Edits to “Statement”, “Solution”, “Answer”
en/11.1.27.md
+6 −4
| @@ -1,7 +1,9 @@ | |||
| ### Statement | |||
| − | $11.1.27.$ | ||
| + | $11.1.27.$ | ||
| + | Why can an electric motor burn out if its rotor is stopped? | ||
| + | |||
| ### Solution | |||
| To solve this, let us first understand the normal operation (motor spinning) | |||
| · The motor has coils that rotate inside a magnetic field. | |||
| · According to Faraday's Law, as the coil rotates in the magnetic field, a voltage is induced in the opposite direction to the current feeding it. This voltage is called back electromotive force (back EMF, or CEMF). | |||
| · The back EMF acts as a "natural regulator": it opposes the source voltage and limits the current flowing through the coils. | |||
| · The resulting current is small (only what is needed to overcome the mechanical load and losses), and the motor runs stably without overheating. | |||
| --- | |||
| Now let us analyze the case of a locked rotor (motor stalled) | |||
| If the rotor stops due to an overload or jam: | |||
| @@ -20,12 +22,12 @@Solution | |||
| · There is no motion → no magnetic flux change → the back EMF disappears. | |||
| · With the opposition gone, the source voltage (V) is applied directly across the ohmic resistance (R) of the coils, which is very small. | |||
| · By Ohm's Law, the current surges to a very high value: | |||
| − | I_{\text{start}} = \frac{V}{R} | ||
| − | · This current is 5 to 10 times higher than the rated current. All this electrical energy is converted into heat (Joule effect: P = I^2 R). | ||
| + | $ I_{\text{start}} = \frac{V}{R}$ | ||
| + | · This current is 5 to 10 times higher than the rated current. All this electrical energy is converted into heat (Joule effect: $P = I^2 R)$. | ||
| As a result: | |||
| The extreme heat melts the insulating enamel on the wires, causes internal short circuits, and burns out the motor. | |||
| #### Answer | |||
| − | |||
| + | so the motor burn because the increment of current tranformed un heat by joule efect . | ||
| @@ -1,7 +1,9 @@ | |||
| ### Statement | ### Statement | ||
| $11.1.27.$ |
$11.1.27.$ | ||
| Why can an electric motor burn out if its rotor is stopped? | |||
| ### Solution | ### Solution | ||
| To solve this, let us first understand the normal operation (motor spinning) | To solve this, let us first understand the normal operation (motor spinning) | ||
| · The motor has coils that rotate inside a magnetic field. | · The motor has coils that rotate inside a magnetic field. | ||
| · According to Faraday's Law, as the coil rotates in the magnetic field, a voltage is induced in the opposite direction to the current feeding it. This voltage is called back electromotive force (back EMF, or CEMF). | · According to Faraday's Law, as the coil rotates in the magnetic field, a voltage is induced in the opposite direction to the current feeding it. This voltage is called back electromotive force (back EMF, or CEMF). | ||
| · The back EMF acts as a "natural regulator": it opposes the source voltage and limits the current flowing through the coils. | · The back EMF acts as a "natural regulator": it opposes the source voltage and limits the current flowing through the coils. | ||
| · The resulting current is small (only what is needed to overcome the mechanical load and losses), and the motor runs stably without overheating. | · The resulting current is small (only what is needed to overcome the mechanical load and losses), and the motor runs stably without overheating. | ||
| --- | --- | ||
| Now let us analyze the case of a locked rotor (motor stalled) | Now let us analyze the case of a locked rotor (motor stalled) | ||
| If the rotor stops due to an overload or jam: | If the rotor stops due to an overload or jam: | ||
| @@ -20,12 +22,12 @@Solution | |||
| · There is no motion → no magnetic flux change → the back EMF disappears. | · There is no motion → no magnetic flux change → the back EMF disappears. | ||
| · With the opposition gone, the source voltage (V) is applied directly across the ohmic resistance (R) of the coils, which is very small. | · With the opposition gone, the source voltage (V) is applied directly across the ohmic resistance (R) of the coils, which is very small. | ||
| · By Ohm's Law, the current surges to a very high value: | · By Ohm's Law, the current surges to a very high value: | ||
| I_{\text{start}} = \frac{V}{R} | $ I_{\text{start}} = \frac{V}{R}$ | ||
| · This current is 5 to 10 times higher than the rated current. All this electrical energy is converted into heat (Joule effect: P = I^2 R). | · This current is 5 to 10 times higher than the rated current. All this electrical energy is converted into heat (Joule effect: $P = I^2 R)$. | ||
| As a result: | As a result: | ||
| The extreme heat melts the insulating enamel on the wires, causes internal short circuits, and burns out the motor. | The extreme heat melts the insulating enamel on the wires, causes internal short circuits, and burns out the motor. | ||
| #### Answer | #### Answer | ||
| so the motor burn because the increment of current tranformed un heat by joule efect . | |||