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Answer:

The voltage at the middle source is [tex](2-4\mathbf{i})\ V[/tex]

Step-by-step explanation:

Voltage Sources in Series

When two or more voltage sources are connected in series, the total voltage is the sum of the individual voltages of each source.

The figure shown has three voltage sources of values:

[tex]2 + 6\mathbf{i}[/tex]

[tex]a + b\mathbf{i}[/tex]

[tex]2 - 5\mathbf{i}[/tex]

The sum of these voltages is:

[tex]V_t=4+a+(6+b-5)\mathbf{i}[/tex]

Operating:

[tex]V_t=4+a+(1+b)\mathbf{i}[/tex]

We know the total voltage is [tex]6-3\mathbf{i}[/tex], thus:

[tex]4+a+(1+b)\mathbf{i}=6-3\mathbf{i}[/tex]

Equating the real parts and the imaginary parts independently:

4+a=6

1+b=-3

Solving each equation:

a = 2

b = -4

The voltage at the middle source is [tex](2-4\mathbf{i})\ V[/tex]

The voltage at the middle source is;

E2 = (2 + 4i) V

From the image, we see the voltage at each point as;

E1 = (2 + 6i) V

E2 = (a + bi) V

E3 = (2 - 5i) V

The given voltages are in series and as such the total voltage will be;

E_t = E1 + E2 + E3

We are told that the total source voltage is (6 - 3i)V. Thus;

(6 - 3i) = (2 + 6i) + (a + bi) + (2 - 5i)

Rearranging gives;

(a + bi) = 6 - 3i - (2 + 6i) - (2 - 5i)

(a + bi) = 6 - 3i - 2 - 6i - 2 + 5i

(a + bi) = 2 - 4i

The middle source voltage is E2 = (a + bi) V. Thus;

E2 = 2 - 4i

Read more at;https://brainly.com/question/8151005

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