1, solution: the secondary current is zero, so the secondary voltage: U2=|U2 (phasor) |=|jωMI 1 (phasor) | = 2π f× m× i1= 2× 3.14×

M=0. 1(H).

2. Solution: When S is closed, the three-phase currents are equal, indicating that the circuit is a three-phase symmetrical circuit. Let the load phase voltage be u and the line voltage be √3U. At this time, the circuit is connected in a triangle, and I 1=I2=I3=√3U/|Z|=5A.

Let UAB (phasor) = ∠ 3U ∠ 0, then: UBC (phasor) = ∠ 3U ∠- 120, UCA (phasor) = ∠ 3U ∠ 120. Let Z = | Z |∞φ.

Then: Ia (phasor) = (∠ 3u/| z |)∞-φ, Ib (phasor) =(∠3u/| z |)∞(- 120-φ), Ic (phasor) = (∠.

Ia (phasor) +Ib (phasor) +Ic (phasor) =0.

I 1 (phasor) =(3U/| Z |)∞(-φ-30)= 5∞(-φ-30), I2 (phasor) = (3u/| z | )∞ (-φ-30).

3U/|Z|=5，U/|Z|=5/3 .

When s is disconnected:

I' 1 (phasor) =Ia (phasor) = ∠ 3u ∠ 0/z = (∠ 3u/| z |) ∞-φ = (5/∠ 3 )∞-φ. I'3 (phasor) =-Ib (phasor) =-UBC (phasor)/z = (-√ 3u/| z |) ∞ (-120-φ) = (-5/√ 3 )∞ (-666)

So: I' 1=I'3=5/√3(A). -the first empty answer.

According to KCl: I' 2 (phasor) =Ia (phasor) -Ib (phasor) = (∠ 3u/| z |) ∞ (-φ)-(∠ 3u/| z | )∞ (-120-φ) =

So: I'2=5A. -the second empty answer.

The words "phase current B remains unchanged" did not appear in the title.

3. Solution: I = p/(√ 3 ucos φ) = 5500/(√ 3× 380× 0.88) = 9.496 (a).