Chapter 28, Problem 23P

(a)

To determine

The current in each branch of the circuit.

Answer to Problem 23P

The current in right hand branch is $1.31\text{A}$, current in middle branch is $0.462\text{A}$ and current in left hand branch is $\text{0}\text{.846}\text{A}$.

Explanation of Solution

Write the expression based on junction rule.

    ${\displaystyle \sum _{\text{junction}}I}=0$

Here, ${\displaystyle \sum _{\text{junction}}I}$ is the sum current at the junction.

Write the expression to obtain the loop rule.

    ${\displaystyle \sum _{\text{closed}\text{loop}}\Delta V}=0$

Here, ${\displaystyle \sum _{\text{closed}\text{loop}}\Delta V}$ is the potential drop across each element in a closed circuit.

The flow of current in the circuit is as shown in the figure below.

Figure-(1)

Here, $I_1,I_2$ and $I_3$ are the current in different section of the circuit.

Writer the equation of Kirchhoff’s voltage rule in loop $BCDEB$.

    $\begin{array}{c}\left(R_4+R_3\right)I_2-\left(R_1+R_2\right)I_1+E_1-E_2=0\\ E_1-E_2=\left(R_1+R_2\right)I_1-\left(R_4+R_3\right)I_2\end{array}$                           (I)

Writer the equation of Kirchhoff’s voltage rule in loop $ABEFA$.

    $\begin{array}{c}{R}_5{I}_1+\left(R_4+R_3+R_5\right)I_2-E_2=0\\ E_2=R_5{I}_1+\left(R_4+R_3+R_5\right)I_2\end{array}$                                          (II)

Write the expression based on junction rule at node $B$.

    $I_3=I_1+I_2$                                                                                                             (III)

Conclusion:

Substitute $12.0\text{V}$ for $E_1$, $4.00\text{V}$ for $E_2$, $1.00\text{Ω}$ for $R_1$, $3.00\text{Ω}$ for $R_2$, $5.00\text{Ω}$ for $R_3$ and $1.00\text{Ω}$ for $R_4$ in equation (I).

    $\begin{array}{l}12.0\text{V}-4.00\text{V}=\left(1.00\text{Ω}+3.00\text{Ω}\right)I_1-\left(1.00\text{Ω}+5.00\text{Ω}\right)I_2\\ 8\text{V}=\left(4.00\text{Ω}\right)I_1-\left(6.00\text{Ω}\right)I_2\\ 4I_1-6I_2=8\text{V}\end{array}$                           (IV)

Substitute $4.00\text{V}$ for $E_2$, $8.00\text{Ω}$ for $R_5$, $3.00\text{Ω}$ for $R_2$, $5.00\text{Ω}$ for $R_3$ and $1.00\text{Ω}$ for $R_4$ in equation (II).

    $\begin{array}{l}4.00\text{V}=\left(8.00\text{Ω}\right)I_1+\left(1.00\text{Ω}+5.00\text{Ω}+8.00\text{Ω}\right)I_2\\ 4.00\text{V}=\left(8.00\text{Ω}\right)I_1+\left(14.00\text{Ω}\right)I_2\\ 2.00\text{V}=\left(4.00\text{Ω}\right)I_1+\left(7.00\text{Ω}\right)I_2\\ 4I_1+7I_2=2.00\text{V}\end{array}$                           (V)

Solve equation (IV) and (V).

    $\begin{array}{c}-6I_2-7I_2=8\text{V}-2\text{V}\\ -13I_2=6\\ I_2=-\frac{6}{13}\text{A}\\ I_2=-0.462\text{A}\end{array}$

Substitute $-0.462\text{A}$ for $I_2$ in equation (V) to calculate $I_1$.

    $\begin{array}{c}4I_1+7\left(-0.4612\text{A}\right)=2.00\text{V}\\ 4I_1=2+3.2284\\ I_1=\frac{5.2284}{4}\text{A}\\ I_1=1.31\text{A}\end{array}$

Substitute $1.31\text{A}$ for $I_1$, $-0.462\text{A}$ for $I_2$ in equation (III) to calculate $I_3$.

    $\begin{array}{c}{I}_3=1.31\text{A}+\left(-0.462\text{A}\right)\\ =\text{0}\text{.846}\text{A}\end{array}$

Therefore, the current in right hand branch is $1.31\text{A}$, current in middle branch is $0.462\text{A}$ and current in left hand branch is $\text{0}\text{.846}\text{A}$.

(b)

To determine

The energy delivered by each battery.

Answer to Problem 23P

The energy delivered by $12.0\text{V}$ battery is $1.88\text{kJ}$ and the energy delivered by $4.00\text{V}$ battery is $-222\text{J}$.

Explanation of Solution

Write the expression to obtain power.

    $P=VI$

Here, $P$ is the power, $I$ is the current and $V$ is the voltage.

Write the expression to obtain energy.

    $E=Pt$                                                                                                       (VI)

Here, $E$ is the energy, $P$ is the power and $t$ is the time for which power is applied.

Substitute $VI$ for $P$ in the above equation.

    $E=VIt$                                                                                                    (VII)

Conclusion:

Substitute $12.0\text{V}$ for $V$, $1.31\text{A}$ for $I$ and $2.00\min$ for $t$ in equation (VII) to calculate $E_1$

    $\begin{array}{c}{E}_1=\left(12.0\text{V}\right)\left(1.31\text{A}\right)\left(2.00\min \right)\\ =\left(12.0\text{V}\right)\left(1.31\text{A}\right)\left(2.00\min \times \frac{60\text{s}}{1\min }\right)\\ =1880\text{J}\times \frac{1\text{kJ}}{{10}^{-3}\text{J}}\\ =1.88\text{kJ}\end{array}$

Here, $E_1$ is the energy delivered by the $12.0\text{V}$ battery.

Substitute $4.00\text{V}$ for $V$, $-0.04612\text{A}$ for $I$ and $2.00\min$ for $t$ in equation (IV) to calculate $E_2$

    $\begin{array}{c}{E}_2=\left(4.00\text{V}\right)\left(-0.04612\text{A}\right)\left(2.00\min \right)\\ =\left(4.00\text{V}\right)\left(-0.04612\text{A}\right)\left(2.00\min \times \frac{60\text{s}}{1\min }\right)\\ =-222\text{J}\end{array}$

Here, $E_2$ is the energy delivered by the $4.00\text{V}$ battery.

Therefore, the energy delivered by $12.0\text{V}$ battery is $1.88\text{kJ}$ and the energy delivered by $4.00\text{V}$ battery is $-222\text{J}$.

(c)

To determine

The energy delivered to each resistor.

Answer to Problem 23P

The energy delivered to the $1\text{Ω}$ resistor connected to $12.0\text{V}$ battery is $205\text{J}$, the energy delivered to the $3\text{Ω}$ resistor is $616\text{J}$, the energy delivered to the $5\text{Ω}$ resistor is $128\text{J}$, the energy delivered to the $8\text{Ω}$ resistor is $687\text{J}$ and the energy delivered to the $1\text{Ω}$ resistor connected to $4.00\text{V}$ battery is $25.6\text{J}$.

Explanation of Solution

Write the expression of power in terms of current and resistance.

    $P=I^{2}R$

Here, $P$ is the power, $I$ is the current and $R$ is the resistance.

Substitute $I^{2}R$ for $P$ in equation (VI) to calculate $E$.

    $E=I^{2}Rt$                                                                                                            (VIII)

Conclusion:

Substitute $1\text{Ω}$ for $R$, $-0.4612\text{A}$ for $I$ and $2.00\min$ for $t$ in equation (VIII) to calculate $E_{1\text{Ω}\text{4}\text{V}}$.

    $\begin{array}{c}{E}_{1\text{Ω}\text{4}\text{V}}={\left(-0.4612\text{A}\right)}^2\left(1\text{Ω}\right)\left(2.00\min \right)\\ ={\left(-0.4612\text{A}\right)}^2\left(1\text{Ω}\right)\left(2.00\min \times \frac{60\text{s}}{1\min }\right)\\ =25.6\text{J}\end{array}$

Here, $E_{1\text{Ω}\text{4}\text{V}}$ is the energy delivered to the $1\text{Ω}$ resistor connected to $4.00\text{V}$ battery.

Substitute $8\text{Ω}$ for $R$, $0.848\text{A}$ for $I$ and $2.00\min$ for $t$ in equation (VIII) to calculate $E_{8\text{Ω}}$.

    $\begin{array}{c}{E}_{8\text{Ω}}={\left(0.848\text{A}\right)}^2\left(8\text{Ω}\right)\left(2.00\min \right)\\ ={\left(0.848\text{A}\right)}^2\left(8\text{Ω}\right)\left(2.00\min \times \frac{60\text{s}}{1\min }\right)\\ =687\text{J}\end{array}$

Here, $E_{8\text{Ω}}$ is the energy delivered to the $8\text{Ω}$ resistor.

Substitute $5\text{Ω}$ for $R$, $-0.4612\text{A}$ for $I$ and $2.00\min$ for $t$ in equation (VIII) to calculate $E_{5\text{Ω}}$.

    $\begin{array}{c}{E}_{5\text{Ω}}={\left(-0.4612\text{A}\right)}^2\left(5\text{Ω}\right)\left(2.00\min \right)\\ ={\left(-0.4612\text{A}\right)}^2\left(5\text{Ω}\right)\left(2.00\min \times \frac{60\text{s}}{1\min }\right)\\ =128\text{J}\end{array}$

Here, $E_{5\text{Ω}}$ is the energy delivered to the $5\text{Ω}$ resistor.

Substitute $3\text{Ω}$ for $R$, $1.31\text{A}$ for $I$ and $2.00\min$ for $t$ in equation (VIII) to calculate $E_{3\text{Ω}}$.

    $\begin{array}{c}{E}_{3\text{Ω}}={\left(1.31\text{A}\right)}^2\left(3\text{Ω}\right)\left(2.00\min \right)\\ ={\left(1.31\text{A}\right)}^2\left(3\text{Ω}\right)\left(2.00\min \times \frac{60\text{s}}{1\min }\right)\\ =616\text{J}\end{array}$

Here, $E_{3\text{Ω}}$ is the energy delivered to the $3\text{Ω}$ resistor.

Substitute $1\text{Ω}$ for $R$, $1.31\text{A}$ for $I$ and $2.00\min$ for $t$ in equation (VIII) to calculate $E_{1\text{Ω}12\text{V}}$.

    $\begin{array}{c}{E}_{1\text{Ω}\text{4}\text{V}}={\left(1.31\text{A}\right)}^2\left(1\text{Ω}\right)\left(2.00\min \right)\\ ={\left(1.31\text{A}\right)}^2\left(1\text{Ω}\right)\left(2.00\min \times \frac{60\text{s}}{1\min }\right)\\ =205\text{J}\end{array}$

Here, $E_{1\text{Ω}12\text{V}}$ is the energy delivered to the $1\text{Ω}$ resistor connected to $12.0\text{V}$ battery.

Therefore, the energy delivered to the $1\text{Ω}$ resistor connected to $12.0\text{V}$ battery is $205\text{J}$, the energy delivered to the $3\text{Ω}$ resistor is $616\text{J}$, the energy delivered to the $5\text{Ω}$ resistor is $128\text{J}$, the energy delivered to the $8\text{Ω}$ resistor is $687\text{J}$ and the energy delivered to the $1\text{Ω}$ resistor connected to $4.00\text{V}$ battery is $25.6\text{J}$.

(d)

To determine

The type of energy storage transformation that produced in the operation of circuit.

Explanation of Solution

The chemical energy of the $12.0\text{V}$ battery is transformed into the internal energy in all the resistors and the chemical energy of the $4.00\text{V}$ battery is not used, rather and its chemical energy is increased which is being charged from the $12.0\text{V}$ battery.

(e)

To determine

The total amount of energy transformed into internal energy in the resistor.

Answer to Problem 23P

The total amount of energy transformed into internal energy in the resistor is $1.66\text{kJ}$.

Explanation of Solution

Write the expression to obtain the total amount of energy transformed into internal energy in the resistor.

    $U=E_{1\text{Ω}12\text{V}}+E_{3\text{Ω}}+E_{5\text{Ω}}+E_{8\text{Ω}}+E_{1\text{Ω}\text{4}\text{V}}$

Here, $U$ is the total amount of energy transformed into internal energy in the resistor.

Conclusion:

Substitute $205\text{J}$ for $E_{1\text{Ω}12\text{V}}$, $616\text{J}$ for $E_{3\text{Ω}}$, $128\text{J}$ for $E_{5\text{Ω}}$, $687\text{J}$ for $E_{8\text{Ω}}$ and $25.6\text{J}$ for $E_{1\text{Ω}\text{4}\text{V}}$ in the above equation to calculate $U$.

    $\begin{array}{c}U=205\text{J}+616\text{J}+128\text{J}+687\text{J}+25.6\text{J}\\ \text{=1661}\text{.4}\text{J}\times \frac{1\text{kJ}}{{10}^3\text{J}}\\ \approx 1.66\text{kJ}\end{array}$

Therefore, the total amount of energy transformed into internal energy in the resistor is $1.66\text{kJ}$.