Chapter 10, Problem 9P

Use nodal analysis to find v_o in the circuit of Fig. 10.58.

Figure 10.58

To determine

Find the voltage $v_o$ in the circuit of Figure 10.58 using nodal analysis and MATLAB.

Answer to Problem 9P

The value of voltage $v_o\left(t\right)$ in the given circuit is $6.154\cos \left(1000t+70.26°\right)\text{V}$.

Explanation of Solution

Given data:

Refer Figure 10.58 in the textbook for nodal 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.

$V=V_m\sin \left(\omega t+\varphi \right)$ (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.

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

Here,

$V_m\sin \left(\omega t+\varphi \right)$ 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 $\left(10\cos 1000t\text{V}\right)$ with equation (3), the magnitude, angular frequency, and phase angle of source voltage are $10\text{V}$, $1000\frac{\text{rad}}{\text{s}}$ and $0°$ respectively.

Substitute $10\text{V}$ for $V_m$ and $0°$ for $\varphi$ in equation (4).

$V_s=10\angle 0°\text{V}$

Substitute $1000\frac{\text{rad}}{\text{s}}$ for $\omega$ and $10\text{mH}$ for $L$ in equation (1) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left(1000\frac{\text{rad}}{\text{s}}\right)\left(10\text{mH}\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot \text{s}\\ \text{1}\text{mH}=1\times {10}^{-3}\text{H}\end{array}\right\}\\ =j\left(1000\frac{\text{rad}}{\text{s}}\right)\left(10\times {10}^{-3}\text{H}\right)\\ =j10\Omega \end{array}$

Substitute $1000\frac{\text{rad}}{\text{s}}$ for $\omega$ and $50\mu \text{F}$ for $C$ in equation (2) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(1000\frac{\text{rad}}{\text{s}}\right)\left(50\mu \text{F}\right)}\left\{\begin{array}{l}\because 1\text{F}=1\frac{\text{s}}{\Omega }\\ 50\mu \text{F}=50\times {10}^{-6}\text{F}\end{array}\right\}\\ =\frac{1}{j\left(1000\frac{\text{rad}}{\text{s}}\right)\left(50\times {10}^{-6}\frac{\text{s}}{\Omega }\right)}\\ =-j20\Omega \end{array}$

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

Apply Kirchhoff’s current law at node $V_1$ in Figure 1.

$\begin{array}{l}\frac{V_1-\left(10\angle 0°\right)}{20}+\frac{V_1-0}{20}+\frac{V_1-V_2}{-j20}=0\\ \frac{V_1}{20}-\frac{\left(10\angle 0°\right)}{20}+\frac{V_1}{20}+\frac{V_1}{-j20}+\frac{V_2}{j20}=0\\ \left(\frac{1}{20}+\frac{1}{20}+\frac{1}{-j20}\right)V_1+\frac{V_2}{j20}=\frac{\left(10\angle 0°\right)}{20}\end{array}$

Simplify the equation as follows.

$\begin{array}{l}\left(0.05+0.05+j0.05\right)V_1-j0.05V_2=0.5\angle 0°\\ \left(0.1+j0.05\right)V_1-j0.05V_2=0.5\angle 0°\end{array}$

$\left(2+j\right)V_1-j{V}_2=10$ (5)

Apply Kirchhoff’s current law at node $V_2$ in Figure 1.

$\frac{V_2-V_1}{-j20}+\frac{V_2-0}{30+j10}+4i_o=0$ (6)

From Figure 1, write the expression for current $i_o$.

$i_o=\frac{V_1}{20\Omega }$ (7)

Substitute equation (7) in (6).

$\begin{array}{l}\frac{V_2-V_1}{-j20}+\frac{V_2-0}{30+j10}+4\left(\frac{V_1}{20}\right)=0\\ \frac{V_2}{-j20}+\frac{V_1}{j20}+\frac{V_2}{30+j10}+\frac{V_1}{5}=0\\ \left(\frac{1}{5}+\frac{1}{j20}\right)V_1+\left(\frac{1}{30+j10}+\frac{1}{-j20}\right)V_2=0\\ \left(0.2-j0.05\right)V_1+\left(0.03-j0.01+j0.05\right)V_2=0\end{array}$

Simplify the equation as follows.

$\left(0.2-j0.05\right)V_1+\left(0.03+j0.04\right)V_2=0$

$\left(4-j1\right)V_1+\left(0.6+j0.8\right)V_2=0$ (8)

MATLAB Code:

Solve the two linear equations (5) and (8) using MATLAB to find the node voltage.

syms v1 v2

eq1 = (2 + 1*1i)*v1 +(-1*1i)*v2 == 10;

eq2 = (4 +(-1*1i))*v1 +(0.6 + 0.8*1i)*v2 == 0;

sol = solve([eq1, eq2], [v1, v2]);

val1 = sol.v1;

val2 = sol.v2;

v1real=real(val1);

v1imag=imag(val1);

v2real=real(val2);

v2imag=imag(val2);

v1=sprintf('%.3f + %.3fi V', v1real, v1imag);

v2=sprintf('%.3f + %.3fi V', v2real, v2imag)

The command window output:

v2 = '0.149 + 6.485i V'

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

$v_o=\left(\frac{30}{30+j10}\right)V_2$

Substitute $0.149+j6.485\text{V}$ for $V_2$.

$\begin{array}{c}{v}_o=\left(\frac{30}{30+j10}\right)\left(0.149+j6.485\text{V}\right)\\ =2.0796+j5.792\\ =6.154\angle 70.26°\text{V}\end{array}$

Represent the voltage in time domain.

$v_o\left(t\right)=6.154\cos \left(1000t+70.26°\right)\text{V}$

Conclusion:

Therefore, the value of voltage $v_o\left(t\right)$ in the given circuit is $6.154\cos \left(1000t+70.26°\right)\text{V}$.