Edits to “Statement”, “Answer”
en/11.1.29.md
+2 −3
| @@ -1,7 +1,6 @@ | |||
| ### Statement | |||
| − | $11.1.29.$ [Insert the problem statement] | ||
| − | |||
| + | $11.1.29.$What rotational frequency will a permanent-magnet DC motor achieve when connected to a circuit with an EMF $\mathcal{E}$ and a total resistance $R$, if, when operating as a dynamo (generator), it develops an EMF $\mathcal{E}_0$ at a frequency $f_0$? The friction torque on the motor shaft is $M$. | ||
| ### Solution | |||
| For the first circuit we have: | |||
| \begin{equation} | |||
| \mathcal{E}+\mathcal{E}_{i1}=IR | |||
| \end{equation} | |||
| where $\mathcal{E}_{i1}$ is an induced emf of a motor in the first case. | |||
| Similarly,for the second one we could write: | |||
| \begin{equation} | |||
| \mathcal{E}_0+\mathcal{E}_{i2}=0 | |||
| \end{equation} | |||
| where $\mathcal{E}_{i2}$ is an induced emf of a motor in the second circuit. | |||
| Using the fact the induced emf is proportional to its frequency we get: | |||
| \begin{equation} | |||
| \frac{\mathcal{E}_{i1}}{f}=\frac{\mathcal{E}_{i2}}{f_0} | |||
| \end{equation} | |||
| Or it is the same as: | |||
| \begin{equation} | |||
| \mathcal{E}_{i1}=-\mathcal{E}_0\frac{f}{f_0} | |||
| \end{equation} | |||
| In the steady state, we know that the net torque is zero,so: | |||
| \begin{equation} | |||
| M+M_m=0 | |||
| \end{equation} | |||
| where $M_m$ is an electromagnetic torque of the motor which we could through it's angular frequency and power $P_m$: | |||
| \begin{equation} | |||
| \omega M_m=P_m | |||
| \end{equation} | |||
| But we could write $P_m$ as $\mathcal{E}_{i1}I$ and $\omega$ as $2\pi f$ for the first case.By plugging them into eq(6) and using eq(5) we get $I$: | |||
| \begin{equation} | |||
| I=\frac{2\pi M f_0}{\mathcal{E}_0} | |||
| \end{equation} | |||
| Finally,plugging last equation into the first one and using eq(4) we get our answer for frequency: | |||
| \begin{equation} | |||
| f=f_0(\frac{\mathcal{E}}{\mathcal{E}_0}-\frac{2\pi MRf_0}{\mathcal{E}_0^2}) | |||
| \end{equation} | |||
| @@ -83,4 +82,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $f=f_0(\frac{\mathcal{E}}{\mathcal{E}_0}-\frac{2\pi MRf_0}{\mathcal{E}_0^2})$ | ||
| @@ -1,7 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $11.1.29.$ [Insert the problem statement] | $11.1.29.$What rotational frequency will a permanent-magnet DC motor achieve when connected to a circuit with an EMF $\mathcal{E}$ and a total resistance $R$, if, when operating as a dynamo (generator), it develops an EMF $\mathcal{E}_0$ at a frequency $f_0$? The friction torque on the motor shaft is $M$. | ||
| ### Solution | ### Solution | ||
| For the first circuit we have: | For the first circuit we have: | ||
| \begin{equation} | \begin{equation} | ||
| \mathcal{E}+\mathcal{E}_{i1}=IR | \mathcal{E}+\mathcal{E}_{i1}=IR | ||
| \end{equation} | \end{equation} | ||
| where $\mathcal{E}_{i1}$ is an induced emf of a motor in the first case. | where $\mathcal{E}_{i1}$ is an induced emf of a motor in the first case. | ||
| Similarly,for the second one we could write: | Similarly,for the second one we could write: | ||
| \begin{equation} | \begin{equation} | ||
| \mathcal{E}_0+\mathcal{E}_{i2}=0 | \mathcal{E}_0+\mathcal{E}_{i2}=0 | ||
| \end{equation} | \end{equation} | ||
| where $\mathcal{E}_{i2}$ is an induced emf of a motor in the second circuit. | where $\mathcal{E}_{i2}$ is an induced emf of a motor in the second circuit. | ||
| Using the fact the induced emf is proportional to its frequency we get: | Using the fact the induced emf is proportional to its frequency we get: | ||
| \begin{equation} | \begin{equation} | ||
| \frac{\mathcal{E}_{i1}}{f}=\frac{\mathcal{E}_{i2}}{f_0} | \frac{\mathcal{E}_{i1}}{f}=\frac{\mathcal{E}_{i2}}{f_0} | ||
| \end{equation} | \end{equation} | ||
| Or it is the same as: | Or it is the same as: | ||
| \begin{equation} | \begin{equation} | ||
| \mathcal{E}_{i1}=-\mathcal{E}_0\frac{f}{f_0} | \mathcal{E}_{i1}=-\mathcal{E}_0\frac{f}{f_0} | ||
| \end{equation} | \end{equation} | ||
| In the steady state, we know that the net torque is zero,so: | In the steady state, we know that the net torque is zero,so: | ||
| \begin{equation} | \begin{equation} | ||
| M+M_m=0 | M+M_m=0 | ||
| \end{equation} | \end{equation} | ||
| where $M_m$ is an electromagnetic torque of the motor which we could through it's angular frequency and power $P_m$: | where $M_m$ is an electromagnetic torque of the motor which we could through it's angular frequency and power $P_m$: | ||
| \begin{equation} | \begin{equation} | ||
| \omega M_m=P_m | \omega M_m=P_m | ||
| \end{equation} | \end{equation} | ||
| But we could write $P_m$ as $\mathcal{E}_{i1}I$ and $\omega$ as $2\pi f$ for the first case.By plugging them into eq(6) and using eq(5) we get $I$: | But we could write $P_m$ as $\mathcal{E}_{i1}I$ and $\omega$ as $2\pi f$ for the first case.By plugging them into eq(6) and using eq(5) we get $I$: | ||
| \begin{equation} | \begin{equation} | ||
| I=\frac{2\pi M f_0}{\mathcal{E}_0} | I=\frac{2\pi M f_0}{\mathcal{E}_0} | ||
| \end{equation} | \end{equation} | ||
| Finally,plugging last equation into the first one and using eq(4) we get our answer for frequency: | Finally,plugging last equation into the first one and using eq(4) we get our answer for frequency: | ||
| \begin{equation} | \begin{equation} | ||
| f=f_0(\frac{\mathcal{E}}{\mathcal{E}_0}-\frac{2\pi MRf_0}{\mathcal{E}_0^2}) | f=f_0(\frac{\mathcal{E}}{\mathcal{E}_0}-\frac{2\pi MRf_0}{\mathcal{E}_0^2}) | ||
| \end{equation} | \end{equation} | ||
| @@ -83,4 +82,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $f=f_0(\frac{\mathcal{E}}{\mathcal{E}_0}-\frac{2\pi MRf_0}{\mathcal{E}_0^2})$ | ||