Question

In slip test on a 3-phase alternator, the maximum and minimum voltage per phase are V₁ and V₂ respectively whereas the maximum and minimum phase currents are found to be I₁ and I₂ respectively. The value of the d-axis synchronous reactance x_d and q-axis synchronous reactance x_q are determined as
1. \({x_d} = \frac{{{V_1}}}{{{I_1}}},\;{x_q} = \frac{{{V_2}}}{{{I_2}}}\)
2. \({x_d} = \frac{{{V_1}}}{{{I_2}}},\;{x_q} = \frac{{{V_2}}}{{{I_1}}}\)
3. \({x_d} = \frac{{{V_1}}}{{\left( {{I_{1 - }}{I_2}} \right)}},\;{x_q} = \frac{{{V_2}}}{{({I_1} + {I_2})}}\)
4. \({x_d} = \frac{{({V_1} + {V_2})}}{{({I_1} - {I_2})}},\;{x_q} = \frac{{\left( {{V_1} - {V_2}} \right)}}{{\left( {{I_2} + {I_1}} \right)}}\)

Answer 1

Correct Answer - Option 2 : \({x_d} = \frac{{{V_1}}}{{{I_2}}},\;{x_q} = \frac{{{V_2}}}{{{I_1}}}\)
\(\begin{array}{l} {x_d} = \frac{{maximum\;vlotage}}{{minimum\;current}} = \frac{{{V_1}}}{{{I_2}}}\;\\ {x_q} = \frac{{minimum\;vlotage}}{{maxmum\;current}} = \frac{{{V_2}}}{{{I_1}}}\; \end{array}\)