Sinusoidal Steady-State Analysis

Demonstrative Video

Steady-State Analysis Methodology

Forced or steady-state response of circuits to sinusoidal inputs can be obtained by using phasor.
Covered Ohm’s and Kirchhoff’s laws for ac circuits.
Nodal analysis, mesh analysis, Thevenin’s theorem, Norton’s theorem, superposition, and source transformations in ac circuits.
Techniques already introduced for dc circuits.
Illustration with examples.

Nodal Analysis

Find \(i_x\) using Nodal analysis?

\[\begin{aligned} 20 \cos 4 t & \Rightarrow 20 \angle 0^{\circ}\\ \omega &= 4~\text{rad/s} \\ 1 \mathrm{H} & \Rightarrow j \omega L=j 4 \\ 0.5 \mathrm{H} & \Rightarrow j \omega L=j 2 \\ 0.1 \mathrm{~F} & \Rightarrow \frac{1}{j \omega C}=-j 2.5 \end{aligned}\]

\[\begin{aligned} &\frac{20-\mathbf{V}_{1}}{10}=\frac{\mathbf{V}_{1}}{-j 2.5}+\frac{\mathbf{V}_{1}-\mathbf{V}_{2}}{j 4} \quad \Rightarrow (1+j 1.5) \mathbf{V}_{1}+j 2.5 \mathbf{V}_{2}=20 \\ &2 \mathbf{I}_{x}+\frac{\mathbf{V}_{1}-\mathbf{V}_{2}}{j 4}=\frac{\mathbf{V}_{2}}{j 2} \quad \Rightarrow \frac{2 \mathbf{V}_{1}}{-j 2.5}+\frac{\mathbf{V}_{1}-\mathbf{V}_{2}}{j 4}=\frac{\mathbf{V}_{2}}{j 2} \quad \Rightarrow 11 \mathbf{V}_{1}+15 \mathbf{V}_{2}=0 \\ &\mathbf{I}_{x}=\mathbf{V}_{1} /-j 2.5 \\ &{\left[\begin{array}{cc} 1+j 1.5 & j 2.5 \\ 11 & 15 \end{array}\right]\left[\begin{array}{c} \mathbf{V}_{1} \\ \mathbf{V}_{2} \end{array}\right]=\left[\begin{array}{c} 20 \\ 0 \end{array}\right]} \Rightarrow \mathbf{V}_{1}=18.97 \angle 18.43^{\circ} \mathrm{V} \\ & \hspace{16em} \mathbf{V}_{2}=13.91 \angle 198.3^{\circ} \mathrm{V} \\ &\mathbf{I}_{x}=\frac{\mathbf{V}_{1}}{-j 2.5}=\frac{18.97 \angle 18.43^{\circ}}{2.5 \angle-90^{\circ}}=7.59 \angle 108.4^{\circ} \mathrm{A}\\ \Rightarrow & i_{x}=7.59 \cos \left(4 t+108.4^{\circ}\right) \mathrm{A} \end{aligned}\]

Super-Nodal Analysis

Compute \(V_1\) and \(V_2\) ?

\[\begin{aligned} &3=\frac{\mathbf{V}_{1}}{-j 3}+\frac{\mathbf{V}_{2}}{j 6}+\frac{\mathbf{V}_{2}}{12} \\ &\mathbf{V}_{1}=\mathbf{V}_{2}+10 \angle 45^{\circ} \end{aligned}\]

\[\begin{aligned} &\mathbf{V}_{1}=25.78 \angle-70.48^{\circ} \mathrm{V} \\ &\mathbf{V}_{2}=31.41 \angle-87.18^{\circ} \mathrm{V} \end{aligned}\]

Mesh Analysis

Determine the current \(I_0\) using Mesh analysis ?

\[\begin{array}{r} (8+j 10-j 2) \mathbf{I}_{1}-(-j 2) \mathbf{I}_{2}-j 10 \mathbf{I}_{3}=0 \\ (4-j 2-j 2) \mathbf{I}_{2}-(-j 2) \mathbf{I}_{1}-(-j 2) \mathbf{I}_{3}+20 \angle 90^{\circ}=0 \end{array}\] For mesh \(3, \mathbf{I}_{3}=5\). \[\begin{gathered} (8+j 8) \mathbf{I}_{1}+j 2 \mathbf{I}_{2}=j 50 \\ j 2 \mathbf{I}_{1}+(4-j 4) \mathbf{I}_{2}=-j 20-j 10 \\ {\left[\begin{array}{cc} 8+j 8 & j 2 \\ j 2 & 4-j 4 \end{array}\right]\left[\begin{array}{c} \mathbf{I}_{1} \\ \mathbf{I}_{2} \end{array}\right]=\left[\begin{array}{c} j 50 \\ -j 30 \end{array}\right]} \\ \mathbf{I}_{2}=6.12 \angle-35.22^{\circ} \mathrm{A} \\ \mathbf{I}_{o}=-\mathbf{I}_{2}=6.12 \angle 144.78^{\circ} \mathrm{A} \end{gathered}\]

Super-Mesh Analysis

Determine \(V_0\) using mesh analysis?

\[\begin{aligned} &-10+(8-j 2) \mathbf{I}_{1}-(-j 2) \mathbf{I}_{2}-8 \mathbf{I}_{3} =0 \\ &\mathbf{I}_{2} =-3 \\ &(8-j 4) \mathbf{I}_{3}-8 \mathbf{I}_{1}+(6+j 5) \mathbf{I}_{4}-j 5 \mathbf{I}_{2} =0 \\ &\mathbf{I}_{4} =\mathbf{I}_{3}+4 \end{aligned}\]

\[\begin{aligned} \mathbf{I}_{1} &=3.618 \angle 274.5^{\circ} \mathrm{A} \\ \mathbf{V}_{o} &=-j 2\left(\mathbf{I}_{1}-\mathbf{I}_{2}\right) \\ &=9.756 \angle 222.32^{\circ} \mathrm{V} \end{aligned}\]