Question 1.55: Determine the node voltages and the currents across all the......
Determine the node voltages and the currents across all the resistors of the circuit shown in Fig. 1, using node method.
Solution of node voltages
The given circuit has four nodes.In this one of the nodes is chosen as the reference node, which is indicated by 0. The voltage of the reference node is zero volt. Let us choose three other nodes and assign node voltages V_1, V_2 and V_3 as shown in Fig. 2. Let the
current delivered by 20 V source be I_s . With reference to Fig. 2, we get,
V_3=20 V
From the above equation we can say that the node voltage V_3 is a known quantity. Now, the number of unknowns in the circuit are three and they are V_1, V_2 and I_s . Therefore, we can write three node equations
using KCL (corresponding to three nodes) and a unique solution for unknowns can be obtained by solving the three equations.
The node basis matrix equation for the circuit shown in Fig. 2 is obtained by inspection as shown below:
\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)
\begin{array}{l|l|}G _{11}=\frac{1}{4}+\frac{1}{10}=0.35 & G _{12}= G _{21}=-\frac{1}{10}=-0.1 \\G _{22}=\frac{1}{10}+\frac{1}{2}+\frac{1}{1}=1.6 & G _{13}= G _{31}=0 \\G _{33}=\frac{1}{1}+\frac{1}{10}=1.1 & G _{23}= G _{32}=-\frac{1}{1}=-1\end{array} V _3=10 VOn substituting the above terms in equation (1), we get,
\left[\begin{array}{ccc}0.35 & -0.1 & 0 \\-0.1 & 1.6 & -1 \\0 & -1 & 1.1\end{array}\right]\left[\begin{array}{c}V _1 \\V _2 \\20\end{array}\right]=\left[\begin{array}{c}25 \\20 \\I _{ s }\end{array}\right] …..(2)
The node equations of the circuit are obtained by multiplying the matrices on the left-hand side of equation (2) and equating to the terms on the right-hand side.
From row-1, we get,
0.35 V_1-0.1 V_2=25 ….(3)
From row-2, we get,
-0.1 V _1+1.6 V _2-20=20 \quad \Rightarrow \quad-0.1 V _1+1.6 V _2=40 …..(4)
On multiplying equation (3) by 16, we get,
5.6 V_1-1.6 V_2=400 …..(5)
On adding equations (4) and (5), we get,
-0.1 V _1+1.6 V _2+5.6 V _1-1.6 V _2=40+400 \Rightarrow 5.5 V _1=440 \therefore V_1=\frac{440}{5.5}=80 VFrom equation (2), we get,
From equation (2), we get V_2=\frac{40+0.1 V_1}{1.6}=\frac{40+0.1 \times 80}{1.6}=30 V
V_1=80 V ; \quad V_2=30 V \text { and } \quad V_3=20 V
To solve branch voltages and currents
The given circuit has five resistance branches. Let us denote the resistance branch voltages as V_{ a }, V_{ b }, V _{ c }\text { and } V _{ e } and resistance branch currents as I_a, I_b, I_c, I_d \text { and } I_e , as shown in Fig. 3. The signs of branch voltages and currents are chosen such that they are all positive. The branch voltages and currents are solved as shown below
\begin{aligned}V_a=V_1=80 V \\ V_b=V_1-V_2=80-30=50 V \\& V_c=V_2=30 V \\& V_d=V_2-V_3=30-20=10 V \\& V_e=V_3=20 V\end{aligned}\\ \begin{aligned}& I _{ a }=\frac{V_{ a }}{4}=\frac{80}{4}=20 A \\& I _{ b }=\frac{ V _{ b }}{10}=\frac{50}{10}=5 A \\& I _{ c }=\frac{ V _{ c }}{2}=\frac{30}{2}=15 A \\& I _{ d }=\frac{ V _{ d }}{1}=\frac{10}{1}=10 A \\& I _e=\frac{ V _{ e }}{10}=\frac{20}{10}=2 A\end{aligned}
