Chapter 3, Problem 63E

Consider the seven-element circuit depicted in Fig. 3.99. (a) How many nodes, loops, and branches does it contain? (b) Calculate the current flowing through each resistor. (c) Determine the voltage across the current source, assuming the top terminal is the positive reference terminal.

■ FIGURE 3.99

To determine

Find the number of nodes, number of loops and number of branches in the circuit.

Answer to Problem 63E

Number of nodes in the circuit is $5$, numbers of loops in the circuit is $6$ and numbers of branches in the circuit is $7$.

Explanation of Solution

The circuit diagram is redrawn as shown in Figure 1.

Refer to the redrawn Figure 1.

A point where two or more branches have common connection is known as node.

In Figure 1 two branches are connected at point $a$, four branches are connected at point $b$, three branches are connected at point $c$, three branches are connected at point $d$ and two branches are connected at point $e$; therefore, number of nodes in the circuit is $5$.

Each electrical element or device present in the circuit is known as branch.

There is $1$ current source and $6$ resistor present in the circuit, therefore, total number of branches is $7$.

The circuit diagram is redrawn as shown in Figure 2,

Refer to the redrawn Figure 2,

If starting and ending node is same for a path then it is known as closed path or loop.

$ABEFA$, $BCDEB$, $ABCDEFA$, $BGHC$, $BGHCDEB$, $ABGHCDEFA$ are the loops present in the circuit; therefore, number of loops in the circuit is $6$.

Conclusion:

Thus, the number of nodes in the circuit is $5$, numbers of loops in the circuit is $6$ and numbers of branches in the circuit is $7$.

To determine

Find the value of current through each resistor.

Answer to Problem 63E

The value of current through $2\Omega$ resistor is $-789.3\text{ mA}$, the value of current through $5\Omega$ resistor is $-1.211\text{ A}$, value of current through $\text{1 }\Omega$ resistor is $-526.2\text{ mA}$, value of current through $2\Omega$ resistor is $-2\text{ A}$ and value of current through $2\Omega$ resistor is $-263.1\text{ mA}$.

Explanation of Solution

Formula used:

The expression for parallel combination of resistance is as follows.

$\frac{1}{R}=\left(\frac{1}{R_1}+{\frac{1}{R}}_2\right)$ (1)

Here,

$R$ is the total resistances and

$R_1$ and $R_2$ are the resistances in the circuit.

Calculation:

The circuit diagram is redrawn as shown in Figure 3.

Refer to the redrawn Figure 3.

The expression for current $i_3$ is as follows.

$i_3=i_1\times \frac{R_2}{R_2+R_5+R_6+R_7}$ (2)

Here,

$i_3$ is the current flowing through series combination of  resistor $R_5$, $R_6$ and $R_7$.

$i_1$ is the total current and

$R_2$, $R_5$, $R_6$ and $R_7$ are the resistances in the circuit.

The expression for current $i_2$ is as follows.

$i_2=i_1\times \frac{R_5+R_6+R_7}{R_2+R_5+R_6+R_7}$ (3)

Here,

$i_2$ is the current flowing through $R_2$ resistor,

$i_1$ is the total current and

$R_2$, $R_5$, $R_6$ and $R_7$ are the resistances in the circuit.

Refer to the redrawn Figure 1.

The expression for current $i_4$ is as follows.

$i_4=i_3\times \frac{R_3}{R_3+R_4}$ (4)

Here,

$i_3$ is the current in the circuit,

$i_4$ is the current flowing through $R_4$ resistor and

$R_3$ and $R_4$ are the resistances in the circuit.

The expression for current $i_5$ is as follows.

$i_5=i_3\times \frac{R_4}{R_3+R_4}$ (5)

Here,

$i_3$ is the current in the circuit,

$i_5$ is the current flowing through $R_3$ resistor and

$R_3$ and $R_4$ are the resistances in the circuit.

Refer to the redrawn Figure 1.

As $R_3$ and $R_4$ are in parallel therefore from equation (1).

$\begin{array}{l}\frac{1}{R_7}=\left(\frac{1}{R_3}+\frac{1}{R_4}\right)\\ \frac{1}{R_7}=\frac{1}{2\text{Ω}}+\frac{1}{1\text{Ω}}\\ \frac{1}{R_7}=\frac{1}{0.67\text{Ω}}\end{array}$

Rearrange for $R_7$.

$R_7=0.67\text{Ω}$

Refer to the redrawn Figure 3.

As current source direction is downward means change sign of current means value of current is $-2\text{ A}$.

Substitute $-2\text{ A}$ for $i_1$, $5\text{Ω}$ for $R_2$, $0.67\text{Ω}$ for $R_7$, $2\text{Ω}$ for $R_5$ and $5\text{Ω}$ for $R_6$ in the equation (2).

$\begin{array}{c}{i}_3=-2\text{ A}\times \frac{5\text{Ω}}{5\text{Ω}+2\text{Ω}+5\text{Ω}+0.67\text{Ω}}\\ =-2\text{ A}\times \frac{5\text{Ω}}{12.67\text{Ω}}\\ =-0.7893\text{ A}\\ =-789.3\text{ mA}\left\{\because 1{\text{A=10}}^3\text{ mA}\right\}\end{array}$

As resistor $R_5$ and $R_6$ are in series means value of current is same for both.

So value of current through resistor $R_5$ is $-789.3\text{ mA}$ and value of current through resistor $R_6$ is $-789.3\text{ mA}$.

Substitute $-2\text{ A}$ for $i_1$, $5\text{Ω}$ for $R_2$, $0.67\text{Ω}$ for $R_7$, $2\text{Ω}$ for $R_5$ and $5\text{Ω}$ for $R_6$ in the equation (3).

$\begin{array}{c}{i}_2=-2\text{ A}\times \frac{2\text{Ω}+5\text{Ω}+0.67\text{Ω}}{5\text{Ω}+2\text{Ω}+5\text{Ω}+0.67\text{Ω}}\\ =-2\text{ A}\times \frac{7.67\text{Ω}}{12.67\text{Ω}}\\ =-1.211\text{ A}\end{array}$

So, the value of current through resistor $R_2$ is $-1.211\text{ A}$.

Refer to the redrawn Figure 1.

Substitute $-789.3\text{ mA}$ for $i_3$, $2\text{Ω}$ for $R_3$ and $1\text{Ω}$ for $R_4$ in the equation (4).

$\begin{array}{c}{i}_4=-789.3\text{ mA}\times \frac{2\text{Ω}}{2\text{Ω}+1\text{Ω}}\\ =-789.3\text{ mA}\times \frac{2\text{Ω}}{3\text{Ω}}\\ =-526.2\text{ mA}\end{array}$

So, the value of current through resistor $R_4$ is $-526.2\text{ mA}$.

Substitute $-789.3\text{ mA}$ for $i_3$, $2\text{Ω}$ for $R_3$ and $1\text{Ω}$ for $R_4$ in the equation (5).

$\begin{array}{c}{i}_5=-789.3\text{ mA}\times \frac{1\text{Ω}}{2\text{Ω}+1\text{Ω}}\\ =-789.3\text{ mA}\times \frac{1\text{Ω}}{3\text{Ω}}\\ =-263.1\text{ mA}\end{array}$

So, the value of current through resistor $R_3$ is $-263.1\text{ mA}$.

As resistor $R_1$ is in series with independent current source $i_1$ and element connected in series have same value of current means value of current through resistor $R_1$ is $-2\text{ A}$.

Conclusion:

Thus, the value of current through $2\Omega$ resistor is $-789.3\text{ mA}$, the value of current through $5\Omega$ resistor is $-1.211\text{ A}$, value of current through $\text{1 }\Omega$ resistor is $-526.2\text{ mA}$, value of current through $2\Omega$ resistor is $-2\text{ A}$ and value of current through $2\Omega$ resistor is $-263.1\text{ mA}$.

To determine

Find value of voltage across current source.

Answer to Problem 63E

The voltage across current source is $-10.055\text{V}$.

Explanation of Solution

Formula used:

The expression for voltage is as follows.

$v=iR$ (6)

Here,

$v$ is the voltage ,

$i$ is the current and

$R$ is value of resistance.

Calculation:

Refer to the redrawn Figure 3.

The expression for voltage $v_1$ is as follows.

$v_1=v_2+v_3$ (7)

Here,

$v_1$ is voltage across independent current source $i_1$,

$v_2$ is voltage across resistor $R_1$ and

$v_3$ is voltage across resistor $R_2$.

Refer to the redrawn Figure 3.

Substitute $-2\text{ A}$ for $i$ and $2\text{Ω}$ for $R$ in equation (6).

$\begin{array}{c}v=\left(-2\text{ A}\right)\left(2\text{Ω}\right)\\ =-4\text{V}\end{array}$

So value of voltage $v_2$ is $-4\text{V}$.

Substitute $-1.211\text{ A}$ for $i$ and $5\text{Ω}$ for $R$ in equation (6).

$\begin{array}{c}v=\left(-1.211\text{ A}\right)\left(5\text{Ω}\right)\\ =-6.055\text{V}\end{array}$

So value of voltage $v_3$ is $-6.055\text{V}$.

Substitute $-4\text{V}$ for $v_2$ and $-6.055\text{V}$ for $v_3$ in equation (7).

$\begin{array}{c}{v}_1=-4\text{V}+-6.055\text{V}\\ =-10.055\text{V}\end{array}$

Conclusion:

Thus, the voltage across current source is $-10.055\text{V}$.