An unbalanced \(3\phi\) system can be resolved into three balanced systems in the sinusoidal steady state.

This method of resolving an unbalanced system into three balanced phasor system has been proposed by C. L. Fortescue.

This method is called resolving symmetrical components of the original phasors or simply .

Symmetrical components transformation will present how unbalanced components like Y-or-\(\varDelta\) -connected loads, transformers, generators and transmission lines can be resolved into symmetrical components.

We can then combine all these components together to form what are called .

Three symmetrical components:

A balanced three-phase system with the same phase sequence as the original sequence

A balanced three-phase system with the opposite phase sequence as the original sequence

Three phasors that are equal in magnitude and phase

The easiest method to analyse unbalanced system operation due to faults is through the use of symmetrical components

let us first define the \(\alpha\)-operator for which the following relations holds \[\begin{aligned} \alpha & =e^{j120^{0}}=-\dfrac{1}{2}+j\dfrac{\sqrt{3}}{2}\\ \alpha^{2} & =e^{j240^{0}}=-\dfrac{1}{2}-j\dfrac{\sqrt{3}}{2}=\alpha^{*}\\ \alpha^{3} & =1\\ \alpha^{4} & =\alpha\\ \alpha^{5} & =\alpha^{2}~\text{and so on} \end{aligned}\] Note that we have \[\boxed{1+\alpha+\alpha^{2}=0}\]

Using the \(\alpha\)-operator we can write \[\begin{aligned} V_{b1} & =\alpha^{2}V_{a1} \quad V_{c1}=\alpha V_{a1}\\ V_{b2} & =\alpha V_{a2} \quad V_{c2}=\alpha^{2}V_{a2} \end{aligned}\] Finally, we get \[\begin{aligned} V_{a0} & =V_{b0}=V_{c0}\\ V_{a} & =V_{a0}+V_{a1}+V_{a2}\\ V_{b} & =V_{a0}+\alpha^{2}V_{a1}+\alpha V_{a2}=V_{b0}+V_{b1}+V_{b2}\\ V_{c} & =V_{a0}+\alpha V_{a1}+\alpha^{2}V_{a2}=V_{c0}+V_{c1}+V_{c2} \end{aligned}\]

Therefore, \[\begin{aligned} \left[\begin{array}{c} V_{a}\\ V_{b}\\ V_{c} \end{array}\right] & =\left[\begin{array}{ccc} 1 & 1 & 1\\ 1 & \alpha^{2} & \alpha\\ 1 & \alpha & \alpha^{2} \end{array}\right]\left[\begin{array}{c} V_{a0}\\ V_{a1}\\ V_{a2} \end{array}\right] \end{aligned}\] The symmetrical component transformation matrix is then given by \[\begin{aligned} \left[\begin{array}{c} V_{a0}\\ V_{a1}\\ V_{a2} \end{array}\right] & =\dfrac{1}{3}\left[\begin{array}{ccc} 1 & 1 & 1\\ 1 & \alpha & \alpha^{2}\\ 1 & \alpha^{2} & \alpha \end{array}\right]\left[\begin{array}{c} V_{a}\\ V_{b}\\ V_{c} \end{array}\right]\\ \Longrightarrow & V_{a012}=C\cdot V_{abc}\\ \Longrightarrow & I_{a012}=C\cdot I_{abc} \end{aligned}\] where \(C\) is the symmetrical component transformation matrix

The original phasor components can be obtained from the inverse symmetrical component transformation, i.e., \[\begin{aligned} V_{abc} & =C^{-1}V_{a012}\\ I_{abc} & =C^{-1}I_{a012} \end{aligned}\]

The three-phase power in the original unbalanced system is given by \[\begin{aligned} P_{abc}+jQ_{abc} & =V_{a}I_{a}^{*}+V_{b}I_{b}^{*}+V_{c}I_{c}^{*}=V_{abc}^{T}I_{abc}^{*}\\ & =\left(C^{-1}V_{a012}\right)^{T}\left(C^{-1}I_{a012}\right)^{*}\\ & =V_{a012}^{T}C^{-T}\left(C^{-1}\right)^{*}I_{a012}^{*} \end{aligned}\] we can deduce, \[C^{-T}\left(C^{-1}\right)^{*}=3\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]\]

Therefore, \[P_{abc}+jQ_{abc}=3\left(V_{a0}I_{a0}^{*}+V_{a1}I_{a1}^{*}+V_{a2}I_{a2}^{*}\right)\] Thus, complex power is three times the summation of the complex power of the three phase sequences.

Balanced \(3\phi\) system consists of \(I_1\) only; \(I_2\) & \(I_0\) being zero

Presence of \(I_2\) or \(I_0\) in a \(3\phi\) system introduces un-symmetry and is indicative of an abnormal condition of the circuit

In an unbalanced 3-phase system \(\overrightarrow{I_{1}}+\overrightarrow{I_{2}} = 0\).

The resultant solely consists of \(I_0\) i.e. Vector sum of all sequence currents in 3-phase unbalanced system \[\begin{aligned} = \vec{I_{R0}} + \vec{I_{Y0}} + \vec{I_{B0}} \end{aligned}\]

In a 3-phase, 4 wire unbalanced system, the magnitude of \[\begin{aligned} \text{Zero sequence current} = \dfrac{1}{3} \cdot \left[\text{Current in neutral wire}\right] \end{aligned}\]

In the absence of path through the neutral of a 3-phase system, \(I_n =0\) and \(I_L\) contain no \(I_0\).

\(\Delta\) load \(\Rightarrow\) no path to \(I_n\) & \(I_L\), and can contain no \(\overrightarrow{I_{0}}\)

In a 3-phase unbalanced system, \(|\overrightarrow{I_{2}}|\) cannot exceed that of \(|\overrightarrow{I_{1}}|\)

If \(|\overrightarrow{I_{2}}|\) were the greater, the phase sequence of the resultant system would be reversed.

The current of a single phase load drawn from a 3-phase system comprises \[I_1=I_2=I_0\]