In the proposed scheme, the EMF of the first element is 2 V, the second is 4 V, the third is 6 V. The resistance

In the proposed scheme, the EMF of the first element is 2 V, the second is 4 V, the third is 6 V. The resistance of the first resistor is 4 ohms, the second is 6 ohms, and the third is 8 ohms. Find the amperage in all sections of the circuit. Disregard the source resistance.

Let’s use 1 Kirchhoff’s law at the node:
I1 – I2 + I3 = 0;
I2 = I1 + I3;
Let’s use 2 Kirchhoff’s law:
U1 + U2 = E1 – E2;
I1 * R1 + I2 * R2 = E1 – E2;
I1 * R1 + (I1 + I3) * R2 = E1 – E2;
I1 * R1 + I1 * R2 + I3 * R2 = E1 – E2;
In this way:
I1 = (E1 – E2 – I3 * R2) / (R1 + R2);
I1 = (2 – 4 – I3 * 6) / (4 + 6);
I1 = – 0.2 – 0.6 * I3;
Otherwise:
U3 + U2 = E3 – E2;
I3 * R3 + I2 * R2 = E3 – E2;
I3 * R3 + (I1 + I3) * R2 = E3 – E2;
I3 * R3 + (-0.2 – 0.6 * I3 + I3) * R2 = E3 – E2;
I3 * 8 + (- 0.2 + 0.4 * I3) * 6 = 6 – 4;
I3 * 8 – 1.2 + 2.4 * I3 = 2;
10.4 * I3 = 2 + 1.2 = 3.2;
I3 = 4/13 = 0.31 A;
We get:
I2 = (E3 – E2 – I3 * R3) / R2 = (6 – 4 – 4/13 * 8) / 6 = – 1/13 = – 0.08 A;
I1 = (E1 – E2 – I2 * R2) / R1 = (2 – 4 – (- 1/13) * 6) / 4 = – 5/13 = -0.38 A;

Answer: I2 = – 0.08 A; I1 = -0.38 A;