Question 1.59: Find the voltages across various elements in the circuit sho......
Find the voltages across various elements in the circuit shown in Fig. 1, using node method.
The graph of the given circuit is shown in Fig. 2. It has six branches and three nodes. Hence, the circuit will have six voltages corresponding to
six branches. The branch voltages depend on the node voltages. In node analysis, the voltage of one of the nodes is chosen as the reference and it is equal to zero volt. In the circuit of Fig. 3, the reference node is denoted as 0. The voltages of the other two nodes are denoted as \bar{V}_1 \text { and } \bar{V}_2 .
The node basis matrix equation of the circuit shown in Fig. 3 is obtained by inspection as shown below:
\left[\begin{array}{ll}\bar{Y}_{11} & \bar{Y}_{12} \\\bar{Y}_{21} & \bar{Y}_{22} \end{array}\right]\left[\begin{array}{l}\bar{V}_1 \\ \bar{V}_2 \end{array}\right]=\left[\begin{array}{l}\bar{I}_{11} \\ \bar{I}_{22} \end{array}\right] …….(1)
\begin{array}{l|l}\bar{Y}_{11}=\frac{1}{3}+\frac{1}{j 3}+\frac{1}{-j 5}=0.333-j 0.133 &\overline{ I }_{11}=5 \angle 90^{\circ}=j 5 \\\bar{Y}_{22}=\frac{1}{j 3}+\frac{1}{-j 5}+\frac{1}{6}=0.167-j 0.133 & \overline{ I }_{22}=10\angle 0^{\circ}=10 \\\bar{Y}_{12}=\bar{Y}_{21}=-\left(\frac{1}{j 3}+\frac{1}{-j 5}\right)=j 0.133 &\end{array}On substituting the above terms in equation (1), we get,
\left[\begin{array}{rr}0.333-j 0.133 & j 0.133 \\j 0.133 & 0.167-j 0.133\end{array}\right] \quad\left[\begin{array}{l}\bar{V}_1 \\ \bar{V}_2\end{array}\right]=\left[\begin{array}{c}j 5 \\10\end{array}\right] …………(2)
To solve the node voltages by Cramer’s rule, let us define three determinants \Delta^{\prime}, \Delta_1^{\prime} \text { and } \Delta_2^{\prime} as shown below:
\Delta^{\prime}=\left|\begin{array}{rr}0.333- j 0.133 & j 0.133 \\j 0.133 & 0.167 j0.133\end{array}\right|;\Delta_1^{\prime}=\left|\begin{array}{rr}j 5 & j 0.133 \\10 & 0.167- j 0.133\end{array}\right| ; \Delta_2^{\prime}=\left|\begin{array}{rr}0.333- j 0.133 & j 5 \\j0.133 & 10\end{array}\right|Now the node voltages are given by,\bar{V}_1=\frac{\Delta_1^{\prime}}{\Delta} \text { and}\bar{V}_2=\frac{\Delta_2^{\prime}}{\Delta}
\begin{aligned}& \Delta^{\prime}=\left|\begin{array}{rr}0.333-j 0.133 & j 0.133 \\j 0.133 & 0.167-j 0.133\end{array}\right|=[(0.333-j 0.133)\times(0.167-j 0.133)]-[j 0.133]^2 \\&=0.0556-j 0.0665 \\&\Delta_1^{\prime}=\left|\begin{array}{rr}j 5 & j 0.133 \\10 & 0.167-j 0.133\end{array}\right|=[j 5 \times(0.167-j 0.133)]-[10 \times j 0.133] \\&=0.665-j 0.495\end{aligned}
\Delta_2^{\prime}=\left|\begin{array}{rr}0.333-j 0.133 & j 5 \\j 0.133 & 10\end{array}\right|=[(0.333-j 0.133) \times 10]-[j 0.133\times j 5] = 3.995 − j1.33
\therefore \overline{ V }_1=\frac{\Delta_1^{\prime}}{\Delta^{\prime}}=\frac{0.665- j 0.495}{0.0556- j 0.0665}=9.302+ j2.2227=9.564 \angle 13.4^{\circ} V
\bar{V}_2=\frac{\Delta_2^{\prime}}{\Delta^{\prime}}=\frac{3.995- j 1.33}{0.0556- j 0.0665}=41.3339+ j25.5163=48.575 \angle 31.7^{\circ} V
To find branch voltages
The branch voltages are denoted by \overline{ V }_{ a }, \overline{ V }_{ b }, \overline{ V }_{ c }, \overline{ V }_{ d }, \overline{ V }_{ e } \text { and } \overline{ V }_{ f } \text {, }as shown in Fig. 4. The polarites of branch voltages are chosen arbitrarily.
The branch voltages depend on the node voltages. The relation between
branch and node voltages are obtained with reference to Fig. 4 as shown
below :
