Chapter 7, Problem 29E

Calculate the voltage labeled v_x in Fig. 7.52, assuming the circuit has been running a very long time, if (a) a 10 Ω resistor is connected between terminals x and y; (b) a 1 H inductor is connected between terminals x and y; (c) a 1 F capacitor is connected between terminals x and y; (d) a 4 H inductor in parallel with a 1 Ω resistor is connected between terminals x and y.

FIGURE 7.52

To determine

Find the voltage $v_x$.

Answer to Problem 29E

The voltage $v_x$ across the $12\Omega$ resistor is $-18.33\text{V}$.

Explanation of Solution

Given data:

Value of resistance connected between terminals $x$ and $y$ is $10\Omega$.

Calculation:

The redrawn circuit is shown in Figure 1 as follows:

Here,

$v_s$ is the voltage supply from branch $AC$,

$L_1$ and $R_1$ connected across branch $AB$,

$R_3$ resistance across branch $BD$,

$C_1$ and $L_3$ is connected across branch $BE$,

$i_s$ is the current supply from branch $EF$,

$R_4$ is the resistance across branch $EG$,

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

$L_2$ and $C_2$ connected across branch $IJ$.

The redrawn circuit is shown in Figure 2.

Refer to the Figure 2:

When steady state is reached, all the inductors connected are short circuited and all the capacitors connected are open circuited.

The equivalent resistance for series combination across branch is as follows:

$R=\left(R_2+R_3\right)$ (1)

Substitute $10\Omega$ for $R_2$ and $20\text{}\Omega$ for $R_3$ in the equation (1):

$R=\left(10\Omega +20\Omega \right)$

$R=30\Omega$ (2)

The redrawn circuit is shown in Figure 3 as follows:

Refer to the redrawn Figure 3:

The expression for the nodal analysis at node voltage $v$ is,

$\frac{v_{}-v_s}{R_1}+\frac{v}{R}+i_s+\frac{v}{R_4+R_5}=0$ (3)

Here,

$v$ is the node voltage across node $B$.

The expression for the current across the resistance $R_4$ and $R_5$ is as follows:

$i=\frac{v}{R_4+R_5}$ (4)

The expression for the voltage across the resistance $R_5$ is as follows:

$v_x=i\times R_5$ (5)

The current across the resistances $R_4$ and $R_5$ is same.

Substitute $5\text{A}$ for $i_s$ and $1\text{V}$ for $v_s$, $20\Omega$ for $R_1$, $30\Omega$ for $R$, $15\Omega$ for $R_4$ and $12\Omega$ for $R_5$ in equation (3):

$\begin{array}{l}\frac{v_{}-1\text{V}}{20\Omega }+\frac{v}{30\Omega }+5\text{A}+\frac{v}{15\Omega +12\Omega }=0\\ v\left(\frac{1}{20\Omega }+\frac{1}{30\Omega }\text{+}\frac{1}{27\text{}\Omega }\right)=-5\text{A+}\frac{1\text{V}}{20\Omega }\\ v\left(0.05+0.033+0.037\right)=-5\text{A+0}\text{.05}\text{A}\end{array}$

Solve for $v_{}$:

$\begin{array}{l}v\left(0.12\right)=-4.95\text{A}\\ v=\frac{-4.95\text{A}}{0.12}\end{array}$

$v=-41.25\text{V}$ (6)

Substitute $-41.25\text{V}$ for $v$ and $15\Omega$ for $R_4$ and $12\Omega$ for $R_5$ in the equation (4):

$\begin{array}{c}i=\frac{-41.25\text{V}}{15\Omega \text{+12 }\Omega }\\ =\frac{-41.25\text{ V}}{27\Omega }\\ =-1.52\text{A}\end{array}$

Substitute $-1.52\text{A}$ for $i$ and $12\Omega$ for $R_5$ in the equation (5):

$\begin{array}{c}{v}_x=\left(-1.52\text{A}\times \text{12}\Omega \right)\\ =-18.33\text{V}\end{array}$

Conclusion:

Thus, the voltage $v_x$ across the $12\Omega$ resistor is $-18.33\text{V}$.

To determine

Find the voltage $v_x$.

Answer to Problem 29E

The voltage $v_x$ across the $12\Omega$ resistor is $-16\text{V}$.

Explanation of Solution

Given Data:

Value of inductor connected between terminals $x$ and $y$ is $1\text{H}$.

Calculation:

The redrawn circuit with inductor connected across branch $x$ and $y$ is shown in Figure 4 as follows:

The redrawn circuit is shown in Figure 4 as follows:

Refer to the Figure 5:

When steady state is reached, the inductor connected across branch $x$ and $y$ and across branches $\text{AB}$, $\text{BE}$ and $\text{IJ}$ are short circuited and capacitor across branchesare open circuited.

The redrawn circuit is shown in Figure 6 as follows:

The expression for the nodal analysis at node voltage $v$ is:

$\frac{v_{}-v_s}{R_1}+\frac{v}{R_3}+i_s+\frac{v}{R_4+R_5}=0$ (7)

The expression for the current across the resistance $R_4$ and $R_5$ is:

$i=\frac{v}{R_4+R_5}$ . (8)

The expression for the current across the resistance $R_5$ is:

$i=\frac{v}{R_5}$ (9)

The current across the resistances $R_4$ and $R_5$ is same.

Substitute $5\text{A}$ for $i_s$ and $1\text{V}$ for $v_s$, $20\Omega$ for $R_1$, $20\Omega$ for $R_3$, $15\Omega$ for $R_4$ and $12\Omega$ for $R_5$ in equation (7):

$\begin{array}{l}\frac{v_{}-1\text{V}}{20\Omega }+\frac{v}{20\Omega }+5\text{A}+\frac{v}{15\Omega +12\Omega }=0\\ v\left(\frac{1}{20\Omega }+\frac{1}{20\Omega }\text{+}\frac{1}{27\text{}\Omega }\right)=-5\text{A}+\frac{1\text{V}}{20\Omega }\\ v\left(0.05+0.05+0.037\right)=-5\text{A}+\text{0}\text{.05}\text{A}\end{array}$

Solve for $v_{}$:

$\begin{array}{l}v\left(0.137\right)=-4.95\text{A}\\ v=\frac{-4.95\text{A}}{0.137}\end{array}$

$v=-36\text{V}$ (10)

Substitute $-36\text{V}$ for $v$ and $15\Omega$ for $R_4$ and $12\Omega$ for $R_5$ in equation (8):

$\begin{array}{c}i=\frac{-36\text{V}}{15\Omega \text{+12 }\Omega }\\ =\frac{-36\text{ V}}{27\Omega }\\ =-1.33\text{A}\end{array}$

Substitute $-1.33\text{A}$ for $i$ and $12\Omega$ for $R_5$ in equation (9):

$\begin{array}{c}{v}_x=\left(-1.33\text{A}\times \text{12}\Omega \right)\\ =-16\text{V}\end{array}$

Conclusion:

Thus, the voltage $v_x$ across the $12\Omega$ resistor is $-16\text{V}$.

To determine

Find the voltage $v_x$.

Answer to Problem 29E

The voltage $v_x$ across the $12\Omega$ resistor is $-25.25\text{V}$.

Explanation of Solution

Given data:

Value of capacitor connected between terminals $x$ and $y$ is $1\text{F}$.

Calculation:

The redrawn circuit of capacitor connected across branch $x$ and $y$ is shown in Figure 7 as follows:

Refer to the redrawn Figure 7:

The redrawn circuit is shown in Figure 8 as follows:

When steady state is reached, thecapacitors connected across branch $x$ and $y$ and other branches are open circuited and inductor connected is short circuited.

The redrawn circuit is shown in Figure 9 as follows:

Refer to the Figure 9:

The expression for the nodal analysis at node voltage $v_{}$ is as follows:

$\frac{v_{}-v_s}{R_1}+i_s+\frac{v}{R_4+R_5}=0$ (11)

The expression for the current across the resistance $R_4$ and $R_5$ is as follows:

$i=\frac{v}{R_4+R_5}$ . (12)

The expression for the voltage across the resistance $R_5$ is,

$v_x=i\times R_5$ (13)

The current across the resistances $R_4$ and $R_5$ is same.

Substitute $5\text{A}$ for $i_s$ and $1\text{V}$ for $v_s$, $20\Omega$ for $R_1$, $15\Omega$ for $R_4$ and $12\Omega$ for $R_5$ in equation (11):

$\begin{array}{l}\frac{v_{}-1\text{V}}{20\Omega }+5\text{A}+\frac{v}{15\Omega +12\Omega }=0\\ v\left(\frac{1}{20\Omega }\text{+}\frac{1}{27\text{}\Omega }\right)=-5\text{A+}\frac{1\text{V}}{20\Omega }\\ v\left(0.05+0.037\right)=-5\text{A+0}\text{.05}\text{A}\end{array}$

Solve for $v$:

$\begin{array}{c}v\left(0.087\right)=-4.95\text{A}\\ v=\frac{-4.95\text{A}}{0.087}\\ =-56.81\text{V}\end{array}$

Substitute $-56.81\text{V}$ for $v$ and $15\Omega$ for $R_4$ and $12\Omega$ for $R_5$ in equation (12):

$\begin{array}{c}i=\frac{-56.81\text{V}}{15\Omega \text{+12 }\Omega }\\ =\frac{-56.81\text{ V}}{27\Omega }\\ =-2.10\text{A}\end{array}$

Substitute $-2.10\text{A}$ for $i$ and $12\Omega$ for $R_5$ in equation (13):

$\begin{array}{c}{v}_x=\left(-2.10\text{A}\times \text{12}\Omega \right)\\ =-25.25\text{V}\end{array}$

Conclusion:

Thus, the voltage $v_x$ across the $12\Omega$ resistor is $-25.25\text{V}$.

To determine

Find the voltage $v_x$.

Answer to Problem 29E

The voltage $v_x$ across the $12\Omega$ resistor is $-16\text{V}$.

Explanation of Solution

Given Data:

Value of inductor connected between terminal $x$ and $y$ is $4\text{H}$ and

Value of resistor connected in parallel with inductor across terminal $x$ and $y$ is $1\Omega$.

Calculation:

The redrawn circuit is shown in Figure 10:

Refer to the redrawn Figure 10:

The redrawn circuit is shown in Figure 11 as follows:

When steady state is reached, the inductor connected across branch $x$ and $y$ and other branches are short circuited, resistance across branch $x$ and $y$ is removed because current flow from less resistance path and capacitors across branches are open circuited.

The redrawn circuit is shown in Figure 12,

Refer to the Figure 12:

The expression for the nodal analysis at node voltage $v_{}$ is as follows:

$\frac{v_{}-v_s}{R_1}+\frac{v}{R_3}+i_s+\frac{v}{R_4+R_5}=0$ (14)

The expression for the current across the resistance $R_4$ and $R_5$ is as follows:

$i=\frac{v}{R_4+R_5}$ . (15)

The expression for the voltage across the resistance $R_5$ is as follows:

$v_x=i\times R_5$ (16)

The current across the resistances $R_4$ and $R_5$ is same.

Substitute $5\text{A}$ for $i_s$ and $1\text{V}$ for $v_s$, $20\Omega$ for $R_1$, $20\Omega$ for $R_3$, $15\Omega$ for $R_4$ and $12\Omega$ for $R_5$ in equation (14):

$\begin{array}{l}\frac{v_{}-1\text{V}}{20\Omega }+\frac{v}{20\Omega }+5\text{A}+\frac{v}{15\Omega +12\Omega }=0\\ v\left(\frac{1}{20\Omega }+\frac{1}{20\Omega }\text{+}\frac{1}{27\text{}\Omega }\right)=-5\text{A}+\frac{1\text{V}}{20\Omega }\\ v\left(0.05+0.05+0.037\right)=-5\text{A}+\text{0}\text{.05}\text{A}\end{array}$

Solve for $v_{}$:

$\begin{array}{c}v\left(0.137\right)=-4.95\text{A}\\ v=\frac{-4.95\text{A}}{0.137}\\ =-36\text{V}\end{array}$

Substitute $-36\text{V}$ for $v$ and $15\Omega$ for $R_4$ and $12\Omega$ for $R_5$ in equation (15):

$\begin{array}{c}i=\frac{-36\text{V}}{15\Omega \text{+12 }\Omega }\\ =\frac{-36\text{ V}}{27\Omega }\\ =-1.33\text{A}\end{array}$

Substitute $-1.33\text{A}$ for $i$ and $12\Omega$ for $R_5$ in equation (16):

$\begin{array}{c}{v}_x=\left(-1.33\text{A}\times \text{12}\Omega \right)\\ =-16\text{V}\end{array}$

Conclusion:

Thus, the voltage $v_x$ across the $12\Omega$ resistor is $-16\text{V}$.