Question 9.11: Use the node-voltage method to find the branch currents Ia ...

We can describe the circuit in terms of two node voltages because it contains three essential nodes. Four branches terminate at the essential node that stretches across the bottom of Fig. 9.34, so we use it as the reference node. The remaining two essential nodes are labeled 1 and 2, and the appropriate node voltages are designated \pmb{V}_{1} and \pmb{V}_{2} Figure 9.35 reflects the choice of reference node and the terminal labels.

Summing the currents away from node 1 yields

-10.6 +\frac{\pmb{V}_{1}}{10}+\frac{\pmb{V}_{1}-\pmb{V}_{2}}{1 + j 2}= 0.

Multiplying by 1+j2and collecting the coefficients of \pmb{V}_{1} and \pmb{V}_{2} generates the expression

\pmb{V}_{1}=\left(1.1 + j0.2\right) – \pmb{V}_{2} = 10.6 + j 21.2.

Summing the currents away from node 2 gives

\frac{\pmb{V}_{2}-\pmb{V}_{1}}{1 + j 2}+\frac{\pmb{V}_{2}}{-j 5}+\frac{\pmb{V}_{2}-20\pmb{I}_{x}}{5}= 0.

The controlling current \pmb{I}_{x} is

\pmb{I}_{x}=\frac{\pmb{V}_{1}- \pmb{V}_{2}}{1 + j 2}.

Substituting this expression for \pmb{I}_{x}  into the node 2 equation, multiplying by 1 + j 2, and collecting coefficients of \pmb{V}_{1} and \pmb{V}_{2} produces the equation

-5\pmb{V}_{1} +\left (4.8 + j0.6\right)\pmb{V}_{2} = 0.

The solutions for \pmb{V}_{1} and \pmb{V}_{2} are

\pmb{V}_{1}= 68.40 – j16.80 V,

\pmb{V}_{2} = 68 – j 26 V.

Hence the branch currents are

\pmb{I}_{a} =\frac{\pmb{V}_{1}}{10 }= 6.84 – j1.68 A,

\pmb{I}_{x}=\frac{\pmb{V}_{1}- \pmb{V}_{2}}{1 + j 2}= 3.76 + j1.68 A,

\pmb{I}_{b} =\frac{\pmb{V}_{2}-20\pmb{I}_{x}}{5}= -1.44 – j11.92 A,

\pmb{I}_{c} =\frac{\pmb{V}_{2}}{-j 5}= 5.2 + j13.6 A.

To check our work, we note that

\pmb{I}_{a} + \pmb{I}_{x} = 6.84 – j1.68 + 3.76 + j1.68= 10.6 A,

\pmb{I}_{x} = \pmb{I}_{b} +\pmb{I}_{c}=-1.44 – j11.92 + 5.2 + j13.6= 3.76 + j1.68 A.