Chapter 4, Problem 43E

To determine

Find the value of $i_x$ and $v_a$ in the circuit.

Answer to Problem 43E

The value of $i_x$ is $1.792\text{ A}$ and the value of $v_a$ is $6.846\text{ V}$.

Explanation of Solution

Calculation:

The circuit diagram is redrawn as shown in Figure 1,

Refer to the redrawn Figure 1,

Apply KVL in the mesh $ODAO$,

$v_3+i_1{R}_1+(i_1-i_2)R_2=0\text{ V}$ (1)

Here,

$v_3$ is the voltage of $\text{9 V}$ independent source,

$i_1$ is the current flowing in the mesh $ODAO$,

$i_2$ is the current flowing in the mesh $ABOA$,

$R_1$ is the resistance across branch $AD$ and

$R_2$ is the resistance across branch $AO$.

Apply KVL in the mesh $ABOA$,

$v_1+\left(i_2-i_4\right)R_3+\left(i_2-i_1\right)R_2=0\text{ V}$ (2)

Here,

$v_1$ is the voltage of $\text{0}\text{.2}{i}_X$ dependent source,

$i_4$ is the current flowing in the mesh $BCOB$ and

$R_3$ is the resistance across branch $BO$.

Apply KVL in the mesh $DOCD$,

$v_3=\left(i_3-i_4\right)R_5+i_3{R}_6$ (3)

Here,

$i_3$ is the current flowing in the mesh $DOCD$,

$R_5$ is the resistance across branch $CO$ and

$R_6$ is the resistance across branch $CD$.

Apply KVL in the mesh $BCOB$,

$v_2=\left(i_4-i_3\right)R_5+\left(i_4-i_2\right)R_3$ (4)

Here,

$v_2$ is the voltage of $0.1v_a$ dependent source.

The expression for the voltage across $7\Omega$ resistor is as follows,

$v_a=\left(-i_1{R}_1\right)$ (5)

Here,

$v_a$ is the voltage across $7\Omega$ resistor.

Refer to the redrawn Figure 1,

Substitute $\text{9 V}$ for $v_3$, $\text{7 }\Omega$ for $R_1$, $\text{7 }\Omega$ for $R_2$ in equation (1),

$\begin{array}{l}9\text{ V}+\left(\text{7 }\Omega \right)i_1+\left(\text{7 }\Omega \right)(i_1-i_2)=0\text{ V}\\ 9\text{ V}+\left(\text{7 }\Omega \right)i_1+\left(\text{7 }\Omega \right)i_1-\left(\text{7 }\Omega \right)i_2=0\text{ V}\end{array}$

$9\text{ V}+\left(\text{14 }\Omega \right)i_1-\left(\text{7 }\Omega \right)i_2=0\text{ V}$ (6)

Substitute $\text{0}\text{.2}{i}_X$ for $v_1$, $\text{7 }\Omega$ for $R_2$ and $\text{4 }\Omega$ for $R_3$ in equation (2),

$\begin{array}{l}\text{0}\text{.2}{i}_X+\left(\text{4 }\Omega \right)\left(i_2-i_4\right)+\left(\text{7 }\Omega \right)\left(i_2-i_1\right)=0\text{ V}\\ \text{0}\text{.2}{i}_X+\left(\text{4 }\Omega \right)i_2-\left(\text{4 }\Omega \right)i_4+\left(\text{7 }\Omega \right)i_2-\left(\text{7 }\Omega \right)i_1=0\text{ V}\\ -\left(\text{7 }\Omega \right)i_1+\left(\text{11 }\Omega \right)i_2+\text{0}\text{.2}{i}_X-\left(\text{4 }\Omega \right)i_4=0\text{ V}\end{array}$

Substitute $i_3$ for $i_x$ in the above equation,

$-\left(\text{7 }\Omega \right)i_1+\left(\text{11 }\Omega \right)i_2+\text{0}\text{.2}{i}_3-\left(\text{4 }\Omega \right)i_4=0\text{ V}$ (7)

Substitute $\text{9 V}$ for $v_3$, $\text{1 }\Omega$ for $R_5$, $\text{4 }\Omega$ for $R_6$ in equation (3),

$\begin{array}{l}\text{9 V}=\left(\text{1 }\Omega \right)\left(i_3-i_4\right)+\left(\text{4 }\Omega \right)i_3\\ \text{9 V}=\left(\text{1 }\Omega \right)i_3-\left(\text{1 }\Omega \right)i_4+\left(\text{4 }\Omega \right)i_3\\ \text{9 V}=\left(\text{5 }\Omega \right)i_3-\left(\text{1 }\Omega \right)i_4\end{array}$

Rearrange the above equation for $i_4$,

$i_4=5i_3-\text{9 V}$ (8)

Substitute $0.1v_a$ for $v_a$, $\text{1 }\Omega$ for $R_5$ and $\text{4 }\Omega$ for $R_3$ in equation (4),

$0.1v_a=\left(\text{1 }\Omega \right)\left(i_4-i_3\right)+\left(\text{4 }\Omega \right)\left(i_4-i_2\right)$

Substitute $-i_1{R}_1$ for $v_a$ and $\text{7 }\Omega$ for $R_1$ in above equation,

$\begin{array}{c}0.1\left(\left(\text{7 }\Omega \right)\left(-i_1\right)\right)=\left(\text{1 }\Omega \right)\left(i_4-i_3\right)+\left(\text{4 }\Omega \right)\left(i_4-i_2\right)\\ -\left(0.7\Omega \right)i_1=-\left(4\Omega \right)i_2-\left(1\Omega \right)i_3+\left(4\Omega \right)i_4+\left(1\Omega \right)i_4\end{array}$

$-\left(0.7\Omega \right)i_1=-\left(4\Omega \right)i_2-\left(1\Omega \right)i_3+\left(\text{5 }\Omega \right)i_4$ (9)

Substitute $5i_3-\text{9 V}$ for $i_4$ in equation (7),

$\begin{array}{l}-\left(\text{7 }\Omega \right)i_1+\left(\text{11 }\Omega \right)i_2+\text{0}\text{.2}{i}_3-\left(\text{4 }\Omega \right)\left(5i_3-\text{9 V}\right)=0\text{ V}\\ -\left(\text{7 }\Omega \right)i_1+\left(\text{11 }\Omega \right)i_2+\text{0}\text{.2}{i}_3-\left(\text{20 }\Omega \right)i_3+\text{36 V}=0\text{ V}\end{array}$

$-\left(\text{7 }\Omega \right)i_1+\left(\text{11 }\Omega \right)i_2-\left(\text{19}\text{.8 }\Omega \right)i_3\text{+36 V}=0\text{ V}$ (10)

Substitute $5i_3-\text{9 V}$ for $i_4$ in equation (9),

$\begin{array}{l}-\left(0.7\Omega \right)i_1=-\left(4\Omega \right)i_2-\left(1\Omega \right)i_3+\left(\text{5 }\Omega \right)\left(5i_3-\text{9 V}\right)\\ -\left(0.7\Omega \right)i_1=-\left(4\Omega \right)i_2-\left(1\Omega \right)i_3+\left(\text{25 }\Omega \right)i_3-45\text{ V}\end{array}$

$-\left(0.7\Omega \right)i_1=-\left(4\Omega \right)i_2+\left(\text{24 }\Omega \right)i_3-45\text{ V}$ (11)

Rearrange the equation (6), (10) and (11),

$\begin{array}{c}14i_1-7i_2+0i_3=-9\\ -7i_1+11i_2-19.8i_3=-\text{36}\\ 0.7i_1-4i_2+24i_3=45\end{array}$

The equations so formed can be written in matrix form as,

$\left(\begin{array}{ccc}14& -7& 0\\ -7& 11& -19.8\\ 0.7& -4& 24\end{array}\right)\left(\begin{array}{c}{i}_1\\ i_2\\ i_2\end{array}\right)=\left(\begin{array}{c}-9\\ -36\\ 45\end{array}\right)$

Therefore, by Cramer’s rule,

The determinant of the coefficient matrix is as follows,

$\begin{array}{c}\Delta =\left|\begin{array}{ccc}14& -7& 0\\ -7& 11& -19.8\\ 0.7& -4& 24\end{array}\right|\\ =1508.22\end{array}$

The 1^st determinant is as follows,

$\begin{array}{c}{\Delta }_1=\left|\begin{array}{ccc}-9& -7& 0\\ -36& 11& -19.8\\ 45& -4& 24\end{array}\right|\\ =-1474.2\end{array}$

The 2^nd determinant is as follows,

$\begin{array}{c}{\Delta }_2=\left|\begin{array}{ccc}14& -9& 0\\ -7& -36& -19.8\\ 0.7& 45& 24\end{array}\right|\\ =-1009.2599\end{array}$

The 3^rd determinant is as follows,

$\begin{array}{c}{\Delta }_3=\left|\begin{array}{ccc}14& -7& -9\\ -7& 11& -36\\ 0.7& -4& 45\end{array}\right|\\ =2702.7\end{array}$

Simplify for $i_1$,

$\begin{array}{c}{i}_1=\frac{{\Delta }_1}{\Delta }\\ =\frac{-1474.2}{1508.22}\\ =-0.978\text{ A}\end{array}$

Simplify for $i_2$,

$\begin{array}{c}{i}_2=\frac{{\Delta }_2}{\Delta }\\ =\frac{-1009.2599}{1508.22}\\ =-0.6692\text{ A}\end{array}$

Simplify for $i_3$,

$\begin{array}{c}{i}_3=\frac{{\Delta }_3}{\Delta }\\ =\frac{2702.7}{1508.22}\\ =1.792\text{ A}\end{array}$

The value of $i_x$ is equal to $i_3$ from Figure 1, therefore,

$i_x=1.792\text{ A}$

Substitute $0.978\text{ A}$ for $i_1$ and $\text{7 }\Omega$ for $R_1$ in equation (5),

$\begin{array}{c}{v}_a=-\left(-0.978\text{ A}\right)\left(\text{7 }\Omega \right)\\ =6.846\text{ V}\end{array}$

Conclusion:

Thus, the value of $i_x$ is $1.792\text{ A}$ and the value of $v_a$ is $6.846\text{ V}$.