Question

In an unbalanced three-phase system, phase current $I_a$ = 1$\angle (-{90}^0)$ pu, negative sequence current $I_{b2}$ = 4$\angle (-{150}^0)$ pu, zero sequence current $I_{c0}$ = 3$\angle {90}^0$pu. The magnitude of phase current $I_b$ in pu is

• A. 1.00
• B. 7.81
• C. 11.53
• D. 13.00

Answer 1

The correct option is C 11.53

Given, $I_a$ = 1$\angle (-{90}^0)$ pu

$I_{b2}$ = 4$\angle (-{150}^0)$ pu

$I_{c0}$ = 3 $\angle \left({90}^0\right)$pu

As zero sequence current in all the three phases of unbalanced 3-$\varphi$ system are equal, therefore,

$I_{a0}$ = $I_{b0}$ = $I_{c0}$ = 3$\angle {90}^0$pu



Also, $I_{b2}$ = 4$\angle -{150}^0$

= Negative phase sequence current of phase b,

Therefore, referring to the negative phase sequence phasor,

$I_{a2}$ = 4$\angle (-{150}^0-{120}^0)=4\angle -{270}^0$ pu

We know that, phase current vectors in matrix form are given by

$\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{I}_{a0}\\ I_{a1}\\ I_{a2}\end{array}\right]$

(where, operator a = 1$\angle {120}^0$)

Here, $I_a$ = $I_{a0}$ + $I_{a1}$ + $I_{a2}$

Given, $I_a$ = 1$\angle -{90}^0$ and

$I_{a0}=3\angle {90}^0$ pu and

$I_{a2}$ = 4 $\angle +{270}^0$ pu (as obtained above)

Therefore,

$I_{a1}$ = $I_a$ - ($I_{a0}$ + $I_{a2}$)

= (1 $\angle -{90}^0$) - (3 $\angle {90}^0+4\angle {270}^0$)

or, $I_{a1}$ = 8$\angle -{90}^0$pu

Therefore,

$I_b$ = $I_{a0}$ + a${}^2$$I_{a1}$ + a $I_{a2}$

= (3$\angle {90}^0$) + (1$\angle {120}^0$)${}^2$ (8 - $\angle {90}^0$) + (1 + $\angle {120}^0$) (4 $\angle -{270}^0$)

= 3$\angle {90}^0$ + 8$\angle {150}^0$ + 4$\angle -{150}^0$

$I_b$ = 11.53 $\angle$ 154.3${}^0$ pu

$\therefore$ Magnitude of phase current,

$I_b$ = 11.53 pu