Chapter 10, Problem 30P

Use mesh analysis to find v_o in the circuit of Fig. 10.78. Let v_s1 = 120 cos (100t + 90°) V, v_s2 = 80 cos 100/ V.

Figure 10.78

To determine

Find the voltage $v_o$ in the circuit of Figure 10.78 using mesh analysis and MATLAB.

Answer to Problem 30P

The value of voltage $v_o\left(t\right)$ in the given circuit is $56.26\cos \left(100t+33.93°\right)\text{V}$.

Explanation of Solution

Given data:

Refer Figure 10.78 in the textbook for mesh analysis.

Formula used:

Write the expression to calculate impedance of the inductor.

$Z_L=j\omega L$ (1)

Here,

$\omega$ is the angular frequency, and

$L$ is the value of inductor.

Write the expression to calculate impedance of the capacitor.

$Z_C=\frac{1}{j\omega C}$ (2)

Here,

$C$ is the value of capacitor.

Write the general representation of sinusoidal function.

$V=V_m\cos \left(\omega t+\varphi \right)$ (3)

Here,

$\varphi$ is the phase angle, and

$V_m$ is the magnitude of source voltage.

Write the general expression to phasor transform of sinusoidal function from time domain to frequency domain.

$v_s=P\left\{V_m\cos \left(\omega t+\varphi \right)\right\}=V_m{e}^{i\varphi }$

Here,

$V_m\cos \left(\omega t+\varphi \right)$ is the time domain representation of source voltage, and

$V_m{e}^{i\varphi }$ is the frequency domain representation of source voltage.

Write the polar form representation of frequency domain.

$v_s=V_m\angle \varphi$ (4)

Calculation:

Comparing given source voltage $\left(v_{s1}\right)$ with equation (3), the magnitude, angular frequency, and phase angle of source voltage are $120\text{V}$, $100\frac{\text{rad}}{\text{s}}$ and $90°$ respectively.

Substitute $120\text{V}$ for $V_m$ and $90°$ for $\varphi$ in equation (4).

$v_{s1}=120\angle 90°\text{V}$

Comparing given source voltage $\left(v_{s2}\right)$ with equation (3), the magnitude, angular frequency, and phase angle of source voltage are $80\text{V}$, $100\frac{\text{rad}}{\text{s}}$ and $0°$ respectively.

Substitute $80\text{V}$ for $V_m$ and $0°$ for $\varphi$ in equation (4).

$v_{s2}=80\angle 0°\text{V}$

Substitute $100\frac{\text{rad}}{\text{s}}$ for $\omega$ and $200\text{mH}$ for $L$ in equation (1) to find $Z_{L\left(200\text{mH}\right)}$.

$\begin{array}{c}{Z}_{L\left(200\text{mH}\right)}=j\left(100\frac{\text{rad}}{\text{s}}\right)\left(200\text{mH}\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot \text{s}\\ \because \text{1}\text{mH}=1\times {10}^{-3}\text{H}\end{array}\right\}\\ =j\left(100\frac{\text{rad}}{\text{s}}\right)\left(200\times {10}^{-3}\text{H}\right)\\ =j20\Omega \end{array}$

Substitute $100\frac{\text{rad}}{\text{s}}$ for $\omega$ and $300\text{mH}$ for $L$ in equation (1) to find $Z_{L\left(300\text{mH}\right)}$.

$\begin{array}{c}{Z}_{L\left(300\text{mH}\right)}=j\left(100\frac{\text{rad}}{\text{s}}\right)\left(300\text{mH}\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot \text{s}\\ \because \text{1}\text{mH}=1\times {10}^{-3}\text{H}\end{array}\right\}\\ =j\left(100\frac{\text{rad}}{\text{s}}\right)\left(300\times {10}^{-3}\text{H}\right)\\ =j30\Omega \end{array}$

Substitute $100\frac{\text{rad}}{\text{s}}$ for $\omega$ and $400\text{mH}$ for $L$ in equation (1) to find $Z_{L\left(400\text{mH}\right)}$.

$\begin{array}{c}{Z}_{L\left(400\text{mH}\right)}=j\left(100\frac{\text{rad}}{\text{s}}\right)\left(400\text{mH}\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot \text{s}\\ \because \text{1}\text{mH}=1\times {10}^{-3}\text{H}\end{array}\right\}\\ =j\left(100\frac{\text{rad}}{\text{s}}\right)\left(400\times {10}^{-3}\text{H}\right)\\ =j40\Omega \end{array}$

Substitute $100\frac{\text{rad}}{\text{s}}$ for $\omega$ and $50\mu \text{F}$ for $C$ in equation (2) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(100\frac{\text{rad}}{\text{s}}\right)\left(50\mu \text{F}\right)}\left\{\begin{array}{l}\because 1\text{F}=1\frac{\text{s}}{\Omega }\\ \because 50\mu \text{F}=50\times {10}^{-6}\text{F}\end{array}\right\}\\ =\frac{1}{j\left(100\frac{\text{rad}}{\text{s}}\right)\left(50\times {10}^{-6}\frac{\text{s}}{\Omega }\right)}\\ =-j200\Omega \end{array}$

The frequency domain representation of given figure with the representation of node voltage is shown in Figure 1.

Apply Kirchhoff’s voltage law in the loop with current $I_1$ in Figure 1.

$\begin{array}{l}20I_1+j30\left(I_1-I_2\right)-120\angle 90°=0\\ 20I_1+j30I_1-j30I_2=120\angle 90°\\ \left(20+j30\right)I_1-j30I_2=120\angle 90°\end{array}$

$\left(2+j3\right)I_1-j3I_2=j12$ (5)

Apply Kirchhoff’s voltage law in the loop with current $I_2$ in Figure 1.

$\begin{array}{l}j40I_2-j200\left(I_2-I_3\right)+j30\left(I_2-I_1\right)=0\\ j40I_2-j200I_2+j200I_3+j30I_2-j30I_1=0\\ -j30I_1+\left(j40-j200+j30\right)I_2+j200I_3=0\\ -j30I_1-j130I_2+j200I_3=0\end{array}$

$-3I_1-13I_2+20I_3=0$ (6)

Apply Kirchhoff’s voltage law in the loop with current $I_3$ in Figure 1.

$\begin{array}{l}\left(10+j20\right)I_3+80-j200\left(I_3-I_2\right)=0\\ \left(10+j20\right)I_3+80-j200I_3+j200I_2=0\\ j200I_2+\left(10+j20-j200\right)I_3=-80\\ j200I_2+\left(10-j180\right)I_3=-80\end{array}$

$j20I_2+\left(1-j18\right)I_3=-8$ (7)

MATLAB Code:

Solve the linear equations (5), (6) and (7) using MATLAB to find the mesh currents.

syms i1 i2 i3

eq1 = (2 + 3*1i)*i1 -3*1i*i2 +0*i3 == 12*1i;

eq2 = -3*i1 -13*i2 +20*i3 == 0;

eq3 = 0*i1 +20*1i*i2 +(1-18*1i)*i3 == -8;

sol = solve([eq1, eq2, eq3], [i1, i2, i3]);

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

i1real=real(val1);

i1imag=imag(val1);

i2real=real(val2);

i2imag=imag(val2);

i3real=real(val3);

i3imag=imag(val3);

i1=sprintf('%.3f + %.3fi A', i1real, i1imag)

i2=sprintf('%.3f + %.3fi A', i2real, i2imag)

i3=sprintf('%.3f + %.3fi A', i3real, i3imag)

The command window output:

i1 = '2.0557 + 3.5651i A'

i2 = '0.4324 + 2.1946i A'

i3 = '0.5894 + 1.9612i A'

From Figure 1, write the expression for $v_o$.

$v_o=-j200\left(I_2-I_3\right)$

Substitute $0.4324+j2.1946\text{A}$ for $I_2$ and $0.5894+j1.9612\text{A}$ for $I_3$.

$\begin{array}{c}{v}_o=-j200\left(\left(0.4324+j2.1946\right)-\left(0.5894+j1.9612\right)\right)\\ =-j200\left(-0.157+j0.2334\right)\\ =46.68+j31.4\text{V}\\ =56.26\angle 33.93°\text{V}\end{array}$

Represent the voltage in time domain.

$v_o\left(t\right)=56.26\cos \left(100t+33.93°\right)\text{V}$

Conclusion:

Therefore, the value of voltage $v_o\left(t\right)$ in the given circuit is $56.26\cos \left(100t+33.93°\right)\text{V}$.