Question 1.62: Determine the voltage Vx in the circuit shown in Fig. 1, us......
Determine the voltage V_x in the circuit shown in Fig. 1, using node analysis.
The given circuit has four nodes. Let us choose one of the nodes as reference. Let voltages of the other nodes with respect to the reference be V_1 ,V_2 and V_3 as shown in Fig. 2. Let I_{s2} be the current delivered by the dependent voltage source. Now, the voltage V_x = V_1.
The general node basis matrix equation of a circuit with three nodes excluding the reference is given by 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. 2, the elements of conductance matrix and source current matrix can be formed as shown below :
\begin{array}{l|l|l} G _{11}=\frac{1}{2}+\frac{1}{4}+\frac{1}{2}=1.25 & G _{12}= G _{21}=-\frac{1}{2}=-0.5 & I _{11}=4+2=6 \\ G _{22}=\frac{1}{2}+\frac{1}{1}=1.5 & G _{13}= G _{31}=-\frac{1}{2}=-0.5 & I _{22}= I _{ s 2}-4 \\ G _{33}=\frac{1}{2}+\frac{1}{1}=1.5 & G _{23}= G _{32}=0 & I _{33}=- I _{ s 2}-2 \end{array}On substituting the above terms in equation (1), we get,
\left[\begin{array}{rrr}1.25 & -0.5 & -0.5 \\-0.5 & 1.5 & 0 \\-0.5 & 0 & 1.5 \end{array}\right]\left[\begin{array}{l}V_1 \\V_2 \\V_3\end{array}\right]=\left[\begin{array}{r}6 \\I _{ s 2}-4 \\- I _{ s 2}-2\end{array}\right] ….. (2)
With reference to Fig. 2, we can write,
V_x = V_1 ……(3)
V_2-V_3=5 V_x ……(4)
From equation (4), we get,
\begin{aligned}V _2 & =5 V _{ x }+ V _3 \\\therefore \quad V _2 & =5 V _1+ V _3\end{aligned} ….. (5)
Using equation (3)
On substituting for V_2 from equation (5) in equation (2), we get,
\left[\begin{array}{rrr} 1.25 & -0.5 & -0.5 \\ -0.5 & 1.5 & 0 \\-0.5 & 0 & 1.5 \end{array}\right]\left[\begin{array}{r} V _1 \\5 V _1+ V _3 \\ V _3 \end{array}\right]=\left[\begin{array}{r}6 \\ I _{ s 2}-4 \\ – I _{ s 2}-2 \end{array}\right] ….. (6)
From row-1 of equation (6), we get,
\begin{aligned} & 1.25 V_1-0.5\left(5 V_1+V_3\right)-0.5 V_3=6 \\ & \therefore-1.25 V_1-V_3=6 \quad \Rightarrow \quad V_3=-1.25 V_1-6 \end{aligned} ………..(7)
From row-2 of equation (6), we get,
\begin{aligned}& -0.5 V_1+1.5\left(5 V_1+V_3\right)=I_{s 2}-4 \\& 7 V_1+1.5 V_3=I_{s 2}-4\end{aligned} ……….(8)
From row-3 of equation (6), we get,
-0.5 V _1+1.5 V _3=- I _{ s 2}-2 …….(9)
On adding equations (8) and (9), we get,
7 V _1+1.5 V _3-0.5 V _1+1.5 V _3= I _{ s 2}-4- I _{ s 2}-26.5 V _1+3 V _3=-6
6.5 V _1+3\left(-1.25 V _1-6\right)=-6
Using equation (7)
\therefore 2.75 V_1-18=-6 \Rightarrow V_1=\frac{-6+18}{2.75}=4.3636 V
\text { Since, } V_x=V_1, \quad V_x=4.3636 V
