Suppose I_A, I_B and I_C are a set of unbalanced current phasors in a three-phase system. The phase-B zero-sequence current I_B0 = 0.1 ∠0° p.u. If phase-A current I_A = 1.1 ∠0° p.u. and phase-C current I_C = (1 ∠120° + 0.1) p.u. then I_B in p.u. is

The correct option is: 1 ∠-120° + 0.1 ∠0°

Step-by-Step Explanation:

Symmetrical component theory allows us to decompose an unbalanced set of three-phase current phasors ($I_A$, $I_B$, and $I_C$) into zero-sequence ($I_0$), positive-sequence ($I_1$), and negative-sequence ($I_2$) components.

By definition, the zero-sequence currents are identical in magnitude and phase for all three phases:
$I_{A0}=I_{B0}=I_{C0}=I_0$

We are given that the phase-B zero-sequence current is:
$I_{B0}=0.1\angle 0°\text{p.u.}$
Thus, the zero-sequence component for all three phases is:
$I_0=0.1\angle 0°=0.1\text{p.u.}$

The phase currents can be expressed in terms of the symmetrical components of phase A ($I_{A1}$ and $I_{A2}$) as follows:
$I_A=I_0+I_{A1}+I_{A2}$
$I_B=I_0+a^2{I}_{A1}+a{I}_{A2}$
$I_C=I_0+a{I}_{A1}+a^2{I}_{A2}$

where the complex operator $a$ represents a phase shift of $120°$:
$a=1\angle 120°$
$a^2=1\angle 240°=1\angle -120°$

Let us substitute the given values of $I_A$ and $I_C$ to find $I_{A1}$ and $I_{A2}$:
$I_A=1.1\angle 0°=1.1$
$I_C=1\angle 120°+0.1=a+0.1$

Using the relation for $I_C$:
$a+0.1=0.1+a{I}_{A1}+a^2{I}_{A2}$
$a=a{I}_{A1}+a^2{I}_{A2}$

Dividing both sides by $a$:
$1=I_{A1}+a{I}_{A2}$

Using the relation for $I_A$:
$1.1=0.1+I_{A1}+I_{A2}$
$1=I_{A1}+I_{A2}$

Comparing the two equations:
1) $I_{A1}+a{I}_{A2}=1$
2) $I_{A1}+I_{A2}=1$

Subtracting equation (2) from equation (1) gives:
$(a-1)I_{A2}=0$

Since $a\ne 1$, we have:
$I_{A2}=0$
Substituting $I_{A2}=0$ back into equation (2) gives:
$I_{A1}=1\angle 0°$

Now, we can compute the current $I_B$:
$I_B=I_0+a^2{I}_{A1}+a{I}_{A2}$
$I_B=0.1\angle 0°+(1\angle -120°)(1\angle 0°)+a\left(0\right)$
$I_B=1\angle -120°+0.1\angle 0°\text{p.u.}$