Edits to “Statement”, “Solution”, “Answer”
en/11.1.28.md
+4 −5
| @@ -1,7 +1,6 @@ | |||
| ### Statement | |||
| − | $11.1.28.$ [Insert the problem statement] | ||
| − | |||
| + | $11.1.28.$ The rotor frequency of a DC motor connected to a battery circuit with an EMF of 24 V and a total resistance of 20 $\Omega$ is 600 rpm at a current of 0.2 A. What EMF will the same motor develop when operating as a dynamo (generator) at a frequency of 1200 rpm? | ||
| ### Solution | |||
| It is known that the induced EMF of an electric motor is directly proportional to the magnetic flux and its frequency. Since only the motor's frequency varies, we could write the following: | |||
| \begin{equation} | |||
| \mathcal{E}_i=k\omega | |||
| \end{equation} | |||
| where $\mathcal{E}_i$ is the induced EMF, $\omega$ is angular frequency of the motor and $k$ is a constant. | |||
| For the first case we have: | |||
| \begin{equation} | |||
| \mathcal{E}_1+k\omega_1=IR | |||
| @@ -26,7 +25,7 @@Solution | |||
| \end{equation} | |||
| − | where $ \mathcal{E}_1=24V $, $I=0.2$A,$R=20$$\Omega$ and $\omega_1=600 | ||
| + | where $ \mathcal{E}_1=24V $, $I=0.2$A,$R=20$$\Omega$ and $\omega_1=600$ rpm | ||
| Similarly,for the second case we could write: | |||
| \begin{equation} | |||
| \mathcal{E}_2+k\omega_2=0 | |||
| @@ -39,7 +38,7 @@Solution | |||
| \end{equation} | |||
| − | where $\omega_2=1200 | ||
| + | where $\omega_2=1200$ rpm | ||
| By substituting $k$ from eq(2) to eq(3) we get the answer for $\mathcal{E}_2$: | |||
| \begin{equation} | |||
| \mathcal{E}_2=(\mathcal{E}_1-IR)\frac{\omega_2}{\omega_1}=40V | |||
| \end{equation} | |||
| @@ -55,4 +54,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\mathcal{E}_2=40V$ | ||
| @@ -1,7 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $11.1.28.$ [Insert the problem statement] | $11.1.28.$ The rotor frequency of a DC motor connected to a battery circuit with an EMF of 24 V and a total resistance of 20 $\Omega$ is 600 rpm at a current of 0.2 A. What EMF will the same motor develop when operating as a dynamo (generator) at a frequency of 1200 rpm? | ||
| ### Solution | ### Solution | ||
| It is known that the induced EMF of an electric motor is directly proportional to the magnetic flux and its frequency. Since only the motor's frequency varies, we could write the following: | It is known that the induced EMF of an electric motor is directly proportional to the magnetic flux and its frequency. Since only the motor's frequency varies, we could write the following: | ||
| \begin{equation} | \begin{equation} | ||
| \mathcal{E}_i=k\omega | \mathcal{E}_i=k\omega | ||
| \end{equation} | \end{equation} | ||
| where $\mathcal{E}_i$ is the induced EMF, $\omega$ is angular frequency of the motor and $k$ is a constant. | where $\mathcal{E}_i$ is the induced EMF, $\omega$ is angular frequency of the motor and $k$ is a constant. | ||
| For the first case we have: | For the first case we have: | ||
| \begin{equation} | \begin{equation} | ||
| \mathcal{E}_1+k\omega_1=IR | \mathcal{E}_1+k\omega_1=IR | ||
| @@ -26,7 +25,7 @@Solution | |||
| \end{equation} | \end{equation} | ||
| where $ \mathcal{E}_1=24V $, $I=0.2$A,$R=20$$\Omega$ and $\omega_1=600 |
where $ \mathcal{E}_1=24V $, $I=0.2$A,$R=20$$\Omega$ and $\omega_1=600$ rpm | ||
| Similarly,for the second case we could write: | Similarly,for the second case we could write: | ||
| \begin{equation} | \begin{equation} | ||
| \mathcal{E}_2+k\omega_2=0 | \mathcal{E}_2+k\omega_2=0 | ||
| @@ -39,7 +38,7 @@Solution | |||
| \end{equation} | \end{equation} | ||
| where $\omega_2=1200 |
where $\omega_2=1200$ rpm | ||
| By substituting $k$ from eq(2) to eq(3) we get the answer for $\mathcal{E}_2$: | By substituting $k$ from eq(2) to eq(3) we get the answer for $\mathcal{E}_2$: | ||
| \begin{equation} | \begin{equation} | ||
| \mathcal{E}_2=(\mathcal{E}_1-IR)\frac{\omega_2}{\omega_1}=40V | \mathcal{E}_2=(\mathcal{E}_1-IR)\frac{\omega_2}{\omega_1}=40V | ||
| \end{equation} | \end{equation} | ||
| @@ -55,4 +54,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\mathcal{E}_2=40V$ | ||