Determine Vo and ID for the network of Fig. 2.163.FIG. 2.163

Determine the value of voltage \(V_{o}\).

Apply Kirchhoff's Current Law at node \(V_{o}\).

\(\frac{V_{o}}{2 \times 10^{3}}+\frac{V_{o}-(-0.7)-10}{2 \times 10^{3}}+\frac{V_{o}-(-1.2)-10}{2 \times 10^{3}}=0\)

\(\frac{V_{o}}{2 \times 10^{3}}+\frac{V_{o}-9.3}{2 \times 10^{3}}+\frac{V_{o}-8.8}{2 \times 10^{3}}=0\)

\(\frac{V_{o}+\left(V_{o}-9.3\right)+\left(V_{o}-8.8\right)}{2 \times 10^{3}}=0\)

\(V_{o}+\left(V_{o}-9.3\right)+\left(V_{o}-8.8\right)=0\)

Further simplification as follows,

\(3 V_{o}-18.1=0\)

\(3 V_{o}=18.1\)

\(V_{o}=\frac{18.1}{3}\)

\(V_{o}=6.03 \mathrm{~V}\)

Thus, the value of voltage, \(V_{o}\) is \(6.03 \mathrm{~V}\).

Determine the value of current \(I_{D}\).

\(I_{D}=\frac{10-(0.7)-V_{o}}{2 \times 10^{3}}\)

Substitute \(6.03 \mathrm{~V}\) for \(V_{o}\) in the equation.

\(\begin{aligned} I_{D} &=\frac{10-(0.7)-(6.03)}{2 \times 10^{3}} \\ &=\frac{3.27}{2 \times 10^{3}} \\ &=1.635 \times 10^{-3} \\ &=1.635 \mathrm{~mA} \end{aligned}\)

Thus, the value of current, \(I_{D}\) is \(1.635 \mathrm{~mA}\).