Chapter 20, Problem 26P

(a)

To determine

The current drawn by the motor when it is first started up.

Answer to Problem 26P

The current drawn by the motor when it is first started up is $6.0\text{A}$.

Explanation of Solution

Write an expression for the current drawn by the motor when it is first started up.

  $I=\frac{{\epsilon }_{ext}}{R}$                                                                                                             (I)

Here, $I$  is the current drawn by the motor when it is first started up, ${\epsilon }_{ext}$ is the external emf and $R$ is the resistance.

Conclusion:

Substitute $12\text{V}$ for ${\epsilon }_{ext}$ and $2.0\Omega$  for $R$ in equation (I) to find $I$.

  $\begin{array}{c}I=\frac{12\text{V}}{2.0\Omega }\\ =6.0\text{A}\end{array}$

Thus, the current drawn by the motor when it is first started up is $6.0\text{A}$.

(b)

To determine

The current draws at normal operating speed.

Answer to Problem 26P

The current draws at normal operating speed is $0.551\text{A}$.

Explanation of Solution

Write an expression for Kirchhoff’s voltage law.

  ${\epsilon }_{ext}-I\text{'}R-{\epsilon }_{back}=0$                                                                                                (II)

Here, $I\text{'}$  is the current draw at normal operating speed and ${\epsilon }_{back}$ is the back emf.

Write an expression for back emf.

  ${\epsilon }_{back}=\frac{P}{I\text{'}}$                                                                                                     (III)

Here, $P$ is the power.

Rewrite the expression (II) by using equation (III).

  $\begin{array}{c}{\epsilon }_{ext}-I\text{'}R-\frac{P}{I\text{'}}=0\\ I\text{'}{\epsilon }_{ext}-{I^{\prime }}^{2}R-P=0\end{array}$                                                                                      (IV)

Rearrange (IV) to calculate $I\text{'}$.

  $I\text{'}=\frac{-{\epsilon }_{ext}\pm \sqrt{{\epsilon }_{ext}^2-4\left(-R\right)\left(-P\right)}}{2\left(-R\right)}$                                                                   (V)

Conclusion:

Substitute $12\text{V}$ for ${\epsilon }_{ext}$, $2.0\Omega$ for $R$ and $6.0\text{W}$ for $P$ in equation (V) to find $I\text{'}$.

  $\begin{array}{c}I\text{'}=\frac{-12\text{V}\pm \sqrt{{\left(12\text{V}\right)}^2-4\left(-2.0\Omega \right)\left(-6.0\text{W}\right)}}{2\left(-2.0\Omega \right)}\\ =\frac{-12\text{V}\pm \sqrt{{\left(12\text{V}\right)}^2-48{\text{V}}^2}}{\left(-4.0\Omega \right)}\\ =0.551\text{A}\end{array}$

Thus, the current draws at normal operating speed is $0.551\text{A}$.

(c)

To determine

The back emf induced in the winding at normal speed.

Answer to Problem 26P

The back emf induced in the winding at normal speed is $10.9\text{V}$.

Explanation of Solution

Write an expression for back emf induced in the winding at normal speed.

  ${{\epsilon }^{\prime }}_{back}={\epsilon }_{ext}-I^{\prime }R$                                                                                          (VI)

Here, ${{\epsilon }^{\prime }}_{ext}$ is the back emf induced in the winding at normal speed.

Conclusion:

Substitute $12\text{V}$ for ${\epsilon }_{ext}$, $0.551\text{A}$ for $I^{\prime }$ and $2.0\Omega$ for $R$ in equation (VI) to find ${{\epsilon }^{\prime }}_{ext}$.

  $\begin{array}{c}{{\epsilon }^{\prime }}_{back}=12\text{V}-\left(0.551\text{A}\right)\left(2.0\Omega \right)\\ =12\text{V}-1.1\text{V}\\ =10.9\text{V}\end{array}$

Thus, the back emf induced in the winding at normal speed $10.9\text{V}$.