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Combination of Resistors

A 6.6 kV, 3-p...

Question

A 6.6 kV, 3-phase, 10 MVA alternator has positive, negative and zero sequence reactances as j0.05 p.u. j0.0375 p.u. and j0.045 p.u. respectively. A double line to ground faults occurs, then the magnitude of positive sequence voltage is

A

0.29 p.u

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B

6.45 p.u.

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C

14.19 p.u.

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D

can not determined

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Solution

The correct option is A 0.29 p.u

$I_{a 1} = \frac{E}{Z_{1} + \frac{Z_{2} Z_{0}}{Z_{2} + Z_{0}}}$

$= \frac{1}{j 0.05 + \frac{j 0.0375 \times j 0.045}{j 0.0375 + j 0.045}}$

$∴ I_{a 1} = - (I_{a 2} + I_{a 0}))$

$= - j 14.19 p . u .$

$I_{a 0} = - I_{a_{1}} \frac{j 0.0375}{j 0.0375 + j 0.045}$

$= j 6.45 p . u .$

$V_{a 1} = V_{a 2} = V_{a 0} = j 0.045 \times j 6.45$

$= - 0.29 p . u .$

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