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 1: Understand the Symmetrical Components
In a three-phase system, any set of unbalanced current phasors $I_A$, $I_B$, and $I_C$ can be decomposed into their symmetrical components: zero-sequence ($I_0$), positive-sequence ($I_1$), and negative-sequence ($I_2$) currents.
The relation between the phase currents and symmetrical components of Phase A is defined as:

$I_A=I_0+I_1+I_2$

$I_B=I_0+a^2{I}_1+a{I}_2$

$I_C=I_0+a{I}_1+a^2{I}_2$

where the operator $a$ is defined as:
$a=1\angle 120°$
and
$a^2=1\angle 240°=1\angle -120°$

For zero-sequence currents, the components are equal in magnitude and phase for all three phases:

$I_{A0}=I_{B0}=I_{C0}=I_0$

Given that the phase-B zero-sequence current is $I_{B0}=0.1\angle 0°$ p.u., we have:

$I_0=0.1\angle 0°\text{ p.u.}$

Step 2: Find positive- and negative-sequence currents
Using the given current for phase A ($I_A=1.1\angle 0°$ p.u.):

$1.1\angle 0°=0.1\angle 0°+I_1+I_2$

Subtracting $0.1\angle 0°$ from both sides gives:

$I_1+I_2=1.0\angle 0°$

Now, using the given current for phase C ($I_C=1\angle 120°+0.1$ p.u.):

$I_C=I_0+a{I}_1+a^2{I}_2$

Substituting $I_C=1\angle 120°+0.1\angle 0°$ and $I_0=0.1\angle 0°$:

$1\angle 120°+0.1\angle 0°=0.1\angle 0°+a{I}_1+a^2{I}_2$

Subtracting $0.1\angle 0°$ from both sides:

$a{I}_1+a^2{I}_2=1\angle 120°$

Since $1\angle 120°=a$, we can rewrite the equation as:

$a{I}_1+a^2{I}_2=a$

Dividing both sides by $a$:

$I_1+a{I}_2=1$

We now have a system of two equations:
1) $I_1+I_2=1$
2) $I_1+a{I}_2=1$
Subtracting equation (1) from equation (2) yields:

$(a-1)I_2=0$

Since $a-1\ne 0$, we find:

$I_2=0$

Substituting this back into equation (1):

$I_1=1\angle 0°$

Step 3: Calculate Phase B Current ($I_B$)
Using the symmetrical component equation for phase B:

$I_B=I_0+a^2{I}_1+a{I}_2$

Substituting the values we obtained:

$I_B=0.1\angle 0°+a^2(1\angle 0°)+a\left(0\right)$

$I_B=0.1\angle 0°+a^2$

Substituting $a^2=1\angle -120°$:

$I_B=1\angle -120°+0.1\angle 0°\text{ p.u.}$