The correct answer is: the current in each phase differs, and the total power must be calculated by adding individual phase powers.

In a balanced three-phase system, the load is equally distributed across all three phases, resulting in equal phase currents and simplified power calculations. The total power can be easily calculated using the formula \(P = \sqrt{3} V_L I_L \cos(\phi)\), where \(V_L\) is the line voltage, \(I_L\) is the line current, and \(\cos(\phi)\) is the power factor.

However, when the load is unbalanced, the current in each phase is different. This is because the impedance in each phase is unequal, leading to an uneven distribution of current. Consequently, the simple formula for balanced systems is no longer applicable. To determine the total power consumed in an unbalanced three-phase system, one must calculate the power in each phase individually using the formula \(P_{phase} = V_{phase} I_{phase} \cos(\phi_{phase})\), and then sum these individual powers to obtain the total power:

\(P_{total} = P_a + P_b + P_c\)

Where \(P_a\), \(P_b\), and \(P_c\) represent the power consumed in phases a, b, and c respectively.

The other options are incorrect because:

• The line and phase voltages are not necessarily equal in an unbalanced system. They are equal only in a wye (star) connection with a balanced load.
• The line current does not become zero; it simply becomes unequal across the phases.
• The power factor does not become unity; it remains dependent on the phase impedances and the load.