Chapter 3, Problem 23E

(a) Determine a numerical value for each current and voltage (i₁, v₁, etc.) in the circuit of Fig. 3.64. (b) Calculate the power absorbed by each element and verify that they sum to zero.

FIGURE 3.64

To determine

Find the value of current and voltage in the circuit.

Answer to Problem 23E

The current $i_1$ is $10.33\text{A}$, $i_2$ is $0.33\text{A}$, $i_3$ is $10\text{A}$, $i_4$ is $9.67\text{A}$ and $i_5$ is $0.33\text{A}$.

The voltage $v_1$ is $2\text{V}$, $v_2$ is $2\text{V}$, $v_3$ is $0.33\text{V}$, $v_4$ is $1.67\text{V}$ and $v_5$ is $1.67\text{V}$.

Explanation of Solution

Calculation:

The redrawn circuit is shown in Figure 1

Refer to the Figure 1.

Since there is no voltage drop across branch $AC$, Therefore the node voltage at $C$ is same as voltage source $v_1$.

$v_2=v_1$ (1)

Here,

$v_1$ is the voltage supply from branch $AB$ and

$v_2$ is the voltage at node $C$.

Refer to the Figure 1.

The expression for KCL at node $C$ is,

$i_1=i_2+i_3$ (2)

Here,

$i_1$ is the current flowing from branch $AB$,

$i_2$ is the current flowing from branch $CD$ and

$i_3$ is the current flowing from branch $CE$.

Refer to the Figure 1.

By ohm’s law the voltage across branch $CD$ is,

$v_2=i_2\times R_1$ (3)

Here,

$R_1$ is the resistance across branch $CD$.

Refer to the Figure 1

The expression for current $i_3$ flowing through current dependent source in branch $CE$ is,

$i_3=5v_1$ (4)

Refer to the Figure 1.

The expression for voltage $v_4$ across the voltage dependent source in branch $EF$ is,

$v_4=5i_2$ (5)

Here,

$v_4$ is the voltage across voltage dependent source across branch $EF$.

Refer to the Figure 1.

Since, the voltage across branch $EF$ is the only voltage across node $E$. Therefore, the expression for voltage $v_5$ across branch $GH$ is as follows,

$v_5=v_4$ (6)

Here,

$v_5$ is the voltage across voltage dependent source across branch $EF$.

By ohm’s law the voltage across branch is $GH$,

$v_5=i_5\times R_2$ (7)

The expression for KCL at node is $E$,

$i_3=i_4+i_5$ (8)

The expression for KVL across the loop $CEFDC$ is,

$v_2-v_3-v_4=0$ (9)

Refer to the Figure 1.

Substitute $2\text{V}$ for $v_1$ in the equation (1).

$v_2=2\text{V}$

Substitute $2\text{V}$ for $v_2$ and $6\Omega$ for $R_1$ in the equation (3).

$2\text{V}=i_2\times 6\Omega$

Rearrange for $i_2$.

$\begin{array}{c}{i}_2=\frac{2\text{V}}{6\Omega }\\ i_2=0.33\text{A}\end{array}$

Substitute $2\text{V}$ for $v_1$ in the equation (4).

$\begin{array}{c}{i}_3=\left(5\Omega \right)2\text{V}\\ \text{=10}\text{A}\end{array}$

Substitute $0.33\text{A}$ for $i_2$ and $10\text{A}$ for $i_3$ in equation (2).

$\begin{array}{c}{i}_1=0.33\text{A+10}\text{A}\\ \text{=10}\text{.33}\text{A}\end{array}$

Substitute $0.33\text{A}$ for $i_2$ in the equation (5).

$\begin{array}{c}{v}_4=5\times 0.33\text{A}\\ \text{=1}\text{.67}\text{V}\end{array}$

Substitute $1.67\text{V}$ for $v_4$ in the equation (6).

$v_5=1.67\text{V}$

Substitute $1.67\text{V}$ for $v_5$ and $5\Omega$ for $R_2$ in the equation (7).

$1.67\text{V}=i_5\times 5\Omega$

Rearrange for $i_5$.

$\begin{array}{c}{i}_5=\frac{1.67\text{V}}{5\Omega }\\ =0.33\text{A}\end{array}$

Substitute $10\text{A}$ for $i_3$ and $0.33\text{A}$ for $i_5$ in the equation (8).

$10\text{A}=i_4+0.33\text{A}$

Rearrange for $i_4$

$i_4=9.67\text{A}$

Substitute $2\text{V}$ for $v_2$ and $1.67\text{V}$ for $v_5$ in the equation (9).

$2\text{V}-v_3-1.67\text{V=0}$

Rearrange for $v_3$.

${\text{v}}_3=0.33\text{V}$

Conclusion:

Thus, the current $i_1$ is $10.33\text{A}$, $i_2$ is $0.33\text{A}$, $i_3$ is $10\text{A}$, $i_4$ is $9.67\text{A}$ and $i_5$ is $0.33\text{A}$.

The voltage $v_1$ is $2\text{V}$, $v_2$ is $2\text{V}$, $v_3$ is $0.33\text{V}$, $v_4$ is $1.67\text{V}$ and $v_5$ is $1.67\text{V}$.

To determine

Calculate the power absorbed by each element and verify total sum of power absorbed is zero.

Answer to Problem 23E

The power absorbed by independent voltage source is $-20.67\text{W}$,

The power absorbed by $6\Omega$ resistor is $0.67\text{W}$,

The power absorbed by dependent current source is $3.3\text{W}$,

The power absorbed by dependent voltage source is $16.15\text{W}$,

The power absorbed by $\text{5 }\Omega$ resistor is $0.55\text{W}$.

Explanation of Solution

Formula used:

Refer to the Figure 1.

The expression for power supply by the voltage source from branch $AB$ is,

$p_1=v_1{i}_1$ (10)

Here,

$p_1$ is the power absorbed by voltage source.

The expression for power absorbed by resistance $R_1$ across branch $CD$ is,

$p_2=i^2{R}_1$ (11)

The expression for power supply by the current dependent source from branch $CE$ is,

$p_3=v_3{i}_3$ (12)

The expression for power supply by the voltage dependent source from branch $EF$ is,

$p_4=v_4{i}_4$ (13)

The expression for power dissipate by the resistance $R_2$ across the branch $GH$ is,

$p_5=v_5{i}_5$ (14)

The expression for sum of total power absorbed in the circuit is,

$p_t=\left(p_1+p_2+p_3+p_4+p_5\right)$ (15)

Calculation:

Refer to the Figure 1.

Substitute $10.33\text{A}$ for $i_1$ and $2\text{V}$ for $v_1$ in the equation (10).

$\begin{array}{c}{p}_1=2\text{V}\times -\text{10}\text{.33}\text{A}\\ \text{=}-\text{20}\text{.66}\text{W}\end{array}$

Substitute $0.33\text{A}$ for $i_2$ and $6\Omega$ for $R_1$ in the equation (11).

$\begin{array}{c}{p}_2={\left(0.33\text{A}\right)}^2\times 6\Omega \\ \text{=}0.67\text{W}\end{array}$

Substitute $10\text{A}$ for $i_3$ and $0.33\text{V}$ for $v_3$ in the equation (12).

$\begin{array}{c}{p}_3=0.33\text{V}\times 10\text{A}\\ \text{=3}\text{.3}\text{W}\end{array}$

Substitute $9.67\text{A}$ for $i_4$ and $1.67\text{V}$ for $v_4$ in the equation (13).

$\begin{array}{c}{p}_4=1.67\text{V}\times 9.67\text{A}\\ \text{=16}\text{.15}\text{W}\end{array}$

Substitute $0.33\text{A}$ for $i_5$ and $5\Omega$ for $R_2$ in the equation (14).

$\begin{array}{c}{p}_5={\left(0.33\right)}^2\text{A}\times 5\Omega \\ \text{=0}\text{.55}\text{W}\end{array}$

Substitute $-20.67\text{W}$ for $p_1$, $0.67\text{W}$ for $p_2$, $3.3\text{W}$ for $p_3$, $16.15\text{W}$ for $p_4$ and $0.55\text{W}$ for $p_5$ in the equation (15).

$\begin{array}{c}{p}_t=-20.67\text{W+0}\text{.67}\text{W+3}\text{.3}\text{W+16}\text{.15}\text{W+0}\text{.55}\text{W}\\ =\text{0}\text{W}\end{array}$

Conclusion:

Thus, the power absorbed by independent voltage source is $-20.67\text{W}$,

The power absorbed by $6\Omega$ resistor is $0.67\text{W}$,

The power absorbed by dependent current source is $3.3\text{W}$,

The power absorbed by dependent voltage source is $16.15\text{W}$,

The power absorbed by $\text{5 }\Omega$ resistor is $0.55\text{W}$.