Chapter 10, Problem 12P

Using Fig. 10.61, design a problem to help other students better understand nodal analysis.

Figure 10.61

For Prob. 10.12.

To determine

Design a problem to provide better understanding of nodal analysis.

Explanation of Solution

Given data:

Refer to Figure 10.61 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.

$i_s=i_m\sin \left(\omega t+\varphi \right)$ (3)

Here,

$\varphi$ is the phase angle, and

$i_m$ is the magnitude of source current.

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

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

Here,

$i_m\sin \left(\omega t+\varphi \right)$ is the time domain representation of source current, and

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

Write the polar form representation of frequency domain.

$i_s=i_m\angle \varphi$ (4)

Calculation:

In Figure 10.61, consider the value of source current $i_s\left(t\right)$ is $10\sin 100t\text{A}$, and the value of resistors $R_1$ and $R_2$ are $10\Omega$ and $10\Omega$ respectively. The value inductor and capacitor are considered as $5\text{H}$, and $20\text{mF}$ respectively.

Comparing assumed source current with equation (3), the magnitude, angular frequency, and phase angle of source current are $10\text{A}$, $100\frac{\text{rad}}{\text{s}}$ and $0°$ respectively.

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

$i_s=10\angle 0°\text{A}$

Substitute $100\frac{\text{rad}}{\text{s}}$ for $\omega$ and $5\text{H}$ for $L$ in equation (1) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left(100\frac{\text{rad}}{\text{s}}\right)\left(5\text{H}\right)\left\{\because 1\text{H}=1\Omega \cdot \text{s}\right\}\\ =j500\Omega \end{array}$

Substitute $100\frac{\text{rad}}{\text{s}}$ for $\omega$ and $20\text{F}$ for $C$ in equation (2) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(100\frac{\text{rad}}{\text{s}}\right)\left(20\text{mF}\right)}\\ =\frac{1}{j\left(100\frac{\text{rad}}{\text{s}}\right)\left(20\times {10}^{-3}\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\text{mF}=1\times {10}^{-3}\text{F}\\ 1\text{F}=1\frac{\text{s}}{\Omega }\end{array}\right\}\\ =-j0.5\Omega \end{array}$

The frequency domain representation of given figure for assumed values is shown in Figure 1.

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

$\frac{V_1-0}{10}+\frac{V_1-V_2}{10}+2i_o-10=0$ (5)

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

$i_o=\frac{V_2}{j500}$ (6)

Substitute equation (6) in (5).

$\begin{array}{l}\frac{V_1-0}{10}+\frac{V_1-V_2}{10}+2\left(\frac{V_2}{j500}\right)-10=0\\ \frac{V_1}{10}+\frac{V_1}{10}-\frac{V_2}{10}+\frac{V_2}{j250}=10\\ \left(\frac{1}{10}+\frac{1}{10}\right)V_1+\left(-\frac{1}{10}+\frac{1}{j250}\right)V_2=10\end{array}$

$0.2V_1+\left(-0.1-j0.004\right)V_2=10$ (7)

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

$\frac{V_2-V_1}{10}+\frac{V_2-0}{-j0.5}+\frac{V_2-0}{j500}-2i_o=0$ (8)

Substitute equation (6) in (8).

$\begin{array}{l}\frac{V_2-V_1}{10}+\frac{V_2-0}{-j0.5}+\frac{V_2-0}{j500}-2\left(\frac{V_2}{j500}\right)=0\\ \frac{V_2}{10}-\frac{V_1}{10}+\frac{V_2}{-j0.5}+\frac{V_2}{j500}-\frac{V_2}{j250}=0\\ -\frac{V_1}{10}+\left(\frac{1}{10}+\frac{1}{-j0.5}+\frac{1}{j500}-\frac{1}{j250}\right)V_2=0\\ -0.1V_1+\left(0.1+j2-j0.002+j0.004\right)V_2=0\end{array}$

$-0.1V_1+\left(0.1+j2.002\right)V_2=0$ (9)

Rearrange equation (7).

$\begin{array}{l}0.2V_1=10-\left(-0.1-j0.004\right)V_2\\ V_1=\frac{10}{0.2}+\frac{\left(0.1+j0.004\right)}{0.2}{V}_2\end{array}$

$V_1=50+\left(0.5+j0.02\right)V_2$ (10)

Substitute equation (7) in (9).

$\begin{array}{l}-0.1\left(50+\left(0.5+j0.02\right)V_2\right)+\left(0.1+j2.002\right)V_2=0\\ -5-\left(0.05+j0.002\right)V_2+\left(0.1+j2.002\right)V_2=0\\ \left(0.05+j2\right)V_2=5\end{array}$

Simplify the equation as follows.

$\begin{array}{c}{V}_2=\frac{5}{0.05+j2}\\ =0.0625-j2.495\text{V}\\ \text{=2}\text{.5}\angle \text{-88}\text{.56}\text{V}\end{array}$

Substitute $\text{2}\text{.5}\angle \text{-88}\text{.56}\text{V}$ for $V_2$ in equation (6) to find $i_o$.

$\begin{array}{c}{i}_o=\frac{\text{2}\text{.5}\angle \text{-88}\text{.56}\text{V}}{j500\Omega }\\ =5\times {10}^{-3}\angle -178.56°\text{A}\\ \text{=}5\angle -178.56°\text{mA}\left\{\because 1\text{mA}=1\times {10}^{-3}\text{A}\right\}\end{array}$

Represent the current in time domain.

$i_o\left(t\right)=5\sin \left(100t-178.56°\right)\text{mA}$

Conclusion:

Thus, a problem has been designed to provide better understanding of nodal analysis.