Question 1.56: Use nodal analysis to solve the circuit shown in Fig.1....

Let the node voltages be V_1 \ and \ V_2. Let us convert the25 V voltage source in series with 5 Ω resistance into an equivalent
current source as shown in Fig. 2.Similarly convert the 50 V voltage source
in series with 2 Ω resistance into an equivalent current source as shown
in Fig. 2.The node basis matrix equation of the given circuit is formed by inspection using the circuit as shown in Fig. 2.

\left[\begin{array}{ll}G_{11} & G_{12} \\G_{21} &G_{22}\end{array}\right]\left[\begin{array}{l}V_1 \\V_2\end{array}\right]=\left[\begin{array}{l}I_{11}\\I_{22}\end{array}\right]                     ………..(1)

With reference to Fig. 2, the elements of conductance matrix and source current matrix are obtained as shown below :

On substituting the above terms in equation (1), we get,

\left[\begin{array}{rr}0.8 & -0.1 \\-0.1 &0.85\end{array}\right]\left[\begin{array}{l}V_1\\V_2\end{array}\right]=\left[\begin{array}{r}5 \\-25\end{array}\right]

In order to solve the node voltages V_1 \ and \ V_2, let us define three determinants \Delta^{\prime}, \Delta_1^{\prime} \text { and } \Delta_2^{\prime} as shown below:

The determinants are evaluated by expanding along the first row and the node voltages are solved by Cramer’s rule.

\Delta^{\prime}=\left|\begin{array}{rr}0.8 & -0.1 \\-0.1 &0.85\end{array}\right|=0.8 \times 0.85-(-0.1)^2=0.67

\Delta_1^{\prime}=\left|\begin{array}{rc}5 & -0.1 \\-25 & 0.85\end{array}\right|=5 \times 0.85-(-25) \times(-0.1)=1.75

\Delta_2^{\prime}=\left|\begin{array}{rr}0.8 & 5 \\-0.1 & -25\end{array}\right|=0.8 \times(-25)-(-0.1) \times 5=-19.5

V_1=\frac{\Delta_1^{\prime}}{\Delta}=\frac{1.75}{0.67}=2.6119 \ V

V_2=\frac{\Delta_2^{\prime}}{\Delta}=\frac{-19.5}{0.67}=-29.1045 V