Find a circuit problem and solve a circuit problem

The current source generates 50W of power. The analysis is as follows:

The current flowing through R and E is equal to IS=1A, and the resistor power pr=i R=10W, (the resistance must be the power consumed).

The power of the voltage source is PE=I E=40W, (the current flows from the positive pole of the voltage source, which also consumes power).

The current source voltage UI=E+UR=40+10=50V, and the power PI=IS UI=50W (the current flows out from the positive and extreme ends of the UI, which is the power generated).

Solution: Let i2 (phasor) = 10 0°a, then: u1 (phasor) = i2 (phasor) (j1) = 10 0° 1 -90° = 10 -90° = -j10 (v). That is: u1 = 10v.

Resistance current: i1 (phasor) = u1 (phasor) r = -j10 1 = -j10 (a).

i (phasor) = i1 (phasor) + i2 (phasor) = -j10 + 10 = 10 - j10 = 10 2 -45 ° (a), so: i = 10 2 a.

Inductance voltage: UL (phasor) = I (phasor) JXL = 10 2 -45° JXL = 10 2XL 45° = 10XL (1 + J).

us=|us (phasor)|=|ul (phasor) = u1 (phasor) |=|-j10+10xl+j10xl|=|10xl+j10（xl-1）|=10/√2。

So: (10xl) +10 (xl-1) = 10

So: ul (phasor) = 10 2 .

us (phasor) = 5 + j5 - j10 = 5 - j5 = 5 2 -45 ° = 10 2 -45 ° (v).

The power in the circuit comes from the current source. Voltage sources and resistors consume power. The current in the circuit is 1A, and the direction is counterclockwise.

Current source output power = voltage source power consumption + resistance consumption power = 30V * 1A + 1A 2 * 10R = 40W

Is this question looking for ux or ix?

KCL equation: UX 8 + ix = 10

KVL equation: ux+4ix-2ix=0

Solve the system of equations and get:

ux=-80/3 (v)

ix=40/3 (a)

kcl：i1+i2=i3；

kvl：i1r1+i3r3=u1，5i1+15i3=180；i2r2+i3r3=u2，10i2+15i3=80。

Solve the system of equations: i1=12, i2=-4, i3=8.

So: va=i3r3=8 15=120(v), vc=u1=180v, vd=u2=80v.

ucd=vc-vd=180-80=100（v）。

Solution: KCL:i1=i2+i3;

kvl：18i1+4i3=15；4i2+10=4i3。

Solve the system of equations: i1=, i2=-1, i3=.