Because it is a three-phase symmetrical resistive load, Ia, Ib and Ic in the figure are all symmetrical three-phase currents. From the amplitude (effective value), Ia = Ib = Ic = 2A (because the ammeter reading is 2A), and the current phase is the same as the phase voltage phase;

Connect a resistor between the line voltage Uab, and the current is Ix=2A, and the phase of the current Ix is the same as the line voltage Uab.

Therefore, the amplitude of current Ix is the same as that of current Ia, and the phase difference is 30 degrees.

Similarly, the amplitude of current Ix is the same as that of current Ib, and the phase difference is 150 degrees.

So,

I'a = 2*Ix*cos(30/2)

=2*2*cos 15

≈ 3.864 (A) (phase advance Ia 15)

I'b = 2*Ix*cos( 150/2)

=2*2*cos75

≈ 1.035 (A) (phase lead Ib 75)

That is, the effective value of current is: I 'a = 3.864ai 'b =1.035a; Ic = 2 A .