Demonstrative Video
SECTION 01
Demonstrative Video
SECTION 01
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.
SECTION 01
Nodal Analysis
Find \(i_x\) using Nodal analysis?
\[\begin{aligned}
20 \cos 4 t & \Rightarrow 20 \angle 0^{\circ}\\
\omega &= 4~\text{rad/s} \\
1 \mathrm{H} & \Rightarrow j \omega L=j 4 \\
0.5 \mathrm{H} & \Rightarrow j \omega L=j 2 \\
0.1 \mathrm{~F} & \Rightarrow \frac{1}{j \omega C}=-j 2.5
\end{aligned}\]
\[\begin{aligned}
&\frac{20-\mathbf{V}_{1}}{10}=\frac{\mathbf{V}_{1}}{-j
2.5}+\frac{\mathbf{V}_{1}-\mathbf{V}_{2}}{j 4} \quad \Rightarrow (1+j
1.5) \mathbf{V}_{1}+j 2.5 \mathbf{V}_{2}=20 \\
&2 \mathbf{I}_{x}+\frac{\mathbf{V}_{1}-\mathbf{V}_{2}}{j
4}=\frac{\mathbf{V}_{2}}{j 2} \quad \Rightarrow \frac{2
\mathbf{V}_{1}}{-j 2.5}+\frac{\mathbf{V}_{1}-\mathbf{V}_{2}}{j
4}=\frac{\mathbf{V}_{2}}{j 2} \quad \Rightarrow 11 \mathbf{V}_{1}+15
\mathbf{V}_{2}=0 \\
&\mathbf{I}_{x}=\mathbf{V}_{1} /-j 2.5 \\
&{\left[\begin{array}{cc}
1+j 1.5 & j 2.5 \\
11 & 15
\end{array}\right]\left[\begin{array}{c}
\mathbf{V}_{1} \\
\mathbf{V}_{2}
\end{array}\right]=\left[\begin{array}{c}
20 \\
0
\end{array}\right]} \Rightarrow
\mathbf{V}_{1}=18.97 \angle 18.43^{\circ} \mathrm{V} \\
& \hspace{16em} \mathbf{V}_{2}=13.91 \angle 198.3^{\circ}
\mathrm{V} \\
&\mathbf{I}_{x}=\frac{\mathbf{V}_{1}}{-j 2.5}=\frac{18.97 \angle
18.43^{\circ}}{2.5 \angle-90^{\circ}}=7.59 \angle 108.4^{\circ}
\mathrm{A}\\
\Rightarrow & i_{x}=7.59 \cos \left(4 t+108.4^{\circ}\right)
\mathrm{A}
\end{aligned}\]
SECTION 01
Super-Nodal Analysis
Compute \(V_1\) and \(V_2\) ?
\[\begin{aligned}
&3=\frac{\mathbf{V}_{1}}{-j 3}+\frac{\mathbf{V}_{2}}{j
6}+\frac{\mathbf{V}_{2}}{12} \\
&\mathbf{V}_{1}=\mathbf{V}_{2}+10 \angle 45^{\circ}
\end{aligned}\]
\[\begin{aligned}
&\mathbf{V}_{1}=25.78 \angle-70.48^{\circ} \mathrm{V} \\
&\mathbf{V}_{2}=31.41 \angle-87.18^{\circ} \mathrm{V}
\end{aligned}\]
SECTION 01
Mesh Analysis
Determine the current \(I_0\) using
Mesh analysis ?
\[\begin{array}{r}
(8+j 10-j 2) \mathbf{I}_{1}-(-j 2) \mathbf{I}_{2}-j 10
\mathbf{I}_{3}=0 \\
(4-j 2-j 2) \mathbf{I}_{2}-(-j 2) \mathbf{I}_{1}-(-j 2)
\mathbf{I}_{3}+20 \angle 90^{\circ}=0
\end{array}\]
SECTION 01
Super-Mesh Analysis
Determine \(V_0\) using mesh
analysis?
\[\begin{aligned}
&-10+(8-j 2) \mathbf{I}_{1}-(-j 2) \mathbf{I}_{2}-8
\mathbf{I}_{3} =0 \\
&\mathbf{I}_{2} =-3 \\
&(8-j 4) \mathbf{I}_{3}-8 \mathbf{I}_{1}+(6+j 5)
\mathbf{I}_{4}-j 5 \mathbf{I}_{2} =0 \\
&\mathbf{I}_{4} =\mathbf{I}_{3}+4
\end{aligned}\]
\[\begin{aligned}
\mathbf{I}_{1} &=3.618 \angle 274.5^{\circ} \mathrm{A} \\
\mathbf{V}_{o} &=-j 2\left(\mathbf{I}_{1}-\mathbf{I}_{2}\right)
\\
&=9.756 \angle 222.32^{\circ} \mathrm{V}
\end{aligned}\]