Nodal analysis is based on Kirchhoff's first (Current) law (KCL)
The value of the voltage V₂ is -6 V
The reason the above value is correct is as follows:
From the given parameters of the circuit are;
[tex]V_s[/tex] = 8 V, R₁ = 100 Ω, R₂ = 200Ω, R₃ = 300Ω, R₄ = 600 [tex]I_s[/tex]Ω, = 10 mA
The circuit simplified the as follows;
R₁ and R₂ which are in series are combined to give;
[tex]R_{series}[/tex] = R₁ + R₂
Therefore;
[tex]R_{series}[/tex] = 100 Ω + 200 Ω = 300 Ω
R₃ and R₄ are combined, given that they are parallel circuits to give;
[tex]R_{parallel} = \dfrac{1}{\dfrac{1}{R3} + \dfrac{1}{R4} }[/tex]
Therefore;
[tex]R_{parallel} = \dfrac{1}{\dfrac{1}{300 \, \Omega} +\dfrac{1}{600 \, \Omega} } = 200 \, \Omega[/tex]
The simplified circuit is as shown in the attached diagram
By Kirchhoff's current law, KCL, at node b, we have;
I₁ + I₂ + I₃ = 0
Where;
[tex]I_1 = \mathbf{\dfrac{V_b - V_s}{300}}[/tex], [tex]I_2 = \mathbf{\dfrac{V_b - 0}{200}}[/tex], and I₃ = [tex]I_s[/tex] = 10 mA
Plugging in the known values gives;
[tex]\mathbf{V_b - V_s}[/tex] = V₂
[tex]\dfrac{V_b - 8}{300} + \dfrac{V_b - 0}{200} +10 \times 10^{-3} = 0[/tex]
2 × ([tex]\mathbf{V_b - 8}[/tex]) + 3 × [tex]\mathbf{V_b}[/tex] = 600 × 10×10⁻³
5·[tex]V_b[/tex] - 16 = -6
[tex]V_b[/tex] = (-6 + 16)/5 = 10/5 = 2
Therefore, [tex]V_b[/tex] = 10 V
V₂ = [tex]\mathbf{V_b - V_s}[/tex]
∴ V₂ = 2 V - 8 V = -6 V
V₂ = -6 V
The value of the voltage V₂ = -6 V
Learn more about nodal analysis here:
https://brainly.com/question/16699518
