Chapter 17, Problem 27E

(a)

To determine

The current and power dissipated in the circuit.

Answer to Problem 27E

The current in the circuit is $\underset{\_}{3000\text{A}}$ and power dissipated is $\underset{\_}{3.6\times {10}^4\text{W}}$.

Explanation of Solution

Given that the voltage is $12\text{V}$, the internal resistance of the battery is $0.004\text{ohm}$, and the resistance of the conductor is negligibly small.

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

  $I=\frac{V}{R}$        (I)

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

Write the expression for power dissipated.

  $P=IV$        (II)

Here, $P$ is the power.

Conclusion:

Substitute $12\text{V}$ for $V$, and $0.004\text{ohm}$ for $R$ in equation (I) to find $I$.

  $\begin{array}{c}I=\frac{12\text{V}}{0.004\text{ohm}}\\ =3000\text{A}\end{array}$

Substitute $3000\text{A}$ for $I$, and $12\text{V}$ for $V$ in equation (II) to find $P$.

  $\begin{array}{c}P=\left(3000\text{A}\right)\left(12\text{V}\right)\\ =3.6\times {10}^4\text{W}\end{array}$

Therefore, the current in the circuit is $\underset{\_}{3000\text{A}}$ and power dissipated is $\underset{\_}{3.6\times {10}^4\text{W}}$.

(b)

To determine

The resistance that required to draw $80\text{A}$ current.

Answer to Problem 27E

The resistance that required to draw $80\text{A}$ current is $\underset{\_}{0.146\text{ohm}}$.

Explanation of Solution

Given that the current drawn is $80\text{A}$.

Equation (I) can be solved for $R$ to find the net resistance of the circuit required to draw the specified amount of current.

  $R=\frac{V}{I}$        (III)

Since the battery has internal resistance, the required resistance can be computed as,

  $R_{\text{required}}=R-R_{\text{internal}}$        (IV)

Here, $R_{\text{required}}$ is the resistance required, $R_{\text{internal}}$ is the internal resistance of the battery.

Conclusion:

Substitute $12\text{V}$ for $V$, and $80\text{A}$ for $I$ in equation (III) to find $R$.

  $\begin{array}{c}R=\frac{12\text{V}}{80\text{A}}\\ =0.15\text{ohm}\end{array}$

Substitute $0.15\text{ohm}$ for $R$, and $0.004\text{ohm}$ for $R_{\text{internal}}$ in equation (IV) to find $R_{\text{required}}$.

  $\begin{array}{c}{R}_{\text{required}}=0.15\text{ohm}-0.004\text{ohm}\\ =0.146\text{ohm}\end{array}$

Therefore, the resistance that required to draw $80\text{A}$ current is $\underset{\_}{0.146\text{ohm}}$.

(c)

To determine

The power dissipated in the motor.

Answer to Problem 27E

The power dissipated in the motor is $\underset{\_}{934\text{W}}$.

Explanation of Solution

Given that the current drawn by the motor is $80\text{A}$, and the resistance is $0.146\text{ohm}$.

Write the expression for the power supplied.

  $P=I^{2}R$        (V)

Here, $P$ is the power.

Conclusion:

Substitute $80\text{A}$ for $I$, and $0.146\text{ohm}$ for $R$ in equation (V) to find $P$.

  $\begin{array}{c}P={\left(80\text{A}\right)}^2\left(0.146\text{ohm}\right)\\ =934\text{W}\end{array}$

Therefore, the power dissipated in the motor is $\underset{\_}{934\text{W}}$.

(d)

To determine

The power dissipated in the battery.

Answer to Problem 27E

The power dissipated in the battery is $\underset{\_}{25.6\text{W}}$.

Explanation of Solution

Given that the internal resistance of the battery is $0.004\text{ohm}$.

Equation (V) can be used to find the power dissipated in the battery, by replacing $R$ by the internal resistance $R_{\text{internal}}$.

  $P=I^2{R}_{\text{internal}}$        (VI)

Conclusion:

Substitute $80\text{A}$ for $I$, and $0.004\text{ohm}$ for $R_{\text{internal}}$ in equation (VI) to find $P$.

  $\begin{array}{c}P={\left(80\text{A}\right)}^2\left(0.004\text{ohm}\right)\\ =25.6\text{W}\end{array}$

Therefore, the power dissipated in the battery is $\underset{\_}{25.6\text{W}}$.