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What is the posit ive sequence currents in a three phase system, when the original phasors at which are Ia = (12 + j6), Ib = (12 – j 12), Ic = (– 15 + j10) ?
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- 14.8 Amp.
- 16.2 Amp.
- 18 Amp.
- 10.2 Amp
Correct Option: A
Ia = (12 + j6)
Ib = (12 – j 12) = 16.97 |45°
Ic = (– 15 + j 10) = 18.02 |46.3°
| Ia1 = | (Ia + αIb + α² Ic) | |
| 3 |
| = | [(12 + j6) + 1|120° 16.97|- 45° + 1|240°. 18.02|146.31° | |
| 3 |
| = | [(12 + j6) + 16.97|75° 18.02|146.31° | |
| 3 |
| = | [(12 + j6) + 16.97|75° 18.02|386.31° | |
| 3 |
| = | [(12 + j6) + 16.97(cos 75° + jsin 75°) + 18.02(cos 386.31° + j sin 386.31°)] | |
| 3 |
| = | [12 + j6 + 4.392 + j 16.392 + 16.16 + j 7.06] | |
| 3 |
| = | (32.55 + j 30.38) = 14.84 |43.02° | |
| 3 |
