Chapter 22, Problem 98A

(a)

To determine

The current drawn by the motor.

Answer to Problem 98A

The current drawn by the motor is $5.0\text{A}$ .

Explanation of Solution

Given:

The volume of the water pumped is $1.0\times {10}^4\text{L}$ .

The height of pumping is $h=8.0\text{m}$ .

The operating resistance of the motor is $R=22.0\Omega$ .

The applied voltage is $V=110\text{V}$ .

Formula used:

The expression for the current $\left(I\right)$ drawn by the motor is,

$I=\frac{V}{R}$

Here, $V$ is the applied voltage and $R$ is the resistance.

Calculation:

The current drawn by the motor is,

$\begin{array}{l}I=\frac{V}{R}\\ I=\frac{110}{22.0}\\ I=5.0\text{A}\end{array}$

Conclusion:

Thus, the current drawn by the motor is $I=5.0\text{A}$ .

(b)

To determine

The efficiency of the motor.

Answer to Problem 98A

The efficiency of the motor is $40\%$ .

Explanation of Solution

Given:

The volume of the water pumped is $1.0\times {10}^4\text{L}$ .

The height of pumping is $h=8.0\text{m}$ .

The time $t=1\text{hr}$ .

The operating resistance of the motor is $R=22.0\Omega$ .

The applied voltage is $V=110\text{V}$ .

Consider the acceleration due to gravity of Earth as $g=9.80{\text{m/s}}^2$

Formula used:

The expression for the work done by the motor in pumping the water is,

  $\begin{array}{c}W=mgh\\ =\left({\rho }_{\text{w}}{V}_{\text{w}}\right)gh\end{array}$

Here, ${\rho }_{\text{w}}$ is the density of the water, and $V_{\text{w}}$ is the volume of the water pumped,

$g$ is the acceleration due to gravity, and $h$ is the height of pumping.

The expression for the energy $\left(E\right)$ drawn by the motor is,

  $E=VIt$

Here, $V$ is the applied voltage, $I$ is the current and $t$ is the time.

The efficiency of the motor is the ratio of the work done $\left(W\right)$ by the motor to the energy $\left(E\right)$ drawn by the motor.

The efficiency $\left(e\right)$ of the motor is,

$e=\frac{W}{E}\times 100\%$

Calculation:

The efficiency of the motor is,

  $\begin{array}{c}e=\frac{W}{E}\times 100\%\\ e=\frac{{\rho }_{\text{w}}{V}_{\text{w}}gh}{VIt}\times 100\%\\ e=\frac{{\rho }_{\text{w}}{V}_{\text{w}}gh}{VIt}\times 100\%\\ e=\frac{\left(1000{\text{kg/m}}^3\right)\times \left(1.0\times {10}^4\text{L}\times \frac{1{\text{m}}^3}{1000\text{L}}\right)\times \left(\times 9.80{\text{m/s}}^2\right)\times 8.0\text{m}}{110\text{V}\times \text{5}\text{.0}\text{A}\times \text{1}\text{hr}\times \left(\frac{3600\text{s}}{1\text{hr}}\right)}\times 100\%\\ e=\frac{\left(1000{\text{kg/m}}^3\times 9.80{\text{m/s}}^2\right)\times \left(10{\text{m}}^3\right)\times 8.0\text{m}}{110\text{V}\times \text{5}\text{.0}\text{A}\times 3600\text{s}}\times 100\%\\ e=\frac{784,000}{1,980,000}\times 100\%\\ e=40\%\end{array}$

Conclusion:

Thus, the efficiency of the motor is $40\%$ .