Network Theorems in the Frequency Domain: AC Circuit Analysis

Demonstrative Video

Superposition Theorem

Determine \(I_0\) using superposition theorem? image

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\[\begin{aligned} \mathbf{Z} &=\frac{-j 2(8+j 10)}{-2 j+8+j 10}=0.25-j 2.25 \\ \mathbf{I}_{o}^{\prime} &=\frac{j 20}{4-j 2+\mathbf{Z}}=\frac{j 20}{4.25-j 4.25} \\ &=-2.353+j 2.353 \end{aligned}\] \[\begin{array}{r} (8+j 8) \mathbf{I}_{1}-j 10 \mathbf{I}_{3}+j 2 \mathbf{I}_{2}=0 \\ (4-j 4) \mathbf{I}_{2}+j 2 \mathbf{I}_{1}+j 2 \mathbf{I}_{3}=0 \\ \mathbf{I}_{3}=5 \\ \mathbf{I}_{2}=2.647-j 1.176 \\ \mathbf{I}_{o}^{\prime \prime}=-\mathbf{I}_{2}=-2.647+j 1.176 \end{array}\]

\[\mathbf{I}_{o}=\mathbf{I}_{o}^{\prime}+\mathbf{I}_{o}^{\prime \prime}=-5+j 3.529=6.12 \angle 144.78^{\circ} \mathrm{A}\]

Source Transformation

Calculate \(V_x\) using source transformation? image

\[\begin{aligned} I_s & = \dfrac{20\angle 90^{\circ}}{5} \\ &= -j4~\mathrm{A} \end{aligned}\] image

\[\begin{aligned} Z_1 & = \dfrac{5(3+j4)}{8+j4} = 2.5+j1.25~\Omega\\ V_s & = I_sZ_1 = -j4(2.5+j1.25)\\ & = 5-j10~\mathrm{V} \end{aligned}\] image

By Voltage division: \[\mathbf{V}_{x}=\frac{10}{10+2.5+j 1.25+4-j 13}(5-j 10)=5.519 \angle-28^{\circ} \mathrm{V}\]

Thevenin’s Theorem

Problem-1

Obtain the Thevenin equivalent at terminals a-b of the circuit. image

\[\begin{aligned} &\mathbf{Z}_{1}=-j 6 \| 8=\frac{-j 6 \times 8}{8-j 6}=2.88-j 3.84 \Omega \\ &\mathbf{Z}_{2}=4 \| j 12=\frac{j 12 \times 4}{4+j 12}=3.6+j 1.2 \Omega \\ &\mathbf{Z}_{\mathrm{Th}}=\mathbf{Z}_{1}+\mathbf{Z}_{2}=6.48-j 2.64 \Omega \end{aligned}\] image

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\[\begin{aligned} & \mathbf{I}_{1}=\frac{120 \angle 75^{\circ}}{8-j 6} \mathrm{~A}, \qquad \mathbf{I}_{2}=\frac{120 \angle 75^{\circ}}{4+j 12} \mathrm{~A} \\ & \mathbf{V}_{\mathrm{Th}}-4 \mathbf{I}_{2}+(-j 6) \mathbf{I}_{1}=0 \\ \Rightarrow & \mathbf{V}_{\mathrm{Th}}=4 \mathbf{I}_{2}+j 6 \mathbf{I}_{1}\\ & = \dfrac{480\angle 75^{\circ}}{4+j12}+\dfrac{720\angle{(75^{\circ}+90^{\circ})}}{8-j6}\\ & = 37.95 \angle 220.31^{\circ} \end{aligned}\]

Problem-2

Find the Thevenin equivalent of the circuit. image

To find \(\mathbf{V}_{\mathrm{Th}}\), we apply \(\mathrm{KCL}\) at node 1 \[15=\mathbf{I}_{o}+0.5 \mathbf{I}_{o} \quad \Rightarrow \quad \mathbf{I}_{o}=10 \mathrm{~A}\] Applying KVL on RHS loop \[\begin{aligned} & -\mathbf{I}_{o}(2-j 4)+0.5 \mathbf{I}_{o}(4+j 3)+\mathbf{V}_{\mathrm{Th}}=0 \\ & \mathbf{V}_{\mathrm{Th}}=10(2-j 4)-5(4+j 3)=-j 55\\ & \mathbf{V}_{\mathrm{Th}}=55 \angle-90^{\circ} \mathrm{V} \end{aligned}\] image

\[\begin{aligned} &3=\mathbf{I}_{o}+0.5 \mathbf{I}_{o} \quad \Rightarrow \quad \mathbf{I}_{o}=2 \mathrm{~A} \\ &\mathbf{V}_{s}=\mathbf{I}_{o}(4+j 3+2-j 4)=2(6-j) \\ &\mathbf{Z}_{\mathrm{Th}}=\frac{\mathbf{V}_{s}}{\mathbf{I}_{s}}=\frac{2(6-j)}{3}=4-j 0.6667 \Omega \end{aligned}\] image

Norton’s Theorem

Obtain current \(I_o\) using Norton’s. image

\[Z_N = 5~\Omega\] image

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\[\begin{aligned} -j 40+(18+j 2) \mathbf{I}_{1}-(8-j 2) \mathbf{I}_{2}-(10+j 4) \mathbf{I}_{3} &=0 \\ (13-j 2) \mathbf{I}_{2}+(10+j 4) \mathbf{I}_{3}-(18+j 2) \mathbf{I}_{1} &=0 \\ \mathbf{I}_{3} &=\mathbf{I}_{2}+3 \end{aligned}\]

On solving: \[\begin{aligned} \mathbf{I}_{2} &=j 8 \qquad \mathbf{I}_{3} =3+j 8 \\ \mathbf{I}_{N} &=\mathbf{I}_{3}=(3+j 8) \mathrm{A} \\ \mathbf{I}_{o} &=\frac{5}{5+20+j 15} \mathbf{I}_{N} \\ &=\frac{3+j 8}{5+j 3}=1.465 \angle 38.48^{\circ} \mathrm{A} \end{aligned}\]