SME
Power Systems · Problem type
Symmetrical components
Use this for “Iₐ = 10∠10° A, Ib = 8∠−100° A and Ic = 6∠140° A. Find the zero-, positive- and negative-sequence currents.”
Diagram
Iₐ = 10∠10° A, Ib = 8∠−100° A, Ic = 6∠140° A.
This is an example. Type your own values to start a new problem.
|Iₐ|
Phase a current, magnitude
example · A
θₐ
Phase a current, angle
example · °
|Ib|
Phase b current, magnitude
example · A
θb
Phase b current, angle
example · °
|Ic|
Phase c current, magnitude
example · A
θc
Phase c current, angle
example · °
Re Iₐ₀
Zero-sequence current, real part
worked out (not typed) · A
Im Iₐ₀
Zero-sequence current, imaginary part
worked out (not typed) · A
Re Iₐ₁
Positive-sequence current, real part
worked out (not typed) · A
Im Iₐ₁
Positive-sequence current, imaginary part
worked out (not typed) · A
Re Iₐ₂
Negative-sequence current, real part
worked out (not typed) · A
Im Iₐ₂
Negative-sequence current, imaginary part
worked out (not typed) · A
Formulas
Re Iₐ₀ = Re(((|Iₐ|∠θₐ) + (|Ib|∠θb) + (|Ic|∠θc)) ÷ 3)
1.2875 = Re(((10∠10) + (8∠−100) + (6∠140)) ÷ 3)
Im Iₐ₀ = Im(((|Iₐ|∠θₐ) + (|Ib|∠θb) + (|Ic|∠θc)) ÷ 3)
−0.7618 = Im(((10∠10) + (8∠−100) + (6∠140)) ÷ 3)
Re Iₐ₁ = Re(((|Iₐ|∠θₐ) + (1∠120°) × (|Ib|∠θb) + (1∠−120°) × (|Ic|∠θc)) ÷ 3)
7.6679 = Re(((10∠10) + (1∠120°) × (8∠−100) + (1∠−120°) × (6∠140)) ÷ 3)
Im Iₐ₁ = Im(((|Iₐ|∠θₐ) + (1∠120°) × (|Ib|∠θb) + (1∠−120°) × (|Ic|∠θc)) ÷ 3)
2.1749 = Im(((10∠10) + (1∠120°) × (8∠−100) + (1∠−120°) × (6∠140)) ÷ 3)
Re Iₐ₂ = Re(((|Iₐ|∠θₐ) + (1∠−120°) × (|Ib|∠θb) + (1∠120°) × (|Ic|∠θc)) ÷ 3)
0.8926 = Re(((10∠10) + (1∠−120°) × (8∠−100) + (1∠120°) × (6∠140)) ÷ 3)
Im Iₐ₂ = Im(((|Iₐ|∠θₐ) + (1∠−120°) × (|Ib|∠θb) + (1∠120°) × (|Ic|∠θc)) ÷ 3)
0.3233 = Im(((10∠10) + (1∠−120°) × (8∠−100) + (1∠120°) × (6∠140)) ÷ 3)
Assumptions
•
j² = −1, and the operator a = 1∠120° turns a phasor 120° ahead; a² = 1∠−120°.
•
Phase sequence abc: a balanced set has only the positive-sequence part, so Iₐ₀ and Iₐ₂ measure the unbalance.
•
The phase currents are rms phasors in amperes, with angles in degrees from one reference.
•
The components found are phase a’s; phase b’s and c’s are the same sizes turned by a and a².
Step-by-step
WE KNOW
|Iₐ| = 10 A, θₐ = 10°, |Ib| = 8 A, θb = −100°, |Ic| = 6 A, θc = 140°
FIND
zero-sequence current, real part (Re Iₐ₀), zero-sequence current, imaginary part (Im Iₐ₀), positive-sequence current, real part (Re Iₐ₁), positive-sequence current, imaginary part (Im Iₐ₁), negative-sequence current, real part (Re Iₐ₂), negative-sequence current, imaginary part (Im Iₐ₂)
Step 1 · Find zero-sequence current, real part (Re Iₐ₀)
Use Re Iₐ₀ = Re(((|Iₐ|∠θₐ) + (|Ib|∠θb) + (|Ic|∠θc)) ÷ 3)
The zero-sequence part is the average of the three phase currents.
Re Iₐ₀ = Re(((10∠10°) + (8∠−100°) + (6∠140°)) ÷ 3)
10∠10° = 10 cos 10° + j10 sin 10° = 9.8481 + j1.7365
8∠−100° = 8 cos(−100°) + j8 sin(−100°) = −1.3892 − j7.8785
6∠140° = 6 cos 140° + j6 sin 140° = −4.5963 + j3.8567
(9.8481 + j1.7365) + (−1.3892 − j7.8785) = (9.8481 + (−1.3892)) + j(1.7365 + (−7.8785)) = 8.4589 − j6.142
(8.4589 − j6.142) + (−4.5963 + j3.8567) = (8.4589 + (−4.5963)) + j(−6.142 + 3.8567) = 3.8626 − j2.2853
3.8626 − j2.2853
3
=
1.2875
−
j0.7618
Re Iₐ₀ = 1.2875 A
Step 2 · Find zero-sequence current, imaginary part (Im Iₐ₀)
Use Im Iₐ₀ = Im(((|Iₐ|∠θₐ) + (|Ib|∠θb) + (|Ic|∠θc)) ÷ 3)
The imaginary part of the same Iₐ₀.
Im Iₐ₀ = Im(((10∠10°) + (8∠−100°) + (6∠140°)) ÷ 3)
Im Iₐ₀ = −0.7618 A → Iₐ₀ = 1.2875 − j0.7618 A = 1.496∠−30.61° A
Step 3 · Find positive-sequence current, real part (Re Iₐ₁)
Use Re Iₐ₁ = Re(((|Iₐ|∠θₐ) + (1∠120°) × (|Ib|∠θb) + (1∠−120°) × (|Ic|∠θc)) ÷ 3)
Turn Ib ahead by 120° and Ic back by 120°, add them to Iₐ and take a third.
Re Iₐ₁ = Re(((10∠10°) + (1∠120°) × (8∠−100°) + (1∠−120°) × (6∠140°)) ÷ 3)
1∠120° × 8∠−100° = (1 × 8)∠(120° + (−100°)) = 8∠20°
1∠−120° × 6∠140° = (1 × 6)∠(−120° + 140°) = 6∠20°
10∠10° = 10 cos 10° + j10 sin 10° = 9.8481 + j1.7365
8∠20° = 8 cos 20° + j8 sin 20° = 7.5175 + j2.7362
6∠20° = 6 cos 20° + j6 sin 20° = 5.6382 + j2.0521
(9.8481 + j1.7365) + (7.5175 + j2.7362) = (9.8481 + 7.5175) + j(1.7365 + 2.7362) = 17.3656 + j4.4726
(17.3656 + j4.4726) + (5.6382 + j2.0521) = (17.3656 + 5.6382) + j(4.4726 + 2.0521) = 23.0038 + j6.5248
23.0038 + j6.5248
3
=
7.6679
+
j2.1749
Re Iₐ₁ = 7.6679 A
Step 4 · Find positive-sequence current, imaginary part (Im Iₐ₁)
Use Im Iₐ₁ = Im(((|Iₐ|∠θₐ) + (1∠120°) × (|Ib|∠θb) + (1∠−120°) × (|Ic|∠θc)) ÷ 3)
The imaginary part of the same Iₐ₁.
Im Iₐ₁ = Im(((10∠10°) + (1∠120°) × (8∠−100°) + (1∠−120°) × (6∠140°)) ÷ 3)
Im Iₐ₁ = 2.1749 A → Iₐ₁ = 7.6679 + j2.1749 A = 7.9704∠15.84° A
Step 5 · Find negative-sequence current, real part (Re Iₐ₂)
Use Re Iₐ₂ = Re(((|Iₐ|∠θₐ) + (1∠−120°) × (|Ib|∠θb) + (1∠120°) × (|Ic|∠θc)) ÷ 3)
Turn Ib back by 120° and Ic ahead by 120°, add them to Iₐ and take a third.
Re Iₐ₂ = Re(((10∠10°) + (1∠−120°) × (8∠−100°) + (1∠120°) × (6∠140°)) ÷ 3)
1∠−120° × 8∠−100° = (1 × 8)∠(−120° + (−100°)) = 8∠−220° = 8∠140°
1∠120° × 6∠140° = (1 × 6)∠(120° + 140°) = 6∠260° = 6∠−100°
10∠10° = 10 cos 10° + j10 sin 10° = 9.8481 + j1.7365
8∠140° = 8 cos 140° + j8 sin 140° = −6.1284 + j5.1423
6∠−100° = 6 cos(−100°) + j6 sin(−100°) = −1.0419 − j5.9088
(9.8481 + j1.7365) + (−6.1284 + j5.1423) = (9.8481 + (−6.1284)) + j(1.7365 + 5.1423) = 3.7197 + j6.8788
(3.7197 + j6.8788) + (−1.0419 − j5.9088) = (3.7197 + (−1.0419)) + j(6.8788 + (−5.9088)) = 2.6778 + j0.9699
2.6778 + j0.9699
3
=
0.8926
+
j0.3233
Re Iₐ₂ = 0.8926 A
Step 6 · Find negative-sequence current, imaginary part (Im Iₐ₂)
Use Im Iₐ₂ = Im(((|Iₐ|∠θₐ) + (1∠−120°) × (|Ib|∠θb) + (1∠120°) × (|Ic|∠θc)) ÷ 3)
The imaginary part of the same Iₐ₂.
Im Iₐ₂ = Im(((10∠10°) + (1∠−120°) × (8∠−100°) + (1∠120°) × (6∠140°)) ÷ 3)
Im Iₐ₂ = 0.3233 A → Iₐ₂ = 0.8926 + j0.3233 A = 0.9494∠19.91° A
Check
Put the numbers back in. Both sides should match.
1.2875 = Re(((10∠10°) + (8∠−100°) + (6∠140°)) ÷ 3) ✓
−0.7618 = Im(((10∠10°) + (8∠−100°) + (6∠140°)) ÷ 3) ✓
7.6679 = Re(((10∠10°) + (1∠120°) × (8∠−100°) + (1∠−120°) × (6∠140°)) ÷ 3) ✓
2.1749 = Im(((10∠10°) + (1∠120°) × (8∠−100°) + (1∠−120°) × (6∠140°)) ÷ 3) ✓
0.8926 = Re(((10∠10°) + (1∠−120°) × (8∠−100°) + (1∠120°) × (6∠140°)) ÷ 3) ✓
0.3233 = Im(((10∠10°) + (1∠−120°) × (8∠−100°) + (1∠120°) × (6∠140°)) ÷ 3) ✓
Topic check
See what you know: 5 questions on this topic
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