Chapter 10, Problem 11P

Using nodal analysis, find i_o(t) in the circuit in Fig. 10.60.

Figure 10.60

For Prob. 10.11.

To determine

Find the current $i_o\left(t\right)$ in the circuit of Figure 10.60 using nodal analysis and MATLAB.

Answer to Problem 11P

The value of current $i_o\left(t\right)$ in the given circuit is $5.515\sin \left(2t+36.78°\right)\text{A}$.

Explanation of Solution

Given data:

Refer Figure 10.60 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 $\left(8\sin \left(2t+30°\right)\text{V}\right)$ with equation (3), the magnitude, angular frequency, and phase angle of source voltage are $8\text{V}$, $2\frac{\text{rad}}{\text{s}}$ and $30°$ respectively.

Substitute $8\text{V}$ for $V_m$ and $30°$ for $\varphi$ in equation (4).

$V_s=8\angle 30°\text{V}$

Convert $\left(\cos 2t\text{A}\right)$ source voltage into frequency domain as follows.

$I_s=1\angle 0°\text{A}$

Substitute $2\frac{\text{rad}}{\text{s}}$ for $\omega$ and $1\text{H}$ for $L$ in equation (1) to find $Z_{L\left(1\text{H}\right)}$.

$\begin{array}{c}{Z}_{L\left(1\text{H}\right)}=j\left(2\frac{\text{rad}}{\text{s}}\right)\left(1\text{H}\right)\left\{\because 1\text{H}=1\Omega \cdot \text{s}\right\}\\ =j2\Omega \end{array}$

Substitute $2\frac{\text{rad}}{\text{s}}$ for $\omega$ and $2\text{H}$ for $L$ in equation (1) to find $Z_{L\left(2\text{H}\right)}$.

$\begin{array}{c}{Z}_{L\left(2\text{H}\right)}=j\left(2\frac{\text{rad}}{\text{s}}\right)\left(2\text{H}\right)\left\{\because 1\text{H}=1\Omega \cdot \text{s}\right\}\\ =j4\Omega \end{array}$

Substitute $2\frac{\text{rad}}{\text{s}}$ for $\omega$ and $0.25\text{F}$ for $C$ in equation (2) to find $Z_{C\left(0.25\text{F}\right)}$.

$\begin{array}{c}{Z}_{C\left(0.25\text{F}\right)}=\frac{1}{j\left(2\frac{\text{rad}}{\text{s}}\right)\left(0.25\frac{\text{s}}{\Omega }\right)}\left\{\because 1\text{F}=1\frac{\text{s}}{\Omega }\right\}\\ =-j2\Omega \end{array}$

Substitute $2\frac{\text{rad}}{\text{s}}$ for $\omega$ and $0.5\text{F}$ for $C$ in equation (2) to find $Z_{C\left(0.5\text{F}\right)}$.

$\begin{array}{c}{Z}_{C\left(0.5\text{F}\right)}=\frac{1}{j\left(2\frac{\text{rad}}{\text{s}}\right)\left(0.5\frac{\text{s}}{\Omega }\right)}\left\{\because 1\text{F}=1\frac{\text{s}}{\Omega }\right\}\\ =-j1\Omega \end{array}$

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.

$\begin{array}{l}\frac{V_1-\left(8\angle 30°\right)}{2}+\frac{V_1-0}{-j1}+\frac{V_1-V_2}{j2}=0\\ \frac{V_1}{2}-\frac{\left(8\angle 30°\right)}{2}+\frac{V_1}{-j1}+\frac{V_1}{j2}-\frac{V_2}{j2}=0\\ \left(\frac{1}{2}+\frac{1}{-j1}+\frac{1}{j2}\right)V_1-\frac{V_2}{j2}=\frac{\left(8\angle 30°\right)}{2}\end{array}$

Simplify the equation as follows.

$\left(0.5+j1-j0.5\right)V_1+j0.5V_2=4\angle 30°$

$\left(0.5+j0.5\right)V_1+j0.5V_2=3.464+j2$        (5)

$\left(0.5+j0.5\right)V_1+j0.5V_2=2-j3.464$

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

$\begin{array}{l}\frac{V_2-V_1}{j2}+\frac{V_2-\left(8\angle 30°\right)}{j4-j2}-1\angle 0°=0\\ \frac{V_2}{j2}-\frac{V_1}{j2}+\frac{V_2}{j2}=\left(1\angle 0°\right)+\frac{8\angle 30°}{j2}\\ -\frac{V_1}{j2}+\left(\frac{1}{j2}+\frac{1}{j2}\right)V_2=3-j3.464\end{array}$

$j0.5V_1-j1V_2=3-j3.464$        (6)

$0.5V_1-j1V_2=-2.464-j2$

MATLAB Code:

Solve the two linear equations (5) and (6) using MATLAB to find the node voltages.

syms v1 v2

eq1 = (0.5 + 0.5*1i)*v1 +(0.5*1i)*v2 == 3.464 + 2*1i;

eq2 = (0.5*1i)*v1 +(-1*1i)*v2 == 3 + (-3.464*1i);

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 + (%.3f)i V', v1real, v1imag)

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

The command window output:

v1 = '3.302 + (-4.417)i V'

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

$i_o=\frac{V_1}{-j1\Omega }$

Substitute $3.302-j4.417\text{V}$ for $V_1$.

$\begin{array}{c}{i}_o=\frac{3.302-j4.417\text{V}}{-j1\Omega }\\ =4.417+j3.302\\ =5.515\angle 36.78°\text{A}\end{array}$

Represent the current in time domain.

$i_o\left(t\right)=5.515\sin \left(2t+36.78°\right)\text{A}$

Conclusion:

Therefore, the value of current $i_o\left(t\right)$ in the given circuit is $5.515\sin \left(2t+36.78°\right)\text{A}$.