Chapter 21, Problem 47P

The circuit shown in Figure P21.47 is connected for 2.00 min. (a) Determine the current in each branch of the circuit. (b) Find the energy delivered by each battery. (c) Find the energy delivered to each resistor. (d) Identify the type of energy storage transformation that occurs in the operation of the circuit. (e) Find the total amount of energy transformed into internal energy in the resistors.

Figure P21.47 Problems 47 and 48.

To determine

The current in each branch of the circuit.

Answer to Problem 47P

The current $I_1$ is $\underset{\_}{0.846\text{A down in the }8.00\Omega \text{ resistor}}$, $I_2$ is $\underset{\_}{0.462\text{A down in the middle branch}}$ and $I_3$ is $\underset{\_}{1.31\text{A up in the right-hand branch}}$ in the circuit.

Explanation of Solution

Write the expression for the Kirchhoff’s loop rule in second loop from the left of the figure $1$.

    $\begin{array}{c}12.0\text{V}-\left(4.00\Omega \right)I_3-\left(6.00\Omega \right)I_2-4.00\text{V}=0\\ \left(4.00\Omega \right)I_3+\left(6.00\Omega \right)I_2=8.00\text{V}\end{array}$        (I)

Here, $I$ is the current.

Write the expression for the Kirchhoff’s loop rule in first loop from the left of the figure $1$.

    $\begin{array}{c}-\left(6.00\Omega \right)I_2-4.00\text{V}+\left(8.00\Omega \right)I_1=0\\ \left(6.00\Omega \right)I_2+4.00\text{V}=\left(8.00\Omega \right)I_1\end{array}$        (II)

Write the expression for the Kirchhoff’s junction rule in the figure $1$.

    $I_3=I_1+I_2$        (III)

Use equation (III) in (I) and solve the equation.

    $\begin{array}{c}\left(4.00\Omega \right)\left(I_1+I_2\right)+\left(6.00\Omega \right)I_2=8.00\text{V}\\ \left(\text{4}\text{.00}\Omega \right)I_1+\left(\text{4}\text{.00}\Omega \right)I_2+\left(6.00\Omega \right)I_2=8.00\text{V}\\ \left(\text{4}\text{.00}\Omega \right)I_1+\left(\text{10}\text{.0}\Omega \right)I_2=8.00\text{V}\end{array}$        (IV)

Use equation (II) to solve for $I_2$.

    $I_2=\left(\frac{4.00\Omega }{3.00\Omega }\right)I_1-\left(\frac{2.00\text{V}}{3.00\Omega }\right)$        (V)

Use equation (IV) in (V) to solve for $I_1$.

    $\left(\text{4}\text{.00}\Omega \right)I_1+\left(\text{10}\text{.0}\Omega \right)\left[\left(\frac{4.00\Omega }{3.00\Omega }\right)I_1-\left(\frac{2.00\text{V}}{3.00\Omega }\right)\right]=8.00\text{V}$        (VI)

Use equation (VI) to solve for $I_1$.

    $\begin{array}{c}{I}_1=\frac{3}{52\Omega }\left(8.00\text{V}+\frac{20}{3}\text{V}\right)\\ =0.846\text{A}\end{array}$        (VII)

Use equation (VII) in (V) to solve for $I_2$.

    $\begin{array}{c}{I}_2=\left(\frac{4.00\Omega }{3.00\Omega }\right)\left(0.846\text{A}\right)-\left(\frac{2.00\text{V}}{3.00\Omega }\right)\\ =0.462\text{A}\end{array}$        (VIII)

Use equation (VIII) and (VII)in (III) to solve for $I_3$.

    $\begin{array}{c}{I}_3=\left(0.846\text{A}\right)+\left(0.462\text{A}\right)\\ =1.31\text{A}\end{array}$        (IX)

Conclusion:

Therefore, the current $I_1$ is $\underset{\_}{0.846\text{A down in the }8.00\Omega \text{ resistor}}$, $I_2$ is $\underset{\_}{0.462\text{A down in the middle branch}}$ and $I_3$ is $\underset{\_}{1.31\text{A up in the right-hand branch}}$ in the circuit.

To determine

The energy delivered to each battery.

Answer to Problem 47P

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

Explanation of Solution

Write the expression for the energy delivered to the battery.

    $\Delta U_{\text{B}}=P\Delta t$        (X)

Here, $\Delta U_{\text{B}}$ is the energy delivered, $P$ is the power, $\Delta t$ is the time interval.

Write the expression for the $P$.

    $P=I\left(\Delta V\right)$        (XI)

Here, $\Delta V$ is the voltage of the battery.

Use equation (XI) in (X) to solve for $\Delta U$.

    $\Delta U_{\text{B}}=\left(\Delta V\right)I\left(\Delta t\right)$        (XII)

Conclusion:

Substitute $4.00\text{V}$ for $\Delta V$, $-0.462\text{A}$ for $I$, $2.00\text{min}$ for $\Delta t$ in equation (XII) to find the power delivered to $4.00\text{V}$ battery.

    $\begin{array}{c}\Delta U_{\text{B}}=\left(4.00\text{V}\right)\left(-0.462\text{A}\right)\left(2.0\text{min}\times \frac{60\text{s}}{1\text{min}}\right)\\ =-222\text{J}\end{array}$

Substitute $12.0\text{V}$ for $\Delta V$, $1.31\text{A}$ for $I$, $2.00\text{min}$ for $\Delta t$ in equation (XII) to find the power delivered to $12.0\text{V}$ battery.

    $\begin{array}{c}\Delta U_{\text{B}}=\left(12.0\text{V}\right)\left(1.31\text{A}\right)\left(2.0\text{min}\times \frac{60\text{s}}{1\text{min}}\right)\\ =1.8\times {10}^3\text{J}\\ \text{=1}\text{.8}\text{kJ}\end{array}$

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

To determine

The energy delivered to each resistor.

Answer to Problem 47P

The energy delivered to $8.00\Omega$ resistor is $\underset{\_}{687\text{J}}$, $5.00\Omega$ resistor is $\underset{\_}{128\text{J}}$, $1.00\Omega$ resistor is $\underset{\_}{25.6\text{J}}$, $3.00\Omega$ resistor is $\underset{\_}{616\text{J}}$ and $1.00\Omega$ resistor in the right hand branch is $\underset{\_}{205\text{J}}$.

Explanation of Solution

Write the expression for the energy delivered to the resistor.

    $\Delta U_{\text{R}}=I^{2}R\Delta t$        (XIII)

Here, $\Delta U_{\text{R}}$ is the energy delivered to the resistor, $R$ is the resistance, $I$ is the current in that branch.

Conclusion:

Substitute $0.846\text{A}$ for $I$, $8.00\Omega$ for $R$, $2.00\text{min}$ for $\Delta t$ in equation (XIII) to find energy delivered to $8.00\Omega$ resistor.

    $\begin{array}{c}\Delta U_{\text{R}}={\left(0.846\text{A}\right)}^2\left(8.00\Omega \right)\left(2.0\text{min}\times \frac{60\text{s}}{1\text{min}}\right)\\ =687\text{J}\end{array}$

Substitute $0.462\text{A}$ for $I$, $5.00\Omega$ for $R$, $2.00\text{min}$ for $\Delta t$ in equation (XIII) to find energy delivered to $5.00\Omega$ resistor.

    $\begin{array}{c}\Delta U_{\text{R}}={\left(0.462\text{A}\right)}^2\left(5.00\Omega \right)\left(2.0\text{min}\times \frac{60\text{s}}{1\text{min}}\right)\\ =128\text{J}\end{array}$

Substitute $0.462\text{A}$ for $I$, $1.00\Omega$ for $R$, $2.00\text{min}$ for $\Delta t$ in equation (XIII) to find energy delivered to $1.00\Omega$ resistor.

    $\begin{array}{c}\Delta U_{\text{R}}={\left(0.462\text{A}\right)}^2\left(1.00\Omega \right)\left(2.0\text{min}\times \frac{60\text{s}}{1\text{min}}\right)\\ =25.6\text{J}\end{array}$

Substitute $1.31\text{A}$ for $I$, $3.00\Omega$ for $R$, $2.00\text{min}$ for $\Delta t$ in equation (XIII) to find energy delivered to $3.00\Omega$ resistor.

    $\begin{array}{c}\Delta U_{\text{R}}={\left(1.31\text{A}\right)}^2\left(3.00\Omega \right)\left(2.0\text{min}\times \frac{60\text{s}}{1\text{min}}\right)\\ =616\text{J}\end{array}$

Substitute $1.31\text{A}$ for $I$, $1.00\Omega$ for $R$, $2.00\text{min}$ for $\Delta t$ in equation (XIII) to find energy delivered to $1.00\Omega$ resistor in the right hand branch.

    $\begin{array}{c}\Delta U_{\text{R}}={\left(1.31\text{A}\right)}^2\left(1.00\Omega \right)\left(2.0\text{min}\times \frac{60\text{s}}{1\text{min}}\right)\\ =205\text{J}\end{array}$

Therefore, the energy delivered to $8.00\Omega$ resistor is $\underset{\_}{687\text{J}}$, $5.00\Omega$ resistor is $\underset{\_}{128\text{J}}$, $1.00\Omega$ resistor is $\underset{\_}{25.6\text{J}}$, $3.00\Omega$ resistor is $\underset{\_}{616\text{J}}$ and $1.00\Omega$ resistor in the right hand branch is $\underset{\_}{205\text{J}}$.

To determine

The type of energy transformation occurs in the operation of the circuit.

Answer to Problem 47P

The chemical energy is transformed to the internal energy in the resistors.

Explanation of Solution

The chemical energy in the $12.0\text{V}$ battery is transformed into internal energy in the resistors. The $4.00\text{V}$ battery is being charged, so its chemical potential energy is increasing at the expense of some of the chemical potential energy in the $12.0\text{V}$ battery.

Conclusion:

Therefore, the chemical energy is transformed to the internal energy in the resistors.

To determine

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

Answer to Problem 47P

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

Explanation of Solution

Write the expression for the total amount of energy transformed into internal energy in the resistors.

    $\Delta U_{\text{T}}=\Delta U_{\text{B1}}+\Delta U_{\text{B2}}$        (XIV)

Here, $\Delta U_{\text{T}}$ is the total amount of energy transformed, $\Delta U_{\text{B1}}$ is the energy delivered to $4.00\text{V}$ battery, $\Delta U_{\text{B2}}$ is the energy delivered to $12.0\text{V}$ battery.

Conclusion:

Substitute $-222\text{J}$ for $\Delta U_{\text{B1}}$, $1.88\text{kJ}$ for $\Delta U_{\text{B2}}$ in equation (XIV) to find $\Delta U_{\text{T}}$.

    $\begin{array}{c}\Delta U_{\text{T}}=-222\text{J}+\left(1.88\text{kJ}\times \frac{{10}^3\text{J}}{1\text{kJ}}\right)\\ =1.66\times {10}^3\text{J}\\ \text{=1}\text{.66}\text{kJ}\end{array}$

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