Question

In the circuit below, V_S = 8 V, R_1=100 ohm, R_2=200 ohm, R_3=300 ohm, R_4=600 ohm, and I_S=10 mA.

Answer 1

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;

$V_s$ = 8 V, R₁ = 100 Ω, R₂ = 200Ω, R₃ = 300Ω, R₄ = 600 $I_s$Ω,  = 10 mA

The circuit simplified the as follows;

R₁ and R₂ which are in series are combined to give;

$R_{series}$ = R₁ + R₂

Therefore;

$R_{series}$ = 100 Ω + 200 Ω = 300 Ω

R₃ and R₄ are combined, given that they are parallel circuits to give;

$R_{parallel} = \dfrac{1}{\dfrac{1}{R3} + \dfrac{1}{R4} }$

Therefore;

$R_{parallel} = \dfrac{1}{\dfrac{1}{300 \, \Omega} +\dfrac{1}{600 \, \Omega} } = 200 \, \Omega$

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;

$I_1 = \mathbf{\dfrac{V_b - V_s}{300}}$, $I_2 = \mathbf{\dfrac{V_b - 0}{200}}$, and I₃ = $I_s$ = 10 mA

Plugging in the known values gives;

$\mathbf{V_b - V_s}$ = V₂

$\dfrac{V_b - 8}{300} + \dfrac{V_b - 0}{200} +10 \times 10^{-3} = 0$

2 × ($\mathbf{V_b - 8}$) + 3 × $\mathbf{V_b}$ = 600 × 10×10⁻³

5·$V_b$ - 16 = -6

$V_b$  = (-6 + 16)/5 = 10/5 = 2

Therefore, $V_b$ = 10 V

V₂ = $\mathbf{V_b - V_s}$

∴ V₂ = 2 V - 8 V = -6 V

V₂ = -6 V

The value of the voltage V₂ = -6 V

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