Chapter 18, Problem 30P

For the circuit shown in Figure P18.30, use Kirchhoff’s rules to obtain equations for (a) the upper loop, (b) the lower loop, and (c) the node on the left side. In Figure P18.29 each case suppresses units for clarity and simplify, combining like terms. (d) Solve the node equation for I₃₆. (e) Using the equation found in (d), eliminate I₃₆ from the equation found in part (b). (f) Solve the equations found in part (a) and part (e) simultaneously for the two unknowns for I₁₈ and I₁₂, respectively. (g) Substitute the answers found in part (f) into the node equation found in part (d), solving for I₃₆. (h) What is the significance of the negative answer for I₁₂?

Figure P18.30

To determine

The equation for the upper loop by using Kirchhoff’s rule

Answer to Problem 30P

The equation for the upper loop by using Kirchhoff’s rule is $\left(13.0\text{Ω}\right)I_{18}+\left(18.0\text{Ω}\right)I_{12}=30.0\text{V}$

Explanation of Solution

Given Info:

Resistance $R_1$ is $8.00\text{Ω}$ Resistance $R_2$ is $5.00\text{Ω}$, Resistance $R_3$ is $11.0\text{Ω}$, Resistance $R_4$ is $7.00\text{Ω}$, Resistance $R_5$ is $5.00\text{Ω}$, Voltage ${\epsilon }_1$ is $18.0\text{V}$, and Voltage ${\epsilon }_2$ is $12.0\text{V}$.

Formula to calculate the equation for upper loop is by Kirchhoff’s voltage law

    ${\epsilon }_1+{\epsilon }_2-(R_1+R_2)I_{18}-(R_3+R_4)I_{12}=0$

Substitute $8.00\text{Ω}$, for $R_1$, $11.0\text{Ω}$,for $R_2$, $5.00\text{Ω}$ for $R_3$, $7.00\text{Ω}$ for $R_4$ $18.0\text{V}$ for ${\epsilon }_1$$12.0\text{V}$ for ${\epsilon }_2$ in above equation, to find the equation for upper loop in the circuit,

    $\begin{array}{c}\left(8.00\text{Ω}+5.00\text{Ω}\right)I_{18}+\left(11.0\text{Ω}+7.00\text{Ω}\right)I_{12}=18.0\text{V}+12.0\text{V}\\ 13.0\text{Ω}{I}_{18}+18.0\text{Ω}{I}_{12}=30.0\text{V}\end{array}$ (I)

• ${\epsilon }_1,{\epsilon }_2,{\epsilon }_3$ is voltage in the circuits,
• $R_1,R_2,R_3,R_4$ is corresponding resistance in the circuits,
• $I_{18},I_{12},I_{36}$ is corresponding current in the circuits,

Conclusion:

The equation for the upper loop by using Kirchhoff’s rule is $\left(13.0\text{Ω}\right)I_{18}+\left(18.0\text{Ω}\right)I_{12}=30.0\text{V}$.

To determine

The equation for the lower loop by using Kirchhoff’s rule.

Answer to Problem 30P

The equation for the lower loop by using Kirchhoff’s rule is $\left(5.0\text{Ω}\right){\text{I}}_{36}-\left(18.0\text{Ω}\right)I_{12}=24.0\text{V}$

Explanation of Solution

Given Info: Resistance $R_2$ is $5.00\text{Ω}$, Resistance $R_3$ is $11.0\text{Ω}$, Resistance $R_4$ is $7.00\text{Ω}$, Resistance $R_5$ is $5.00\text{Ω}$., Voltage ${\epsilon }_2$ is $12.0\text{V}$ Voltage ${\epsilon }_3$ is $30.0\text{V}$,

Formula to calculate the equation for the lower loop is by Kirchhoff’s voltage law

    ${\epsilon }_3-{\epsilon }_2-R_3{I}_{36}-(R_2+R_4)I_{12}=0$

Substitute $11.0\text{Ω}$, for $R_2$, $5.00\text{Ω}$, for $R_3$, $7.00\text{Ω}$ for $R_4$, $12.0\text{V}$ for ${\epsilon }_2$, $36.0\text{V}$ for ${\epsilon }_3$ in above equation, to find the equation for lower loop in the circuit,

  $\begin{array}{c}36.0\text{V}-12.0\text{V}-5.00\Omega I_{36}-(11.0\Omega +7.00\Omega )I_{12}=0\\ 5.00\Omega I_{36}-18.0\Omega I_{12}=24.0\text{V}\end{array}$ (II)

Conclusion:

The equation for the lower loop by using Kirchhoff’s rule is $\left(5.00\Omega \right)I_{36}-\left(18.0\Omega \right)I_{12}=\left(24.0\right)\text{V}$

To determine

The equation of node on the left side by using Kirchhoff’s rule.

Answer to Problem 30P

The equation of node on the left side by using Kirchhoff’s rule is $I_{18}=I_{12}+I_{36}$

Explanation of Solution

Given Info: $I_{18},I_{12},I_{36}$ is corresponding current in the circuits

Formula to calculate node on the left side is by Kirchhoff’s junction rule

  $I_1=I_2+I_3$

  $I_{18}=I_{12}+I_{36}$ (III)

Conclusion:

The equation of node on the left side by using Kirchhoff’s rule is $I_{18}=I_{12}+I_{36}$

To determine

The equation of node for current $I_{36}$ by using Kirchhoff’s rule.

Answer to Problem 30P

The equation of node for current $I_{36}$ by using Kirchhoff’s rule is $I_{36}=I_{18}-I_{12}$

Explanation of Solution

Given Info:$I_{18},I_{12},I_{36}$ is corresponding current in the circuits.

Formula to calculate node on the left side is by Kirchhoff’s junction rule

  $\begin{array}{c}{I}_1=I_2+I_3\\ I_{18}=I_{12}+I_{36}\end{array}$

Rewrite for $I_{36}$

  $I_{36}=I_{18}-I_{12}$ (IV)

Conclusion:

The equation of node for current $I_{36}$ by using Kirchhoff’s rule is $I_{36}=I_{18}-I_{12}$

To determine

The equation after eliminating the current $I_{36}$ from the lower loop equation by using Kirchhoff’s rule.

Answer to Problem 30P

The equation after eliminating the current $I_{36}$ from the lower loop is $\left(5.00\text{Ω}\right)I_{18}-\left(23.0\text{Ω}\right)I_{12}=24.0\text{V}$

Explanation of Solution

Given Info: Equation of node for current $I_{36}$ by using Kirchhoff’s rule $I_{36}=I_{18}-I_{12}$.

Formula to calculate the equation after eliminating the current $I_{36}$ is by lower loop equation in the circuit

  $5.00\Omega I_{36}-18.0\Omega I_{12}=24.0\text{V}$ (II)

Substitute $I_{18}-I_{12}$ for $I_{36}$ to find equation after eliminating the current $I_{36}$

    $\begin{array}{c}5.00\text{Ω}(I_{18}-I_{12})-18.0\text{Ω}{I}_2=24.0\text{V}\\ 5.00\text{Ω}{I}_{18}-5.00\text{Ω}{I}_{{}_{12}}-18.0\text{Ω}{I}_{12}=24.0\text{V}\end{array}$ (V)

Conclusion:

The equation after eliminating the current $I_{36}$ from the lower loop equation by using Kirchhoff’s rule is $\left(5.00\text{Ω}\right)I_{18}-\left(23.0\text{Ω}\right)I_{12}=24.0\text{V}$

To determine

To determine the values of the current $I_{18}$.and $I_{12}$

Answer to Problem 30P

The values of the current in the circuit $I_{18}$ is $2.88\text{A}$ and $I_{12}$ is $-0.413\text{A}$

Explanation of Solution

Given Info: $I_{18},I_{12}$ is corresponding current in the circuits

Rewrite (I) and (V).

  $13.0\text{Ω}{I}_{18}+18.0\text{Ω}{I}_{12}=30.0\text{V}$ (I)

    $5.00\text{Ω}{I}_{18}-23.0\text{Ω}{I}_{12}=24.0\text{V}$ (V)

Solve the above simultaneous equation to obtain $I_{18}$.

  $I_{18}=2.88\text{A}$

Substitute $2.88\text{A}$ for $I_{18}$ in (I) to find $I_{12}$.

  $\begin{array}{c}13.0\Omega (2.88\text{A})+18.0\Omega I_{12}=30.0\text{V}\\ 37.44\text{V}+18.0\Omega I_{12}=30.0\text{V}\\ I_{12}=\frac{-7.44\text{V}}{18.0\Omega }\\ =-0.413\text{A}\end{array}$

Conclusion:

The values of the current in the circuit $I_{18}$ is $2.88\text{A}$ and $I_{12}$ is $-0.413\text{A}$

To determine

To determine the values of the current $I_{36}$

Answer to Problem 30P

The values of the current $I_{36}$ is $3.30A$

Explanation of Solution

Given Info: $I_{18},I_{12},I_{36}$ is corresponding current in the circuits

Formula to calculate the current is by the equation of node for current $I_{36}$ by using Kirchhoff’s rule

  $I_{36}=I_{18}-I_{12}$

Substitute $2.88\text{A}$ for $I_{18}$ and $-0.413\text{A}$ for $I_{12}$ to find $I_{36}$.

  $\begin{array}{c}{I}_{36}=I_{18}-I_{12}\\ =2.88\text{A}-\left(-0.413\text{A}\right)\\ =3.30\text{A}\end{array}$

Conclusion:

The values of the current $I_{36}$ is $3.30A$.

To determine

To significance of the negative answer for $I_{12}$

Answer to Problem 30P

The negative Sign in the value implies that the current is flowing in the reverse direction.

Explanation of Solution

Current $I_{12}$ has a $-0.413\text{A}$ in the given circuit.

When the circuit is solved, a negative value for the variable means that the actual direction of current through that circuit element is opposite. Accordingly, the current flowing in the circuit is reverse to other current in the same circuit

Conclusion:

The negative Sign in the value implies that the current is flowing in the reverse direction.