Question 1.43: Determine the voltages across various elements of the circui......
Determine the voltages across various elements of the circuit shown in Fig. 1, using the node method.
The graph of the given circuit is shown in Fig. 2. It has seven
branches and four nodes.
Let us choose one of the node as reference as shown in Fig. 2.
Let the voltages of other three nodes be V_1,V_2 and V_3.
The reference node is denoted by 0 to indicate that its voltage is zero volt. The circuit with chosen node voltages is shown in Fig. 3.
Method I : Formation of node basis matrix equation by applying KCL
In this method, the node equations are formed using Kirchhoff’s Current Law. The node equation for a node is formed by equating the sum of currents leaving that node to the sum of currents entering that node.
While writing the node equation for a node it is assumed that all the resistances connected to that node will draw current from that node. Hence, the current in the resistances will always leave the node.
With reference to Fig. 4, the node equation for node-1 is formed as shown below:
Currents leaving node-1 : \frac{V_1}{1 / 2}, \frac{V_1-V_2}{1 / 2}, 2 A
Current entering node-1 : Nil
\therefore \frac{V_1}{1 / 2}+\frac{V_1-V_2}{1 / 2}+2=0 \\ 2 V _1+2 V _1-2 V _2+2=0 \\ 4 V_1-2 V_2=-2 ….. (1)
With reference to Fig. 5, the node equation for node-2 is formed as shown below:
Currents leaving node-2 : \frac{V_2-V_1}{1 / 2}, \frac{V_2-V_3}{1 / 4}, \frac{V_2}{1 / 3}
Current entering node-2 : 9A
\therefore \frac{ V _2- V _1}{1 / 2}+\frac{ V _2- V _3}{1 / 4}+\frac{ V _2}{1 / 3}=92 V_2-2 V_1+4 V_2-4 V_3+3 V_2=9
-2 V_1+9 V_2-4 V_3=9 ….. (2)
With reference to Fig. 6, the node equation for node-3 is formed as shown below:
Currents leaving node-3 : \frac{V_3-V_2}{1 / 4}, \frac{V_3}{1 / 4}
Currents leaving node-3 : 2 A
\therefore \frac{ V _3- V _2}{1 / 4}+\frac{ V _3}{1 / 4}=2 \\ 4 V _3-4 V _2+4 V _3=2 \\ -4 V _2+8 V _3=2 ….. (3)
Equations (1), (2) and (3) are node equations of the circuit shown in Fig. 3. The node equations are summarised below for convenience.
4 V_1-2 V_2=-2 \\-2 V_1+9 V_2-4 V_3=9 \\ -4 V_2+8 V_3=2The node equations can be arranged in a matrix form as shown below and then solved by Cramer’s rule.
\left[\begin{array}{rrr}4 & -2 & 0 \\-2 & 9 & -4 \\0 & -4 & 8\end{array}\right]\left[\begin{array}{l}V_1 \\V_2 \\V_3\end{array}\right]=\left[\begin{array}{r}-2 \\ 9 \\2 \end{array}\right] ….. (4)
Method II : Formation of node basis matrix equation by inspection
In this method, the node basis matrix equation is formed by inspection using the circuit shown in Fig. 3. The general node basis matrix equation for a circuit with three nodes excluding the reference is shown in equation (5).
\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] ….. (5)
The elements of conductance matrix and source current matrix are formed as shown below:
\begin{array}{l|l|l}G _{11}=2+2=4 & G _{12}= G _{21}=-2 & I _{11}=-2 \\G _{22}=2+3+4=9 & G _{13}= G _{31}=0 & I _{22}=9 \\G _{33}=4+4=8 & G _{23}= G _{32}=-4 & I _{33}=2\end{array}On substituting the above terms in equation (5), we get,
\left[\begin{array}{rrr} 4 & -2 & 0 \\ -2 & 9 & -4 \\ 0 & -4 & 8 \end{array}\right]\left[\begin{array}{l} V_1 \\ V_2 \\ V_3 \end{array}\right]=\left[\begin{array}{r} -2 \\ 9 \\ 2 \end{array}\right] ….. (6)
Solution of node voltages
It is observed that the node basis matrix equations obtained in methods-I and II are the same. In equation (6) the unknowns are V_1, V_2 \text { and } V_3 . In order to solve 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:
\Delta^{\prime}=\left|\begin{array}{rrr}4 & -2 & 0 \\-2 & 9 &-4 \\0 & -4 & 8\end{array}\right| ; \quad \Delta_1^{\prime}=\left|\begin{array}{rrr}-2 & -2 & 0 \\9 & 9 & -4\\2 & -4 & 8\end{array}\right| ; \quad \Delta_2^{\prime}=\left|\begin{array}{rrr} 4 & -2 & 0 \\-2 & 9 & -4 \\0 & 2 & 8 \end{array}\right| ; \quad \Delta_3^{\prime}=\left|\begin{array}{rrr} 4 & -2 & -2 \\ -2 & 9 & 9 \\0 & -4 & 2 \end{array}\right|
The determinants are evaluated by expanding along first row and the node voltages are solved by Cramer’s rule.
\Delta^{\prime}=\left|\begin{array}{rrr} 4 & -2 & 0 \\ -2 & 9 & -4 \\ 0 & -4 & 8 \end{array}\right|=4 \times\left[9 \times 8-(-4)^2\right]-(-2) \times[-2 \times 8-0]+0 \\ =224-32=192
\Delta_1^{\prime}=\left|\begin{array}{rrr} -2 & -2 & 0 \\ 9 & 9 & -4 \\ 2 & -4 & 8 \end{array}\right| =-2 \times\left[9 \times 8-(-4)^2\right]-(-2) \times[9 \times 8-2 \times(-4)]+0 \\ =-112+160=48
\Delta_2^{\prime}=\left|\begin{array}{rrr} 4 & -2 & 0 \\ -2 & 9 & -4 \\0 & 2 & 8 \end{array}\right|=4 \times[9 \times 8-2 \times(-4)]-(-2) \times[-2 \times 8-0]+0 \\=320-32=288
\Delta_3^{\prime}=\left|\begin{array}{rrr}4 & -2 & -2 \\-2 & 9 & 9 \\0 & -4 & 2\end{array}\right|=4 \times[9 \times 2-(-4) \times 9]-(-2) \times[-2 \times 2-0]+(-2) \times[-2 \times(-4)-0] \\=216-8-16=192
V_1=\frac{\Delta_1^{\prime}}{\Delta^{\prime}}=\frac{48}{192}=0.25 V \\V_2=\frac{\Delta_2^{\prime}}{\Delta^{\prime}}=\frac{288}{192}=1.5 V \\ V_3=\frac{\Delta_3^{\prime}}{\Delta^{\prime}}=\frac{192}{192}=1 V
To solve branch voltages
The given circuit has seven branches. Let us denote the branch voltages as V_a, V_b, V_c, V_d, V_e, V_f \text { and } V_g as shown in Fig. 7. The sign of branch voltages are chosen such that they are all positive. The relation between the branch and node voltages are obtained using the circuit shown in Fig. 7 and the branch voltages are solved as shown below:
V_b=V_2=1.5 V
V _{ c }= V _2=1.5 V
V_d=V_3=1 V
V_e=V_2-V_1=1.5-0.25=1.25 V
V_f=V_2-V_3=1.5-1=0.5 V
V_g=V_3-V_1=1-0.25=0.75 V
Note : The branch voltages are voltages across various elements in the circuit.
