Importance and derivation of current components

Symmetrical components transform an unbalanced set of three phasors into three balanced sets (zero, positive and negative sequence), so an unbalanced three-phase problem can be solved as three simple single-phase problems.

Importance for unbalanced system analysis

• Decoupling: for symmetrical (balanced) equipment such as lines, transformers and machines, the sequence impedance matrix is diagonal. So the zero, positive and negative sequence networks are independent and can be solved separately, instead of solving three coupled phase equations.
• Unsymmetrical fault calculation: L-G, L-L and L-L-G faults are solved by connecting the sequence networks at the fault point in a simple way (series, parallel). This is the standard method for fault level and relay setting.
• Per phase analysis is kept: each sequence network is a single-phase circuit, so per unit methods and Thevenin equivalents still apply.
• Physical meaning: negative sequence current causes rotor heating in machines; zero sequence current indicates earth faults and flows only through neutral/ground paths. Relays (earth fault, negative sequence) are built on these quantities.
• Measurable: sequence components can be measured with sequence filters, which is useful for protection.

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 $I_{a0} = 0$.