Question 1.60: Determine the node voltages of the circuit shown in Fig. 1....
Determine the node voltages of the circuit shown in Fig. 1.
The given circuit has three nodes excluding the reference. The general node basis matrix equation of a circuit with three nodes excluding the reference is shown in equation (1).
\left[\begin{array}{lll}G_{11} & G_{12} & G_{13} \\G_{21} & G_{22} & G_{23} \\G_{31} & G_{32} &G_{33}\end{array}\right]\left[\begin{array}{l}V_1 \\V_2 \\V_3\end{array}\right]=\left[\begin{array}{l}I_{11} \\I_{22} \\I_{33}\end{array}\right] …………(1)
With reference to Fig. 1, the elements of conductance matrix and source current matrix are obtained as shown below :
\begin{array}{l|l|l}G _{11}=\frac{1}{2}+\frac{1}{10}=0.6 & G _{12}= G _{21}=-\frac{1}{10}=-0.1 & I _{11}=4 \\G _{22}=\frac{1}{10}+\frac{1}{4}=0.35 & G _{13}= G _{31}=0 & I _{22}=-3 I _{ x } \\G _{33}=\frac{1}{4}+\frac{1}{5}=0.45 & G _{23}= G _{32}=-\frac{1}{4}=-0.25 & I _{33}=3 I _{ x }\end{array}On substituting the above terms in equation (1), we get,
\left[\begin{array}{rrr}0.6 & -0.1 & 0 \\-0.1 & 0.35 & -0.25 \\0 & -0.25 & 0.45\end{array}\right]\left[\begin{array}{l}V_1 \\V_2 \\V_3\end{array}\right]=\left[\begin{array}{r}4 \\-31_x \\31_x\end{array}\right] ……….(2)
Let us express the value of dependent current source in terms of node voltages. With reference to Fig. 1, we can write,
I _{ x }=\frac{ V _3}{5} ; \quad \therefore 3 I _{ x }=3 \times \frac{ V _3}{5}=0.6 V _3 …..(3)
On substituting for 3I_x from equation (3) in equation (2), we get,
\left[\begin{array}{rrr}0.6 & -0.1 & 0 \\-0.1 & 0.35 & -0.25 \\0 & -0.25 & 0.45\end{array}\right] \quad\left[\begin{array}{l}V_1 \\V_2 \\V_3\end{array}\right]=\left[\begin{array}{r}4 \\-0.6 V_3 \\0.6 V_3\end{array}\right] …..(4)
The terms -0.6 V _3 \text { and }+0.6 V _3 in the source matrix on the right-hand side of equation (4) can be transferred to the left-hand side with the opposite sign as shown in equation (5).
\left[\begin{array}{rrr}0.6 & -0.1 & 0 \\-0.1 & 0.35 & -0.25+0.6 \\0 & -0.25 & 0.45-0.6\end{array}\right]\left[\begin{array}{l}V_1 \\V_2 \\V_3\end{array}\right]=\left[\begin{array}{l}4 \\0 \\0\end{array}\right] …..(5)
\left[\begin{array}{rrr}0.6 & -0.1 & 0 \\-0.1 & 0.35 & 0.35 \\0 & -0.25 & -0.15\end{array}\right]\left[\begin{array}{l}V_1 \\V_2 \\V_3\end{array}\right]=\left[\begin{array}{l}4 \\0 \\0\end{array}\right] ….. (6)
To solve the node voltages V _1, V _2 \text { and } V _3 let us define four determinants \Delta^{\prime}, \Delta_1^{\prime}, \Delta_2^{\prime} \text { and } \Delta_3^{\prime} as shown below:
\begin{array}{ll}\Delta^{\prime}=\left|\begin{array}{rrr}0.6 & -0.1 & 0 \\-0.1 & 0.35 & 0.35 \\0 & -0.25 & -0.15\end{array}\right| & \Delta_1^{\prime}=\left|\begin{array}{rrr}4 & -0.1 & 0 \\0 & 0.35 & 0.35 \\0 & -0.25 & -0.15\end{array}\right| \\\Delta_2^{\prime}=\left|\begin{array}{rrr}0.6 & 4 & 0 \\-0.1 & 0 &0.35 \\0 & 0 & -0.15\end{array}\right| &\Delta_3^{\prime}=\left|\begin{array}{rrr}0.6 & -0.1 & 4 \\-0.1 & 0.35 & 0 \\0 & -0.25 & 0\end{array}\right|\end{array}The determinants are evaluated by expanding along the first row and node voltages are solved by Cramer’s rule.
\Delta^{\prime}=\left|\begin{array}{rrr}0.6 & -0.1 & 0 \\-0.1 & 0.35 & 0.35 \\0 & -0.25 & -0.15\end{array}\right|=0.6 \times[0.35 \times(-0.15)-(-0.25) \times 0.35]
-(-0.1) \times[-0.1 \times(-0.15)-0]+0
=0.021+0.0015=0.0225
\Delta_2^{\prime}=\left|\begin{array}{rrr}0.6 & 4 & 0 \\-0.1 & 0 & 0.35 \\0 & 0 & -0.15\end{array}\right|=0-4 \times[-0.1 \times(-0.15)-0]+0=-0.06
\Delta_3^{\prime}=\left|\begin{array}{crr}0.6 & -0.1 & 4 \\-0.1 & 0.35 & 0 \\0 & -0.25 & 0\end{array}\right|=0-0+4 \times[-0.1 \times(-0.25)-0]=0.1
Now, the node voltages are,
\begin{aligned}&V_1=\frac{\Delta_1^{\prime}}{\Delta^{\prime}}=\frac{0.14}{0.0225}=6.2222V\\&V_2=\frac{\Delta_2^{\prime}}{\Delta^{\prime}}=\frac{-0.06}{0.0225}=-2.6667 V\\&V_3=\frac{\Delta_3^{\prime}}{\Delta^{\prime}}=\frac{0.1}{0.0225}=4.4444 V\end{aligned}