Chapter 3, Problem 15P

Apply nodal analysis to find i_o and the power dissipated in each resistor in the circuit of Fig. 3.64.

Figure 3.64

For Prob. 3.15.

To determine

Find the current $i_o$ and the power dissipated in each resistor in the circuit of Figure 3.64.

Answer to Problem 15P

The value of current $i_o$ is $29.45\text{A}$, and the power dissipated in $6\text{S}$, $5\text{S}$, and $3\text{S}$ resistors are $144.6\text{W}$, $129.6\text{W}$, and $12\text{W}$ respectively.

Explanation of Solution

Given data:

Refer to Figure 3.64 in the textbook for the nodal analysis. In the given circuit, $i_o$ is the current flows through $6\text{S}$ resistor.

Calculation:

Modify the given figure with the representation of supernode and current direction as shown in Figure 1.

Apply Kirchhoff’s current law to the supernode in Figure 1.

$\begin{array}{l}6v_1+5v_2+2=\left(v_3-v_2\right)3\\ 6v_1+5v_2+2=3v_3-3v_2\\ 6v_1+5v_2-3v_3+3v_2=-2\end{array}$

$6v_1+8v_2-3v_3=-2$ (1)

Apply Kirchhoff’s current law to node $v_3$ in Figure 1.

$\begin{array}{l}\left(v_3-v_2\right)3=4+2\\ 3v_3-3v_2=6\\ -3v_2+3v_3=6\end{array}$

$-v_2+v_3=2$ (2)

Reduce the given figure to provide voltage relation as shown in Figure 2.

Apply Kirchhoff’s voltage law to the circuit in Figure 2 to get relationship between voltages.

$10+v_2-v_1=0$

$-v_1+v_2=-10$ (3)

Represent the equations (1), (2), and (3) in matrix form.

$\left[\begin{array}{ccc}6& 8& -3\\ 0& -1& 1\\ -1& 1& 0\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right]=\left[\begin{array}{c}-2\\ 2\\ -10\end{array}\right]$

Obtain the value of determinants as follows.

$\begin{array}{c}\Delta =\left[\begin{array}{ccc}6& 8& -3\\ 0& -1& 1\\ -1& 1& 0\end{array}\right]\\ =6\left(0-1\right)-8\left(0+1\right)-3\left(0-1\right)\\ =-6-8+3\\ =-11\end{array}$

$\begin{array}{c}{\Delta }_1=\left[\begin{array}{ccc}-2& 8& -3\\ 2& -1& 1\\ -10& 1& 0\end{array}\right]\\ =-2\left(0-1\right)-8\left(0+10\right)-3\left(2-10\right)\\ =2-80+24\\ =-54\end{array}$

$\begin{array}{c}{\Delta }_2=\left[\begin{array}{ccc}6& -2& -3\\ 0& 2& 1\\ -1& -10& 0\end{array}\right]\\ =6\left(0+10\right)+2\left(0+1\right)-3\left(0+2\right)\\ =60+2-6\\ =56\end{array}$

$\begin{array}{c}{\Delta }_3=\left[\begin{array}{ccc}6& 8& -2\\ 0& -1& 2\\ -1& 1& -10\end{array}\right]\\ =6\left(10-2\right)-8\left(0+2\right)-2\left(0-1\right)\\ =48-16+2\\ =34\end{array}$

Write the expression for the voltages $v_1$, $v_2$, and $v_3$ using Cramer’s rule.

$v_1=\frac{{\Delta }_1}{\Delta }$ (4)

$v_2=\frac{{\Delta }_2}{\Delta }$ (5)

$v_3=\frac{{\Delta }_3}{\Delta }$ (6)

Substitute $-54$ for ${\Delta }_1$ and $-11$ for $\Delta$ in equation (4) to find $v_1$.

$\begin{array}{c}{v}_1=\frac{-54}{-11}\\ =4.909\text{V}\end{array}$

Substitute $56$ for ${\Delta }_2$ and $-11$ for $\Delta$ in equation (5) to find $v_2$.

$\begin{array}{c}{v}_2=\frac{56}{-11}\\ =-5.091\text{V}\end{array}$

Substitute $34$ for ${\Delta }_3$ and $-11$ for $\Delta$ in equation (6) to find $v_3$.

$\begin{array}{c}{v}_3=\frac{34}{-11}\\ =-3.091\text{V}\end{array}$

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

$i_o=v_1\left(6\text{S}\right)$ (7)

Substitute $4.909\text{V}$ for $v_1$ in equation (7) to find $i_o$.

$\begin{array}{c}{i}_o=\left(4.909\text{V}\right)\left(6\text{S}\right)\\ =29.45\text{A}\end{array}$

Write the expression for power dissipated in $6\text{S}$ resistor.

$P_{\left(6\text{S}\right)}=\left(6\text{S}\right)v_1{}^2$ (8)

Substitute $4.909\text{V}$ for $v_1$ in equation (8) to find $P_{\left(6\text{S}\right)}$.

$\begin{array}{c}{P}_{\left(6\text{S}\right)}=\left(6\text{S}\right){\left(4.909\text{V}\right)}^2\\ =144.589\text{W}\left\{\because \text{W}={\text{V}}^{\text{2}}\cdot \text{S}\right\}\\ \cong 144.6\text{W}\end{array}$

Write the expression for power dissipated in $5\text{S}$ resistor.

$P_{\left(5\text{S}\right)}=\left(5\text{S}\right)v_2{}^2$ (9)

Substitute $-5.091\text{V}$ for $v_2$ in equation (9) to find $P_{\left(5\text{S}\right)}$.

$\begin{array}{c}{P}_{\left(5\text{S}\right)}=\left(5\text{S}\right){\left(-5.091\text{V}\right)}^2\\ =129.59\text{W}\\ \cong 129.6\text{W}\end{array}$

Write the expression for power dissipated in $3\text{S}$ resistor.

$P_{\left(3\text{S}\right)}=\left(3\text{S}\right){\left(v_3-v_2\right)}^2$ (10)

Substitute $-5.091\text{V}$ for $v_2$ and $-3.091\text{V}$ for $v_3$ in equation (10) to find $P_{\left(3\text{S}\right)}$.

$\begin{array}{c}{P}_{\left(3\text{S}\right)}=\left(3\text{S}\right){\left(-3.091\text{V}+5.091\text{V}\right)}^2\\ =\left(3\text{S}\right){\left(2\text{V}\right)}^2\\ =12\text{W}\end{array}$

Conclusion:

Therefore, the value of current $i_o$ is $29.45\text{A}$, and the power dissipated in $6\text{S}$, $5\text{S}$, and $3\text{S}$ resistors are $144.6\text{W}$, $129.6\text{W}$, and $12\text{W}$ respectively.