Solved Problem of AC Circuit Using Theorems in Frequency Domain

Demonstrative Video

Ohm’s Law, KVL and KCL
Voltage Division and Current Division
Mesh Analysis
Nodal Analysis
Thevenin’s Theorem
Nortons Theorem
Source Transformation
Maximum Power Transfer Theorem

Fundamentals
Voltage Division
Current Division

Problem-1

If \(i_s = 5\cos(10t+40^{\circ})\) A then determine \(i_0\).

Solution-1

\[\begin{aligned} \mathrm{i}_{\mathrm{s}}=& 5 \cos \left(10 \mathrm{t}+40^{\circ}\right)\\ \rightarrow & \mathbf{I}_{\mathrm{s}}=5 \angle 40^{\circ} \\ & 0.1 \mathrm{~F} \longrightarrow \frac{1}{j \omega \mathrm{C}}\\ &=\frac{1}{j(10)(0.1)}=-j \\ 0.2 \mathrm{H} & \longrightarrow \quad j \omega \mathrm{L}\\ &=j(10)(0.2)=j 2 \end{aligned}\]

\[\begin{aligned} &\mathbf{Z}_{1}=4 \| \mathrm{j} 2=\frac{\mathrm{j} 8}{4+\mathrm{j} 2}=0.8+\mathrm{j} 1.6 \qquad \mathbf{Z}_{2}=3-\mathrm{j} \\ &\mathbf{I}_{\mathrm{o}}=\frac{\mathbf{Z}_{1}}{\mathbf{Z}_{1}+\mathbf{Z}_{2}} \mathbf{I}_{\mathrm{s}}=\frac{0.8+\mathrm{j} 1.6}{3.8+\mathrm{j} 0.6}\left(5 \angle 40^{\circ}\right) \\ &\mathbf{I}_{\mathrm{o}}=\frac{\left(1.789 \angle 63.43^{\circ}\right)\left(5 \angle 40^{\circ}\right)}{3.847 \angle 8.97^{\circ}}=2.325 \angle 94.46^{\circ} \\ &\mathrm{i}_{\mathrm{o}}(\mathrm{t})=\underline{\mathbf{2 . 3 2 5} \cos \left(\mathbf{1 0} \mathbf{t}+\mathbf{9 4 . 4 6}^{\circ}\right) \mathbf{A}} \end{aligned}\]

Problem-2

Calculate \(V_0(t)\) in the given figure?

Solution-2

\[\begin{aligned} & \omega=200 \\ & 50 \mu \mathrm{F} \longrightarrow \frac{1}{\mathrm{j} \omega \mathrm{C}}=\frac{1}{\mathrm{j}(200)\left(50 \times 10^{-6}\right)}\\ & \quad =-\mathrm{j} 100 \\ & 0.1~\mathrm{H} \longrightarrow j\omega L \\ & \quad = j(200)(0.1) = j20 \end{aligned}\]

\[\begin{gathered} 50 \|-\mathrm{j} 100=\frac{(50)(-\mathrm{j} 100)}{50-\mathrm{j} 100}=\frac{-\mathrm{j} 100}{1-\mathrm{j} 2}=40-\mathrm{j} 20 \\ \mathbf{V}_{\mathrm{o}}=\frac{\mathrm{j} 20}{\mathrm{j} 20+30+40-\mathrm{j} 20}\left(60 \angle 0^{\circ}\right)=\frac{\mathrm{j} 20}{70}\left(60 \angle 0^{\circ}\right)=17.14 \angle 90^{\circ} \\ \mathrm{v}_{\mathrm{o}}(\mathrm{t})=\underline{\mathbf{1 7 . 1 4} \sin \left(\mathbf{2 0 0} \mathbf{t}+\mathbf{9 0}^{\circ}\right) \mathbf{V}} \\ \text { or } \mathrm{v}_{\mathrm{o}}(\mathrm{t})=\underline{\mathbf{1 7 . 1 4} \cos (\mathbf{2 0 0 t}) \mathbf{V}} \end{gathered}\]

KCL and KVL
Mesh Analysis
Nodal Analysis

Problem-3

Use mesh analysis to find \(v_0\) if \(V_{s1} = 120\cos (100t+90)\) V and \(V_{s2} =80 \cos 100t\) V.

Solution-3

\[\begin{array}{ll} 300 \mathrm{mH} & \longrightarrow j \omega L=j 100 \times 300 \times 10^{-3}=j 30 \\ 200 \mathrm{mH} & \longrightarrow j \omega L=j 100 \times 200 \times 10^{-3}=j 20 \\ 400 \mathrm{mH} & \longrightarrow j \omega L=j 100 \times 400 \times 10^{-3}=j 40 \\ 50 \mu \mathrm{F} & \longrightarrow 1 / j \omega \mathrm{C}=\frac{1}{j 100 \times 50 \times 10^{-6}}=-j 200 \end{array}\]

\[\begin{aligned} \text{Mesh-1} \Rightarrow & -120\angle 90^{\circ}+(20+j30)I_1-j30I_2=0 \\ \text{Mesh-2} \Rightarrow & -j30+(j30+j40-j200)I_2+j200I_3=0\\ \text{Mesh-3} \Rightarrow & 80+j200I_2+(10-j180)I_3=0 \end{aligned}\]

\[\left[\begin{array}{ccc} 2+\mathrm{j} 3 & -\mathrm{j} 3 & 0 \\ -3 & -13 & 20 \\ 0 & \mathrm{j} 20 & 1-\mathrm{j} 18 \end{array}\right]\left[\begin{array}{c} \mathrm{I}_{1} \\ \mathrm{I}_{2} \\ \mathrm{I}_{3} \end{array}\right]=\left[\begin{array}{c} \mathrm{j} 12 \\ 0 \\ -8 \end{array}\right]\] \[\begin{aligned} \mathrm{V}_{\mathrm{o}}&=-\mathrm{j} 200\left(\mathrm{I}_{2}-\mathrm{I}_{3}\right)=-\mathrm{j} 200(-0.157+\mathrm{j} 0.2334)\\ &=46.68+\mathrm{j} 31.4=56.26 \angle 33.93^{\circ} \\ \mathrm{v}_{\mathrm{o}}&=\mathbf{5 6 . 2 6} \cos \left(\mathbf{1 0 0 t}+\mathbf{3 3 . 9 3}^{\circ}\right) \mathbf{V} \end{aligned}\]

Problem-4

Find \(V_x\) if \(V_1(t)=20\cos 1000t\) V and \(V_2(t)=20\sin 1000t\)

Solution-4

\[\begin{aligned} \omega & = 10^3~\text{rad/s} \\ X_L & = j\omega L = j10~\Omega \\ X_C & = \dfrac{-j}{\omega C} = -j10~\Omega\\ V_1 & = 20\angle 0^{\circ} \quad V_2 = 20\angle -90^{\circ} \end{aligned}\]

KCL at node \((x), \quad \frac{V_{x}-20 \angle 0^{\circ}}{j 10}+\frac{V_{x}-20 \angle-90^{\circ}}{-j 10}+\frac{V_{x}}{25}=0\) \[\begin{aligned} &\Rightarrow V_{x}\left[\frac{1}{j 10}-\frac{1}{j 10}+\frac{1}{25}\right]=\frac{20}{j 10} \angle 0^{\circ}-\frac{20}{j 10} \angle-90^{\circ} \\ &\Rightarrow \frac{V_{x}}{25}=(2-j 2) \Rightarrow V_{x}=50-j 50=70.71 \angle-45^{\circ}(\mathrm{V}) \\ &\therefore v_{x}(t)=70.71 \cos \left(1000 t-45^{\circ}\right)(\mathrm{V}) \end{aligned}\]

Problem-5

Determine \(I_{R2}(t)\) and \(I_s(t)\), if both sources operate at \(\omega = 2\) rad/sec and \(I_c=2\angle 28^{\circ}\).

Solution-5

\[\begin{aligned} V_c & = \dfrac{1}{j\omega C} I_c = 1\angle -62^{\circ}~\mathrm{V}\\ I_{R2} & = \dfrac{V_c}{2} = 0.5 \angle -62^{\circ}~\mathrm{A}\\ I_s & = I_{R2}+I_c = 1.5\angle -62^{\circ}~\mathrm{A}\\ \end{aligned}\]

\[\begin{aligned} i_s(t) &= 1.5 \cos (2t-62^{\circ})~\mathrm{A} \end{aligned}\]

Problem-6

Determine \(v_1\) ?

Solution-6

\[\begin{aligned} \omega & =2~\text{rad/sec},~L_1 = 0.5~\text{H}~L_2 = 1~\text{H}~C = 0.5~\text{F}\\ X_{L1} & = j1~\Omega, ~X_{L2} = j2~\Omega, ~X_{c} = -j1~\Omega, ~ \end{aligned}\] \[\begin{aligned} \text{KCL at Node (1)},~ & \frac{V_{1}}{j 1}-4 \angle 0^{\circ}-1 \angle-90^{\circ}+\frac{V_{1}-V_{2}}{-j}=0\\ \text{KCL at Node (2)}, ~& \frac{V_{2}-V_{1}}{-j 1}-2 V_{1}+\frac{V_{2}}{1+j 2}+1 \angle-90^{\circ}=0 \end{aligned}\] On solving: \[\begin{aligned} &V_{1}=1 \angle-36.87^{\circ} \\ &V_{1}=\cos \left(2 t-36.87^{\circ}\right)=\cos \left(2 t+143.13^{\circ}\right)(\mathrm{V}) \end{aligned}\]

Thevenin’s Theorem
Norton’s Theorem

Problem-7

Find Thevenin’s equivalence with respect to terminals A-B as shown in the below circuit?

Solution-7

\[\begin{aligned} I & = \dfrac{10\angle 0^{\circ}}{100+j10} \\ &= 0.09995\angle -5.7^{\circ}~\mathrm{A}\\ V_{Th}&=V_{AB} \\ &= \left(I\times j 10\right)-\left(5I\times (-j5)\right)\\ & = 3.48\angle 84.3^{\circ} ~\mathrm{V} \end{aligned}\] \[\begin{aligned} 10\angle 0^{\circ} & = (100+j10)I-j10I_N \cdots (1) \\ -(-j25I) & = -j10I+I_N(j10-j5) \cdots (2)\\ \text{Solving} ~& I_N = 0.6\angle 31^{\circ}~\mathrm{A} \\ Z_{Th} & = \dfrac{V_{Th}}{I_N} = 5.8 \angle 53.3^{\circ} ~\Omega \end{aligned}\]

Problem-8

Find \(V_0\) using Thevenin’s theorem ?

Solution-8

\[\begin{aligned} 2~\Omega & \Rightarrow \text{open} \\ i_0 & = I_1 \\ \text{KVL-1} & ~(4-j4)I_1+j4I_2 = -12 \\ \text{KVL-2} & ~-j2I_1+(-j6)I_2 = 0\\ \text{Solving} ~ I_2 & = 0.6\angle 53.13^{\circ} ~\mathrm{A} \\ V_{Th} & = I_2 \times (-j8) \\ & = 4.8\angle -36.87^{\circ} ~\mathrm{V} \end{aligned}\]

\[\begin{aligned} 2~\Omega & \Rightarrow \text{short and dep. source transformation} \\ \text{KVL} ~& (4-j4)I_1+j4I_2 = -12\\ \text{Solving} ~I_2 & = I_N = 1.34\angle 63.43^{\circ} \\ Z_{TH} & = \dfrac{V_{TH}}{I_N} = 3.58\angle -100.3^{\circ}~\Omega \\ v_0 & = \left(\dfrac{V_{TH}}{Z_{TH}+2}\right) \times 2 = 1.27 \angle 32^{\circ} \end{aligned}\]

Problem-9

Find voltage \(V_0\) across \(1~\Omega\) resistor ?

Solution-9

\[\begin{aligned} \mathbf{V}_{3}+12 \angle 0^{\circ} &=\mathbf{V}_{1} \\ \frac{\mathbf{V}_{2}-\mathbf{V}_{1}}{-j 1}+\frac{\mathbf{V}_{2}-\mathbf{V}_{3}}{1}-2 \mathbf{I}_{x}^{\prime \prime \prime} &=0 \\ \frac{\mathbf{V}_{1}-\mathbf{V}_{2}}{-j 1}+\frac{\mathbf{V}_{3}-\mathbf{V}_{2}}{1}-4 \angle 0^{\circ}+\frac{\mathbf{V}_{3}}{j 1}+\mathbf{I}_{x}^{\prime \prime \prime} &=0 \\ \mathbf{I}_{x}^{\prime \prime \prime} &=\frac{\mathbf{V}_{3}}{1} \end{aligned}\]

The matrix equation is \[\left[\begin{array}{cccc} -1 & 0 & 1 & 0 \\ -\mathrm{j} & (1+\mathrm{j}) & -1 & -2 \\ \mathrm{j} & -(1+\mathrm{j}) & 1-\mathrm{j} & 1 \\ 0 & 0 & -1 & 1 \end{array}\right]\left[\begin{array}{c} \mathbf{V}_{1} \\ \mathbf{V}_{2} \\ \mathbf{V}_{3} \\ \mathbf{I}_{x}^{\prime \prime \prime} \end{array}\right]=\left[\begin{array}{r} -12 \\ 0 \\ 4 \\ 0 \end{array}\right]\] \[\begin{aligned} \mathbf{I_x^{'''}}& = 2.8284\angle 135^{\circ}~\mathrm{A} \\ \mathbf{I_x^{'''}}& = 4\angle 0^{\circ} + \mathbf{I_{sc}}\\ \mathbf{I_{sc}}& = 6.3245\angle 161.57^{\circ}~\mathrm{A} \end{aligned}\]

\[\begin{aligned} Z_{TH} & = (1-j)~\Omega \\ V_0 & = I_{sc} \left[(1)(1-j)/(1+1-j)\right]~\mathrm{V}\\ & = 4 \angle 143.13^{\circ} ~\mathrm{V} \end{aligned}\]

Source Transformation

Problem-10

Calculate \(V_x\) using the method of source Transformation.

Solution-10

Transform the voltage source to a current source \[\mathbf{I}_{s}=\frac{20 \angle -90^{\circ}}{5}=4 \angle-90^{\circ}=-j 4 \mathrm{~A}\] The parallel combination of \(5-\Omega\) resistance and \((3+j 4)\) impedance gives \[\mathbf{Z}_{1}=\frac{5(3+j 4)}{8+j 4}=2.5+j 1.25 \Omega\]

Converting the current source to a voltage source \[\mathbf{V}_{s}=\mathbf{I}_{s} \mathbf{Z}_{1}=-j 4(2.5+j 1.25)=5-j 10 \mathrm{~V}\] By voltage division, \[\mathbf{V}_{x}=\frac{10}{10+2.5+j 1.25+4-j 13}(5-j 10)=5.519 \angle-28^{\circ} \mathrm{V}\]

Maximum Power Transfer Theorem

Problem-11

Choose a load resistance so that maximum power is transferred to it and calculate the actual power transferred?

Solution-11

Since \(v_{\pi}=0\), the dependent current source is an open circuit, and \(R_{T H}=1 \mathrm{k} \Omega\). This can be verified by connecting an independent \(1 \mathrm{~A}\) current source across the \(1 \mathrm{k} \Omega\) resistor; \(v_{\pi}\) will still be zero, so the dependent source remains inactive and hence contributes nothing to \(R_{T H}\).

In order to obtain maximum power delivered into the load, \(R_{L}\) should be set to \(R_{T H}=1 \mathrm{k} \Omega\). \[v_{\mathrm{oc}}=-0.03 v_{\pi}(1000)=-30 v_{\pi}\] where the voltage \(v_{\pi}\) may be found from simple voltage division: \[v_{\pi}=\left(2.5 \times 10^{-3} \sin 440 t\right)\left(\frac{3864}{300+3864}\right)\] so that our Thévenin equivalent is a voltage \(-69.6 \sin 440 t \mathrm{mV}\) in series with \(1 \mathrm{k} \Omega\). The maximum power is given by \[p_{\max }=\frac{v_{T H}^{2}}{4 R_{T H}}=1.211 \sin ^{2} 440 t \mu \mathrm{W}\]

Problem-12

A loudspeaker is connected across terminals A and B of the network. What should its impedance be to obtain maximum power dissipation in it?

Solution-12

Equivalent impedance with respect to the terminals \(A\) and \(B\) is \[Z_{\text {th }}=\frac{(3+j 4)(-j 5)}{3+j 4-j 5}=7.9 \angle-18.43^{\circ} \Omega=(7.5-j 2.5) \Omega\] For maximum power transfer, \(Z_{L}=Z_{\text {th }} *=(7.5+j 2.4) \Omega\)

Equivalent impedance with respect to the terminals \(A\) and \(B\) is \[\begin{aligned} Z_{\mathrm{th}} &=\left[\frac{(10+j 8) j 5}{10+j 8+j 5}+4+j 6\right]|| 10\\ &=\left(\frac{-40+j 50+40+j 52+j 60-78}{10+j 13}\right)|| 10 \\ &=6.14 \angle 30^{\circ} \Omega=(5.316+j 3.07) \Omega \end{aligned}\] For maximum power transfer, \(Z_{L}=Z_{\text {th }} *=6.14 \angle-30^{\circ} \Omega=(5.316-j 3.07) \Omega\)

Problem-13

In the network shown, find the value of \(Z_L\) to which the maximum power can be delivered. Hence, find the value of the maximum power?

Solution-13

With respect to terminals \(A\) and \(B\), the Thevenin voltage is \[V_{\mathrm{th}}=\frac{5 \angle 0^{\circ}}{3+\frac{j 3(3-j 3)}{3-j 3+j 3}} \times\left(\frac{j 3}{3+j 3-j 3}\right)=\frac{45 \angle 0^{\circ}}{18+j 9}=2.236 \angle-26.56^{\circ}(\mathrm{V})\]
Thevenin impedance, \(Z_{\mathrm{h}}=\frac{\left(3+\frac{3 \times j 3}{3+j 3}\right) \times(-j 3)}{3+\frac{3 \times j 3}{3+j 3}-j 3}=3 \angle-53.12^{\circ} \Omega=(1.8-j 2.4) \Omega\)
For maximum power transfer, \(Z_{L}=Z_{\text {th }}{ }^{*}=(1.8+j 2.4) \Omega\)
Current, \(I=\frac{2.236 \angle-26.56^{\circ}}{1.8 \times 2}=0.621 \angle-26.56^{\circ} \mathrm{A}\)
The value of the maximum power is, \(P_{\max }=\frac{\left(V_{\text {th }}\right)^{2}}{4 R}=\frac{(2.236)^{2}}{4 \times 1.8}=0.694 \mathrm{~W}\)