Chapter 28, Problem 28.23P

The circuit shown in Figure P27.17 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.

To determine

The current in each branch of the circuit.

Answer to Problem 28.23P

The current in each branch of the circuit is $0.846\text{A}$ down in the $8.00\Omega$ resistor, $0.462\text{A}$ down in the middle branch, $1.31\text{A}$ up in the right hand branch.

Explanation of Solution

Given info: The circuit connected for $2.00\text{min}$.

Consider the figure given below,

Figure (1)

From the Kirchhoff law,

$\begin{array}{c}{I}_3=I_2+I_1\\ I_3-I_2-I_1=0\end{array}$ . (1)

In the circuit given above there are two loops $\text{abcda}$.

$\begin{array}{c}\left(-4.00\text{V}\right)+I_1\left(8.00\Omega \right)-I_2\left(5.00\Omega \right)-I_2\left(1.00\Omega \right)=0\\ -4.00\text{V}+I_1\left(8.00\Omega \right)-I_2\left(6.00\Omega \right)=0\end{array}$ (2)

Apply Kirchhoff current law for the loop $\text{cdefc}$.

$\begin{array}{c}\left[\begin{array}{c}\left(12.00\text{V}\right)-I_2\left(1.00\Omega \right)-I_2\left(5.00\Omega \right)\\ -I_2\left(3.00\Omega \right)-I_3\left(1.00\Omega \right)-\left(4.00\text{V}\right)\end{array}\right]=0\\ \left(8.00\text{V}\right)-I_2\left(6.00\Omega \right)-I_3\left(4.00\Omega \right)=0\end{array}$

Substitute $\left(I_2+I_1\right)$ for $I_3$ in above equation.

$\begin{array}{c}\left(8.00\text{V}\right)-I_2\left(6.00\Omega \right)-\left(I_2+I_1\right)\left(4.00\Omega \right)=0\\ \left(8.00\text{V}\right)-I_2\left(6.00\Omega \right)-I_2\left(4.00\Omega \right)-I_1\left(4.00\Omega \right)=0\\ \left(8.00\text{V}\right)-I_2\left(10.00\Omega \right)-I_1\left(4.00\Omega \right)=0\end{array}$

Multiply $2$ in the above equation.

$16.00\text{V}-I_1\left(8.00\Omega \right)-I_2\left(20.00\Omega \right)=0$ (3)

Add equation (3) and equation (2).

$\begin{array}{c}12.00V-I_2\left(26.00\Omega \right)=0\\ I_2=\frac{26.00\Omega }{12.00V}\\ =0.462\text{A}\end{array}$

Substitute $0.46\text{A}$ for $I_2$ in the equation (3).

$\begin{array}{c}\left(8.00\text{V}\right)-\left(0.46\text{A}\right)\left(10.00\Omega \right)-I_1\left(4.00\Omega \right)=0\\ I_1\left(4.00\Omega \right)=\left(3.4\text{V}\right)\\ I_1=\frac{3.4\text{V}}{4.00\Omega }\\ =0.846\text{A}\end{array}$

Substitute $0.85\text{A}$ for $I_1$ and $0.46\text{A}$ for $I_2$ in the equation (1).

$\begin{array}{c}{I}_3-0.46\text{A}-0.85\text{A}=0\\ I_3=1.31\text{A}\end{array}$

Conclusion:

Therefore, the current in each branch of the circuit is $0.846\text{A}$ down in the $8.00\Omega$ resistor, $0.462\text{A}$ down in the middle branch, $1.31\text{A}$ up in the right hand branch.

To determine

The energy delivered by each battery.

Answer to Problem 28.23P

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

Explanation of Solution

Given info: The circuit connected for $2.00\text{min}$.

Formula to calculate energy for the battery is,

$\Delta U=\left(\Delta V\cdot I\right)\Delta t$ (4)

Here,

$\Delta U$ is the energy delivered by the battery.

$\Delta V$ is the voltage produced by the battery.

$I$ is the current transferred by the battery.

$\Delta t$ is the time taken to store energy.

For $4.00\text{V}$ battery:

Substitute $4.00\text{V}$ for $\Delta V$, $0.462\text{A}$ for $I$ and $120\text{s}$ for $\Delta t$ in the equation (4).

$\begin{array}{c}\Delta U_1=\left(4.00\text{V}\right)\left(0.462\text{A}\right)120\text{s}\\ =221.76\text{J}\\ \approx \text{222}\text{J}\end{array}$

For $12.0\text{V}$ battery:

Substitute $12.0\text{V}$ for $\Delta V$, $1.31\text{A}$ for $I$ and $120\text{s}$ for $\Delta t$ in the equation (4).

$\begin{array}{c}\Delta U_2=\left(12.0\text{V}\right)\left(1.31\text{A}\right)120\text{s}\\ =1886.4\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ \approx \text{1}\text{.88}\text{kJ}\end{array}$

Conclusion:

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

To determine

The energy delivered to each resistors.

Answer to Problem 28.23P

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

Explanation of Solution

Given info: The circuit connected for $2.00\text{min}$.

Formula to calculate energy transformed into internal energy in the resistors is,

$\Delta U=I^{2}R\Delta t$ (5)

Here,

$R$ is the resistance of the resistance.

For $8.00\Omega$ resistor:

Substitute $8.00\Omega$ for $R$, $0.846\text{A}$ for $I$ and $120\text{s}$ for $\Delta t$ in the equation (5).

$\begin{array}{c}\Delta U_{\text{r1}}={\left(0.846\text{A}\right)}^2\left(8.00\Omega \right)120\text{s}\\ =687\text{J}\end{array}$

For $5.00\Omega$ resistor:

Substitute $5.00\Omega$ for $R$, $0.462\text{A}$ for $I$ and $120\text{s}$ for $\Delta t$ in the equation (5).

$\begin{array}{c}\Delta U_{\text{r2}}={\left(0.462\text{A}\right)}^2\left(5.00\Omega \right)120\text{s}\\ =128\text{J}\end{array}$

For $1.00\Omega$ resistor in the center branch:

Substitute $1.00\Omega$ for $R$, $0.462\text{A}$ for $I$ and $120\text{s}$ for $\Delta t$ in the equation (5).

$\begin{array}{c}\Delta U_{\text{r3}}={\left(0.462\text{A}\right)}^2\left(1.00\Omega \right)120\text{s}\\ =25.6\text{J}\end{array}$

For $3.00\Omega$ resistor:

Substitute $3.00\Omega$ for $R$, $1.31\text{A}$ for $I$ and $120\text{s}$ for $\Delta t$ in the equation (5).

$\begin{array}{c}\Delta U_{\text{r4}}={\left(1.31\text{A}\right)}^2\left(3.00\Omega \right)120\text{s}\\ =616\text{J}\end{array}$

For $1.00\Omega$ resistor in the right hand branch;

Substitute $1.00\Omega$ for $R$, $1.31\text{A}$ for $I$ and $120\text{s}$ for $\Delta t$ in the equation (5).

$\begin{array}{c}\Delta U_{\text{r5}}={\left(1.31\text{A}\right)}^2\left(1.00\Omega \right)120\text{s}\\ =205\text{J}\end{array}$

Conclusion:

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

To determine

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

Answer to Problem 28.23P

The potential energy in the $12.0\text{V}$ battery is transformed into internal energy in the resistor. The $4.00\text{V}$ battery is being charged so; its chemical energy is increasing.

Explanation of Solution

Given info: The circuit connected for $2.00\text{min}$.

For the $4.00\text{V}$ battery the chemical potential energy is increasing at the expense of the chemical energy for the $12.0\text{V}$ battery. So, the $4.00\text{V}$ battery is charging.

For the $12.0\text{V}$ battery the internal energy is converted from the chemical potential energy present in the resistors.

Conclusion:

Therefore, the potential energy in the $12.0\text{V}$ battery is transformed into internal energy in the resistor. The $4.00\text{V}$ battery is being charged so; its chemical energy is increasing.

To determine

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

Answer to Problem 28.23P

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

Explanation of Solution

Given info: The diagram is given above in Figure (1).

Formula for the total amount of energy transformed into internal energy in the resistors is,

$U=\Delta U_{\text{r1}}+\Delta U_{\text{r2}}+\Delta U_{\text{r3}}+\Delta U_{\text{r4}}+\Delta U_{\text{r5}}$

Substitute $687\text{J}$ for $\Delta U_{\text{r1}}$, $128\text{J}$ for $\Delta U_{\text{r2}}$, $25.6\text{J}$ for $\Delta U_{\text{r3}}$, $616\text{J}$ for $\Delta U_{\text{r4}}$, $205\text{J}$ for $\Delta U_{\text{r5}}$ in the above equation.

$\begin{array}{c}U=687\text{J}+128\text{J}+25.6\text{J}+616\text{J}+205\text{J}\\ =1661.6\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ =1.66\text{kJ}\end{array}$

Conclusion:

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