Chapter 17, Problem 23E

(a)

To determine

The current in the circuit.

Answer to Problem 23E

The current in the circuit is $\underset{\_}{6\text{A}}$.

Explanation of Solution

Given that the voltage is $12\text{V}$, and the resistance is $2\text{ohm}$.

Write the expression for the current according to Ohm’s law.

  $I=\frac{\Delta V}{R}$        (I)

Here, $I$ is the current, $\Delta V$ is the voltage, and $R$ is the resistance.

Conclusion:

Substitute $12\text{V}$ for $\Delta V$, and $2\text{ohm}$ for $R$ in equation (I) to find $I$.

  $\begin{array}{c}I=\frac{12\text{V}}{2\text{ohm}}\\ =6\text{A}\end{array}$

Therefore, the current in the circuit is $\underset{\_}{6\text{A}}$.

(b)

To determine

The charge transported through the circuit in $10\text{s}$.

Answer to Problem 23E

The charge transported through the circuit in $10\text{s}$ is $\underset{\_}{60\text{C}}$.

Explanation of Solution

It is obtained that the current in the circuit is $6\text{A}$. Given that the time is $10\text{s}$.

Write the expression for current in terms of charge transported.

  $I=\frac{\Delta Q}{\Delta t}$        (II)

Here, $\Delta Q$ is the charge transported, and $\Delta t$ is the time.

Solve equation (II) for $\Delta Q$.

  $\Delta Q=I\Delta t$        (III)

Conclusion:

Substitute $6\text{A}$ for $I$, and $10\text{s}$ for $\Delta t$ in equation (III) to find $\Delta Q$.

  $\begin{array}{c}\Delta Q=\left(6\text{A}\right)\left(10\text{s}\right)\\ =60\text{C}\end{array}$

Therefore, the charge transported through the circuit in $10\text{s}$ is $\underset{\_}{60\text{C}}$.

(c)

To determine

The work done on the charge by the electric fields in the battery.

Answer to Problem 23E

The work done on the charge by the electric fields in the battery is $\underset{\_}{-720\text{J}}$.

Explanation of Solution

The work done on the charge by the electric field in the battery can be found out by the formula,

  $W_{\text{by battery}}=-IVt$        (IV)

Here, $W_{\text{by battery}}$ is the work done by the electric field in the battery.

The negative sign indicates that the electric field direction is opposite to the direction of motion of charge (electron).

Conclusion:

Substitute $6\text{A}$ for $I$, and $12\text{V}$ for $V$, and $10\text{s}$ for $t$ in equation (IV) to find $W_{\text{by battery}}$.

  $\begin{array}{c}{W}_{\text{by battery}}=-\left(6\text{A}\right)\left(12\text{V}\right)\left(10\text{s}\right)\\ =-720\text{J}\end{array}$

Therefore, the work done on the charge by the electric fields in the battery is $\underset{\_}{-720\text{J}}$.

(d)

To determine

The work done on the charge by the electric fields in the resistor.

Answer to Problem 23E

The work done on the charge by the electric fields in the resistor is $\underset{\_}{720\text{J}}$.

Explanation of Solution

The work done on the charge by the electric fields in the resistor can be computed as,

  $W_{\text{by resistor}}=I^{2}Rt$        (V)

Here, $W_{\text{by resistor}}$ is the work done by the electric fields in the resistor.

Conclusion:

Substitute $6\text{A}$ for $I$, and $2\text{ohm}$ for $R$, and $10\text{s}$ for $t$ in equation (V) to find $W_{\text{by resistor}}$.

  $\begin{array}{c}{W}_{\text{by resistor}}={\left(6\text{A}\right)}^2\left(2\text{ohm}\right)\left(10\text{s}\right)\\ =720\text{J}\end{array}$

Therefore, the work done on the charge by the electric fields in the resistor is $\underset{\_}{720\text{J}}$.

(e)

To determine

The total work done by the electric fields on the charge.

Answer to Problem 23E

The total work done by the electric fields on the charge is zero.

Explanation of Solution

It is obtained that the work done on the charge by the electric fields in the battery is $-720\text{J}$ and the work done on the charge by the electric fields in the resistor is $720\text{J}$.

The total work done by the fields is the sum of the two works done obtained above.

  $W_{\text{total}}=W_{\text{by battery}}+W_{\text{by resistor}}$        (VI)

Conclusion:

Substitute $720\text{J}$ for $W_{\text{by resistor}}$, and $-720\text{J}$ for $W_{\text{by battery}}$ in equation (VI) to find $W_{\text{total}}$.

  $\begin{array}{c}{W}_{\text{total}}=-720\text{J}+720\text{J}\\ =0\end{array}$

Therefore, the total work done by the electric fields on the charge is zero.

(f)

To determine

The energy converted to heat.

Answer to Problem 23E

The energy converted to heat is $\underset{\_}{720\text{J}}$.

Explanation of Solution

The energy converted to heat is equivalent to the work done on the charge by the electric field in the resistor, which is given by the equation (V) and obtained in part (d) as $720\text{J}$.

Conclusion:

Therefore, the energy converted to heat is $\underset{\_}{720\text{J}}$.

(g)

To determine

The source of heat energy in the circuit.

Answer to Problem 23E

The source of heat energy in the circuit is the chemical energy in the battery.

Explanation of Solution

In the circuit, the source of electromotive force is the battery. The chemical reactions taking place inside the battery results the conversion of chemical energy to electrical energy and which in turn converted to heat energy as the charges moves through the resistor.

The charges are driven by the electromotive force and these charges experiences resistance of the conductor or resistance and as a result the electrical energy to heat energy conversion takes place. Fundamentally the source of heat energy in the circuit is the chemical energy in the battery.

Conclusion:

Therefore, the source of heat energy in the circuit is the chemical energy in the battery.