In a Balanced$\Delta - \Delta$ Circuit Shown Below, the Source Voltage

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The Answer is Mesh equations:

\begin{array} { l } - V _ { a b } + Z _ { w } I _ { 1 } + Z _ { \Delta } \left( I _ { 1 } - I _ { 3 } ight) + Z _ { w } \left( I _ { 1 } - I _ { 2 } ight) = 0 \- V _ { b c } + Z _ { w } \left( I _ { 2 } - I _ { 1 } ight) + Z _ { \Delta } \left( I _ { 2 } - I _ { 3 } ight) + Z _ { w } I _ { 2 } = 0 \Z _ { \Delta } \left( I _ { 3 } - I _ { 2 } ight) + Z _ { \Delta } \left( I _ { 3 } - I _ { 1 } ight) + Z _ { \Delta } I _ { 3 } = 0 \I _ { A } = I _ { 1 } = 8.738 \angle - 70.3141 ^ { \circ } \mathrm { A } \\mathrm { I } _ { B } = I _ { 2 } - I _ { 1 } = 8.738 \angle 169.6859 ^ { \circ } \mathrm { A } \\mathrm { I } _ { C } = - \mathrm { I } _ { 2 } = 8.738 \angle 49.6859 ^ { \circ } \mathrm { A } \\mathrm { I } _ { \mathrm { AB } } = \mathrm { I } _ { 1 } - \mathrm { I } _ { 3 } = 5.0449 \angle - 40.3141 ^ { \circ } \mathrm { A } \\mathrm { I } _ { \mathrm { BC } } = \mathrm { I } _ { 2 } - \mathrm { I } _ { 3 } = 5.0449 \angle - 160.3141 ^ { \circ } \mathrm { A } \\mathrm { I } _ { \mathrm { CA } } = - \mathrm { I } _ { 3 } = 5.0449 \angle 79.6859 ^ { \circ } \mathrm { A }\end{array}

$Mesh equations: \begin{array} { l } - V _ { a b } + Z _ { w } I _ { 1 } + Z _ { \Delta } \left( I _ { 1 } - I _ { 3 } ight) + Z _ { w } \left( I _ { 1 } - I _ { 2 } ight) = 0 \ - V _ { b c } + Z _ { w } \left( I _ { 2 } - I _ { 1 } ight) + Z _ { \Delta } \left( I _ { 2 } - I _ { 3 } ight) + Z _ { w } I _ { 2 } = 0 \ Z _ { \Delta } \left( I _ { 3 } - I _ { 2 } ight) + Z _ { \Delta } \left( I _ { 3 } - I _ { 1 } ight) + Z _ { \Delta } I _ { 3 } = 0 \ I _ { A } = I _ { 1 } = 8.738 \angle - 70.3141 ^ { \circ } \mathrm { A } \ \mathrm { I } _ { B } = I _ { 2 } - I _ { 1 } = 8.738 \angle 169.6859 ^ { \circ } \mathrm { A } \ \mathrm { I } _ { C } = - \mathrm { I } _ { 2 } = 8.738 \angle 49.6859 ^ { \circ } \mathrm { A } \ \mathrm { I } _ { \mathrm { AB } } = \mathrm { I } _ { 1 } - \mathrm { I } _ { 3 } = 5.0449 \angle - 40.3141 ^ { \circ } \mathrm { A } \ \mathrm { I } _ { \mathrm { BC } } = \mathrm { I } _ { 2 } - \mathrm { I } _ { 3 } = 5.0449 \angle - 160.3141 ^ { \circ } \mathrm { A } \ \mathrm { I } _ { \mathrm { CA } } = - \mathrm { I } _ { 3 } = 5.0449 \angle 79.6859 ^ { \circ } \mathrm { A } \end{array} ext { ICAbp=vpa (ICAbp, 7) }$

ext { ICAbp=vpa (ICAbp, 7) }

In a Balanced $\Delta - \Delta$ Circuit Shown Below, the Source Voltage

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In a Balanced Δ−Δ\Delta - \DeltaΔ−Δ Circuit Shown Below, the Source Voltage

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In a Balanced $\Delta - \Delta$ Circuit Shown Below, the Source Voltage

Correct Answer:
Verified