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Circuit Theory

Three phase circuits

In the given ...

Question

In the given network $V_{1} = 100 ∠ 0^{0} V, V_{2} = 100 ∠ - 120^{0} V, V_{3} = 100 ∠ + 120^{0} V .$ The phasor current i (in Ampere) is

173.2 ∠ - 60^{0}

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173.2 ∠ 120^{0}

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100.0 ∠ - 60^{0}

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100.0 ∠ 120^{0}

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Solution

The correct option is B $173.2 ∠ 120^{0}$
'i' is taken as output

Apply KCL at node

Node voltage is $100 ∠ 120^{0}$

$i = \frac{100 ∠ 120^{0} - 100 ∠ 0^{0}}{- j 1} + \frac{100 ∠ 120^{0} - 100 ∠ - 120^{0}}{j 1}$

$i = 173.2 ∠ - 60^{0} A$

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In the given network V1=100∠00V,V2=100∠−1200V,V3=100∠+1200V. The phasor current i (in Ampere) is

The correct option is B 173.2∠1200'i' is taken as output Apply KCL at node Node voltage is 100∠1200 i=100∠1200−100∠00−j1+100∠1200−100∠−1200j1 i=173.2∠−600A

In the given network $V_{1} = 100 ∠ 0^{0} V, V_{2} = 100 ∠ - 120^{0} V, V_{3} = 100 ∠ + 120^{0} V .$ The phasor current i (in Ampere) is

The correct option is B $173.2 ∠ 120^{0}$
'i' is taken as output

Apply KCL at node

Node voltage is $100 ∠ 120^{0}$

$i = \frac{100 ∠ 120^{0} - 100 ∠ 0^{0}}{- j 1} + \frac{100 ∠ 120^{0} - 100 ∠ - 120^{0}}{j 1}$

$i = 173.2 ∠ - 60^{0} A$