Chapter 5, Problem 63E

(a)

To determine

Find the two-component Thevenin equivalent of the network

Answer to Problem 63E

The Thevenin equivalent resistance is $6.098\Omega$ and the Thevenin voltage is $-0.6067\text{ V}$.

Explanation of Solution

Formula used:

The expression for the equivalent resistor when resistors are connected in series is as follows:

$R_{eq}=R_1+R_2+\mathrm{......}+R_N$ (1)

Here,

$R_1$, $R_2$,…, $R_N$ are the resistances.

The expression for the equivalent resistor when resistors are connected in parallel is as follows:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\mathrm{......}+\frac{1}{R_N}$ (2)

Here,

$R_1$, $R_2$,…, $R_N$ are the resistances.

The $Y$-network and $\Delta$-network circuit diagram is given in Figure 1.

o

Refer to the redrawn Figure 1:

The expression for the conversion of $\Delta$-network into $Y$-network is as follows:

$R_1=\frac{R_b{R}_c}{R_a+R_b+R_c}$ (3)

$R_2=\frac{R_a{R}_c}{R_a+R_b+R_c}$ (4)

$R_3=\frac{R_a{R}_b}{R_a+R_b+R_c}$ (5)

Here,

$R_1$, $R_2$ and $R_3$ are the resistors connected in $Y$-network and

$R_a$, $R_b$ and $R_c$ are the resistors connected in $\Delta$-network.

Calculation:

To find equivalent resistance of a circuit the independent voltage source is replaced by short circuit

The redrawn circuit diagram is given in Figure 2:

Refer to the redrawn Figure 2:

$\text{11 }\Omega$ and $\text{2 }\Omega$ resistor are connected in parallel.

Substitute $\text{11 }\Omega$ for $R_1$ and $\text{2 }\Omega$ for $R_2$ equation (2):

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{11 }\Omega }+\frac{1}{\text{2 }\Omega }\\ =\frac{13}{\text{22 }\Omega }\end{array}$

Rearrange the equation for $R_{eq}$:

$\begin{array}{c}{R}_{eq}=\frac{\text{22}}{13}\Omega \\ =1.69\Omega \end{array}$

The simplified circuit diagram is given in Figure 3:

Refer to the redrawn Figure 3:

$22\Omega$, $1\Omega$ and $\text{2 }\Omega$ resistors are connected in $\Delta$-network.

Substitute $22\Omega$ for $R_a$, $1\Omega$ for $R_b$ and $\text{1}\text{.69 }\Omega$ for $R_c$ in equation (3):

$\begin{array}{c}{R}_1=\frac{\left(\text{1}\text{.69 }\Omega \right)\left(\text{1 }\Omega \right)}{22\Omega +1\Omega +\text{1}\text{.69 }\Omega }\\ =\frac{1.69}{24.69}\Omega \\ =0.0684\Omega \end{array}$

Substitute $22\Omega$ for $R_a$, $1\Omega$ for $R_b$ and $\text{1}\text{.69 }\Omega$ for $R_c$ in equation (3):

$\begin{array}{c}{R}_2=\frac{\left(\text{1}\text{.69 }\Omega \right)\left(\text{22 }\Omega \right)}{22\Omega +1\Omega +\text{1}\text{.69 }\Omega }\\ =\frac{37.18}{24.69}\Omega \\ =1.51\Omega \end{array}$

Substitute $22\Omega$ for $R_a$, $1\Omega$ for $R_b$ and $\text{1}\text{.69 }\Omega$ for $R_c$ in equation (3):

$\begin{array}{c}{R}_3=\frac{\left(22\Omega \right)\left(\text{1 }\Omega \right)}{22\Omega +1\Omega +\text{1}\text{.69 }\Omega }\\ =\frac{22}{24.69}\Omega \\ =0.891\Omega \end{array}$

The simplified circuit diagram is given in Figure 4.

Refer to the redrawn Figure 4:

$1.51\Omega$ and $12\Omega$ resistor are connected in series.

Substitute $1.51\Omega$ for $R_1$ and $12\Omega$ for $R_2$ equation (1):

$\begin{array}{c}{R}_{eq}=1.51\Omega +12\Omega \\ =13.51\Omega \end{array}$

$0.891\Omega$ and $10\Omega$ resistor are connected in series.

Substitute $0.891\Omega$ for $R_1$ and $10\Omega$ for $R_2$ equation (1):

$\begin{array}{c}{R}_{eq}=0.891\Omega +10\Omega \\ =10.891\Omega \end{array}$

$\text{13}\text{.51 }\Omega$ and $\text{10}\text{.891 }\Omega$ resistor are connected in parallel.

Substitute $\text{13}\text{.51 }\Omega$ for $R_1$ and $\text{10}\text{.891 }\Omega$ for $R_2$ equation (2).

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{13}\text{.51 }\Omega }+\frac{1}{\text{10}\text{.891 }\Omega }\\ =0.16584{\Omega }^{-}\end{array}$

Rearrange the equation for $R_{eq}$.

$\begin{array}{c}{R}_{eq}=\frac{1}{0.16584}\Omega \\ =6.0299\Omega \end{array}$

The simplified circuit diagram is given in Figure 5.

Refer to the redrawn Figure 5:

$6.0299\Omega$ and $0.0684\Omega$ resistor are connected in series.

Substitute $6.0299\Omega$ for $R_1$ and $0.0684\Omega$ for $R_2$ equation (1):

$\begin{array}{c}{R}_{eq}=6.0299\Omega +0.0684\Omega \\ =6.098\Omega \end{array}$

The simplified circuit diagram is given in Figure 6.

Refer to the redrawn Figure 6:

So, the Thevenin equivalent resistance is $6.098\Omega$.

The redrawn circuit diagram is given in Figure 7.

Refer to the redrawn Figure 7:

Apply KCL at node 1:

$\frac{v_1+v_o}{R_1}+\frac{v_1}{R_2}+\frac{v_1-v_2}{R_3}+\frac{v_1-v_3}{R_4}=0\text{ A}$ (6)

Here,

$v_1$ is the voltage at node 1,

$v_2$ is the voltage at node 2,

$v_3$ is the voltage at node 3,

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

$R_1$ is the resistance of $11\Omega$ resistor,

$R_2$ is the resistance of $2\Omega$ resistor,

$R_3$ is the resistance of $12\Omega$ resistor and

$R_4$ is the resistance of $22\Omega$ resistor.

Substitute $9\text{ V}$ for $v_o$, $11\Omega$ for $R_1$, $2\Omega$ for $R_2$, $12\Omega$ for $R_3$ and $22\Omega$ for $R_4$ in equation (7):

$\begin{array}{l}\frac{v_1+9\text{ V}}{11\Omega }+\frac{v_1}{\text{2 }\Omega }+\frac{v_1-v_2}{12\Omega }+\frac{v_1-v_3}{\text{22 }\Omega }=0\text{ A}\\ \frac{12v_1+108\text{ V}+66v_1+11v_1-11v_2+6v_1-6v_3}{132\Omega }=0\text{ A}\\ \frac{95v_1+108\text{ V}-11v_2-6v_3}{132\Omega }=0\text{ A}\end{array}$

Rearrange for $v_1$, $v_2$ and $v_3$.

$95v_1+108\text{ V}-11v_2-6v_3=0\text{ V}$

$95v_1-11v_2-6v_3=-108\text{ V}$ (7)

Apply KCL at node 2:

$\frac{v_2-v_1}{R_3}+\frac{v_2-v_3}{R_5}=0\text{ A}$ (8)

Here,

$R_5$ is the resistance of $10\Omega$ resistor.

Substitute $12\Omega$ for $R_3$ and $10\Omega$ for $R_5$ in equation (8):

$\begin{array}{l}\frac{v_2-v_1}{12\Omega }+\frac{v_2-v_3}{10\Omega }=0\text{ A}\\ \frac{5v_2-5v_1+6v_2-6v_3}{60\Omega }=0\text{ A}\\ \frac{11v_2-5v_1-6v_3}{60\Omega }=0\text{ A}\end{array}$

Rearrange for $v_1$, $v_2$ and $v_3$:

$-5v_1+11v_2-6v_3=0\text{ V}$ (9)

Apply KCL at node 3:

$\frac{v_3-v_1}{R_4}+\frac{v_3-v_2}{R_5}+\frac{v_3}{R_6}=0\text{ A}$ (10)

Here,

$R_6$ is the resistance of $1\Omega$ resistor.

Substitute $22\Omega$ for $R_4$, $10\Omega$ for $R_5$ and $1\Omega$ for $R_6$ in equation (10):

$\begin{array}{l}\frac{v_3-v_1}{22\Omega }+\frac{v_3-v_2}{\text{10 }\Omega }+\frac{v_3}{\text{1 }\Omega }=0\text{ A}\\ \frac{5v_3-5v_1+11v_3-11v_2+110v_3}{\text{110 }\Omega }=0\text{ A}\\ \frac{-5v_1-11v_2+126v_3}{\text{110 }\Omega }=0\text{ A}\end{array}$

Rearrange for $v_1$, $v_2$ and $v_3$,

$-5v_1-11v_2+126v_3=0\text{ V}$ (11)

The equations (7), (9) and (11) can be written in matrix form as:

$\left(\begin{array}{ccc}95& -11& -6\\ -5& 11& -6\\ -5& -11& 126\end{array}\right)\left(\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right)=\left(\begin{array}{c}-108\\ 0\\ 0\end{array}\right)$

Therefore, by Cramer’s rule,

The determinant of the coefficient matrix is as follows:

$\begin{array}{c}\Delta =\left|\begin{array}{ccc}95& -11& -6\\ -5& 11& -6\\ -5& -11& 126\end{array}\right|\\ =117480\end{array}$

The 1^st determinant is as follows:

$\begin{array}{c}{\Delta }_1=\left|\begin{array}{ccc}-108& -11& -6\\ 0& 11& -6\\ 0& -11& 126\end{array}\right|\\ =-142560\end{array}$

The 2^nd determinant is as follows:

$\begin{array}{c}{\Delta }_2=\left|\begin{array}{ccc}95& -108& -6\\ -5& 0& -6\\ -5& 0& 126\end{array}\right|\\ =-71280\end{array}$

The 3^rd determinant is as follows:

$\begin{array}{c}{\Delta }_3=\left|\begin{array}{ccc}95& -11& -108\\ -5& 11& 0\\ -5& -11& 0\end{array}\right|\\ =-11880\end{array}$

Simplify for $v_1$.

$\begin{array}{c}{v}_1=\frac{{\Delta }_1}{\Delta }\\ =\frac{-142560}{117480}\\ =-1.213\end{array}$

Simplify for $v_2$.

$\begin{array}{c}{v}_2=\frac{{\Delta }_2}{\Delta }\\ =\frac{-71280}{117480}\\ =-0.6067\text{ V}\end{array}$

Simplify for $v_3$.

$\begin{array}{c}{v}_3=\frac{{\Delta }_3}{\Delta }\\ =\frac{-11880}{117480}\\ =-0.1011\text{ V}\end{array}$

So, the Thevenin voltage $v_{TH}$ is $-0.6067\text{ V}$.

Conclusion:

Thus, the Thevenin equivalent resistance is $6.098\Omega$ and the Thevenin voltage is,

$-0.6067\text{ V}$.

(b)

To determine

Find the power dissipated by a $1\Omega$ resistor connected between the open terminals.

Answer to Problem 63E

Thepower dissipated by a $1\Omega$ resistor connected between the open terminals is $7.306\text{ mW}$.

Explanation of Solution

Given Data:

The load resistance is $1\Omega$.

Formula used:

The expression for the power dissipated by a resistor is as follows:

$p={\left(i_L\right)}^2{R}_L$ (11)

Here,

$p$ is the power dissipated by a  resistor,

$i_L$ is the current flowing through the load resistor and

$R_L$ is the load resistance

Calculation:

The redrawn circuit diagram is given in Figure 8.

Refer to the redrawn Figure 8:

The expression for the current flowing in the circuit is as follows:

$i_L=\frac{v_{TH}}{R_{TH}+R_L}$ (12)

Here,

$v_{TH}$ is the Thevenin voltage of the circuit and

$R_{TH}$ is the Thevenin resistance of the circuit.

Substitute $6.098\Omega$ for $R_{TH}$, $-0.6067\text{ V}$ for $v_{Th}$ and $1\Omega$ for $R_L$ in equation (12):

$\begin{array}{c}{i}_L=\frac{-0.6067\text{ V}}{6.098\Omega +1\Omega }\\ =-0.08547\text{ A}\end{array}$

Substitute $-0.08547\text{ A}$ for $i_L$ and $1\Omega$ for $R_L$ in equation (11):

$\begin{array}{c}p={\left(-0.08547\text{ A}\right)}^2\left(1\Omega \right)\\ =7.306\times {10}^{-3}\text{ W}\\ \text{=}7.306\text{ mW }\left\{\because {10}^{-3}\text{ W}=1\text{ mW}\right\}\end{array}$

Conclusion:

Thus, the power dissipated by a $1\Omega$ resistor connected between the open terminals is $7.306\text{ mW}$.