\ \mathrm{are}\mathrm{verified}. \end{gathered}$$