--- title: "Derive symmetrical components of currents" subject: "Power System Analysis II" chapter: 4 key: "semester_5/power_system_analysis_ii:symmetrical-components-currents-derivation" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/symmetrical-components-currents-derivation site: BCT Engineering --- # Derive symmetrical components of currents Power System Analysis II · Chapter 4: Unbalanced System Analysis - Asked in: 2072 Kartik, 5 marks - Video: [SEQUENCE COMPONENTS OF LINE CURRENTS](https://www.youtube.com/watch?v=-OU2G6Z5RJ0) — P SUMALATHA - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/symmetrical-components-currents-derivation - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-4 ## Question Derive the expression to calculate symmetrical components of 3-phase un-balance currents. ## Short answer Express each phase current as the sum of zero, positive and negative sequence components. Using a = 1∠120°, relate phase b and c components to phase a. Solving three equations yields I_{a0} = (I_a+I_b+I_c)/3, I_{a1} = (I_a+aI_b+a^2 I_c)/3, I_{a2} = (I_a+a^2 I_b+aI_c)/3. ## Answer Any unbalanced set of three-phase currents can be expressed as the sum of a zero sequence, a positive sequence and a negative sequence balanced set; the components of phase a are $I_{a0} = \frac13(I_a+I_b+I_c)$, $I_{a1} = \frac13(I_a+aI_b+a^2I_c)$ and $I_{a2} = \frac13(I_a+a^2I_b+aI_c)$. #### Derivation for currents Let $I_a, I_b, I_c$ be an unbalanced set of line currents. By Fortescue's theorem each is the sum of zero, positive and negative sequence components: $$ \begin{aligned} I_a &= I_{a0} + I_{a1} + I_{a2} \\ I_b &= I_{b0} + I_{b1} + I_{b2} \\ I_c &= I_{c0} + I_{c1} + I_{c2} \end{aligned} $$ Using operator $a = 1\angle120^\circ$ (so $a^2 = 1\angle240^\circ$, $1 + a + a^2 = 0$, $a^3 = 1$), all components are written in terms of phase a: - Zero sequence: $I_{b0} = I_{c0} = I_{a0}$ - Positive sequence (a-b-c): $I_{b1} = a^2 I_{a1}$, $I_{c1} = a I_{a1}$ - Negative sequence (a-c-b): $I_{b2} = a I_{a2}$, $I_{c2} = a^2 I_{a2}$ Therefore $$ \begin{bmatrix} I_a \\ I_b \\ I_c \end{bmatrix} = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix} \begin{bmatrix} I_{a0} \\ I_{a1} \\ I_{a2} \end{bmatrix} \quad\Rightarrow\quad I_p = A I_s $$ **Solving for the components:** 1. Add the three equations: $I_a + I_b + I_c = 3I_{a0} + (1+a^2+a)I_{a1} + (1+a+a^2)I_{a2} = 3I_{a0}$ 2. Multiply the second by $a$ and the third by $a^2$, then add: $I_a + aI_b + a^2I_c = (1+a+a^2)I_{a0} + (1+a^3+a^3)I_{a1} + (1+a^2+a^4)I_{a2} = 3I_{a1}$ 3. Multiply the second by $a^2$ and the third by $a$, then add: $I_a + a^2I_b + aI_c = 3I_{a2}$ $$ \begin{aligned} I_{a0} &= \tfrac13\left(I_a + I_b + I_c\right) \\ I_{a1} &= \tfrac13\left(I_a + aI_b + a^2I_c\right) \\ I_{a2} &= \tfrac13\left(I_a + a^2I_b + aI_c\right) \end{aligned} $$ or in matrix form $$ I_s = A^{-1}I_p, \qquad A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix} $$ Components of phases b and c follow from those of phase a using the relations above. **Notes:** $I_n = I_a + I_b + I_c = 3I_{a0}$, so zero sequence current exists only if there is a neutral or ground return. In a 3-wire or delta system $I_{a0} = 0$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The zero sequence component of unbalanced currents is: A. Ia0 = (Ia+aIb+a^2Ic)/3 B. Ia0 = (Ia+Ib+Ic)/3 C. Ia0 = Ia+Ib+Ic D. Ia0 = (Ia+a^2Ib+aIc)/3 Answer: B. Adding the three phase equations leaves 3Ia0. (Key point: Zero sequence formula) 2. The positive sequence component is: A. Ia1 = (Ia + a^2Ib + aIc)/3 B. Ia1 = (Ia+Ib+Ic)/3 C. Ia1 = (Ia + aIb + a^2Ic)/3 D. Ia1 = (Ia + aIb + a^2Ic) Answer: C. Multiply Ib by a and Ic by a^2 and add: 3Ia1; negative sequence uses a^2 and a. (Key point: Positive sequence formula) 3. In step 2, the Ia0 and Ia2 terms disappear because: A. a^3 = 0 B. Ia0 and Ia2 are zero C. They cancel with Ia1 D. Their coefficients become 1+a+a^2 = 0 Answer: D. Multiplying by a and a^2 gives coefficient sums 1+a+a^2 and 1+a^2+a^4=1+a^2+a, both zero. 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