Chapter 10, Problem 26P

Use mesh analysis to find current i_o in the circuit of Fig. 10.74.

Figure 10.74

For Prob. 10.26.

To determine

Find the current $i_o\left(t\right)$ in the circuit of Figure 10.74 using mesh analysis.

Answer to Problem 26P

The value of current $i_o\left(t\right)$ in the given circuit is $39.5\cos \left({10}^{3}t-18.43°\right)\text{mA}$.

Explanation of Solution

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 $\left(10\cos {10}^{3}t\text{V}\right)$ 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).

$V_s=10\angle 0°\text{V}$

Convert $\left(20\sin {10}^{3}t\text{V}\right)$ source voltage into frequency domain as follows.

$\begin{array}{l}20\sin {10}^{3}t=20\cos \left({10}^{3}t-90°\right)\\ =20\angle -90°\\ =-j20\end{array}$

Substitute $1000\frac{\text{rad}}{\text{s}}$ for $\omega$ and $0.4\text{H}$ for $L$ in equation (1) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left(1000\frac{\text{rad}}{\text{s}}\right)\left(0.4\text{H}\right)\left\{\because 1\text{H}=1\Omega \cdot \text{s}\right\}\\ =j400\Omega \end{array}$

Substitute $1000\frac{\text{rad}}{\text{s}}$ for $\omega$ and $1\mu \text{F}$ for $C$ in equation (2) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(1000\frac{\text{rad}}{\text{s}}\right)\left(1\times {10}^{-6}\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\text{F}=1\frac{\text{s}}{\Omega }\\ 1\mu \text{F}=1\times {10}^{-6}\text{F}\end{array}\right\}\\ =-j1000\Omega \end{array}$

The frequency domain representation of given figure with the assumed current direction is shown in Figure 1.

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

$\begin{array}{l}\left(2\text{k}\Omega \right)I_1+j400\left(I_1-I_2\right)-10=0\left\{\because 1\text{k}\Omega =1000\Omega \right\}\\ 2000I_1+j400I_1-j400I_2=10\\ \left(2000+j400\right)I_1-j400I_2=10\end{array}$

$\left(200+j40\right)I_1-j40I_2=1$ (5)

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

$\begin{array}{l}-j1000I_2-j20+j400\left(I_2-I_1\right)=0\\ -j1000I_2-j20+j400I_2-j400I_1=0\\ -j400I_1-j600I_2=j20\end{array}$

$40I_1+j60I_2=-2$ (6)

Represent equation (5) and (6) in matrix form.

$\left[\begin{array}{cc}200+j40& -j40\\ 40& 60\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{c}1\\ -2\end{array}\right]$

Obtain the value of determinants as follows.

$\begin{array}{c}\Delta =\left[\begin{array}{cc}200+j40& -j40\\ 40& 60\end{array}\right]\\ =60\left(200+j40\right)+40\left(j40\right)\\ =12000+j4000\\ =12649.1\angle 18.435°\end{array}$

$\begin{array}{c}{\Delta }_1=\left[\begin{array}{cc}1& -j40\\ -2& 60\end{array}\right]\\ =60-2\left(j40\right)\\ =60-j80\\ =100\angle -53.13°\end{array}$

$\begin{array}{c}{\Delta }_2=\left[\begin{array}{cc}200+j40& 1\\ 40& -2\end{array}\right]\\ =-2\left(200+j40\right)-40\\ =-440-j80\\ =447.21\angle -169.69°\end{array}$

Write the expression for mesh currents using Cramer’s rule.

$I_1=\frac{{\Delta }_1}{\Delta }$ (7)

$I_2=\frac{{\Delta }_2}{\Delta }$                                                                                                                      (8)

Substitute $100\angle -53.13°$ for ${\Delta }_1$ and $12649.1\angle 18.435°$ for $\Delta$ in equation (7) to find $I_1$.

$\begin{array}{c}{I}_1=\frac{100\angle -53.13°}{12649.1\angle 18.435°}\\ =\left(2.5-j7.5\right)\times {10}^{-3}\text{A}\\ =\text{7}\text{.91}\times {\text{10}}^{-3}\angle -71.565°\text{A}\\ \text{=7}\text{.91}\angle -71.565°\text{mA}\left\{\because 1\text{mA}=1\times {10}^{-3}\text{A}\right\}\end{array}$

Substitute $447.21\angle -169.69°$ for ${\Delta }_2$ and $12649.1\angle 18.435°$ for $\Delta$ in equation (8) to find $I_2$.

$\begin{array}{c}{I}_2=\frac{447.21\angle -169.69°}{12649.1\angle 18.435°}\\ =\left(-35+j5\right)\times {10}^{-3}\text{A}\\ =\text{35}\text{.35}\times {10}^{-3}\angle \text{171}\text{.875}°\text{A}\\ =\text{35}\text{.35}\angle \text{171}\text{.875}°\text{mA}\end{array}$

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

$i_o=I_1-I_2$

Substitute $\text{7}\text{.91}\angle -71.565°\text{mA}$ for $I_1$ and $\text{35}\text{.35}\angle \text{171}\text{.875}°\text{mA}$ for $I_2$.

$\begin{array}{c}{i}_o=\left(\text{7}\text{.91}\angle -71.565°\text{mA}\right)-\left(\text{35}\text{.35}\angle \text{171}\text{.875}°\text{mA}\right)\\ =0.0375-j0.0125\text{mA}\\ =39.5\angle -18.43°\text{mA}\end{array}$

Represent the current in time domain.

$i_o\left(t\right)=39.5\cos \left({10}^{3}t-18.43°\right)\text{mA}$

Conclusion:

Therefore, the value of current $i_o\left(t\right)$ in the given circuit is $39.5\cos \left({10}^{3}t-18.43°\right)\text{mA}$.