Question 5.6: Steady-State AC Node-Voltage Analysis Use the node-voltage t...

The transformed network is shown in Figure 5.16(b). We obtain two equations by applying KCL at node 1 and at node 2. This yield

\frac{\mathrm{V}_1}{10}+\frac{\mathrm{V}_1-\mathrm{V}_2}{-j5}=2\angle -90^\circ  

\frac{\mathrm{V}_2}{j10}+\frac{\mathrm{V}_2-\mathrm{V}_1}{-j5}=1.5\angle 0^\circ  

These equations can be put into the standard form

\left(0.1+j0.2\right)\mathrm{V}_1-j0.2\mathrm{V}_2=-j2

-j0.2\mathrm{V}_1+j0.1\mathrm{V}_2=1.5

Now, we solve for V₁ yielding

\mathrm{V}_1=16.1\angle 29.7^\circ

Then, we convert the phasor to a time function and obtain

v_1(t) = 16.1 \cos{(100t + 29.7^\circ  )}