Правка разделов «Statement», «Answer»
en/11.3.21.md
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| @@ -1,7 +1,10 @@ | |||
| ### Statement | |||
| − | $11.3.21.$ | ||
| + | $11.3.21.$ | ||
| + | Why does a loaded transformer hum? What is the fundamental frequency of | ||
| + | sound if the transformer is connected to an industrial network? | ||
| + | |||
| ### Solution | |||
| The transformer hums due to magnetostriction: the ferromagnetic core slightly contracts and expands when magnetized by the alternating current. | |||
| The deformation$ \epsilon $is proportional to the square of the magnetic induction B | |||
| $\epsilon(t) \propto B(t)^2 = B_0^2 \sin^2(\omega t) = \frac{B_0^2}{2}\bigl[1 - \cos(2\omega t)\bigr], \qquad \omega = 2\pi f_{\text{grid}}$ | |||
| The oscillatory component has twice the frequency of the mains. | |||
| For an industrial mains frequency of $f_{\text{grid}} = 50\ \text{Hz}$ | |||
| $\boxed{\nu = 2 \times 50\ \text{Hz} = 100\ \text{Hz}}$ | |||
| @@ -17,4 +20,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{\nu = 2 \times 50\ \text{Hz} = 100\ \text{Hz}}$ | ||
| @@ -1,7 +1,10 @@ | |||
| ### Statement | ### Statement | ||
| $11.3.21.$ |
$11.3.21.$ | ||
| Why does a loaded transformer hum? What is the fundamental frequency of | |||
| sound if the transformer is connected to an industrial network? | |||
| ### Solution | ### Solution | ||
| The transformer hums due to magnetostriction: the ferromagnetic core slightly contracts and expands when magnetized by the alternating current. | The transformer hums due to magnetostriction: the ferromagnetic core slightly contracts and expands when magnetized by the alternating current. | ||
| The deformation$ \epsilon $is proportional to the square of the magnetic induction B | The deformation$ \epsilon $is proportional to the square of the magnetic induction B | ||
| $\epsilon(t) \propto B(t)^2 = B_0^2 \sin^2(\omega t) = \frac{B_0^2}{2}\bigl[1 - \cos(2\omega t)\bigr], \qquad \omega = 2\pi f_{\text{grid}}$ | $\epsilon(t) \propto B(t)^2 = B_0^2 \sin^2(\omega t) = \frac{B_0^2}{2}\bigl[1 - \cos(2\omega t)\bigr], \qquad \omega = 2\pi f_{\text{grid}}$ | ||
| The oscillatory component has twice the frequency of the mains. | The oscillatory component has twice the frequency of the mains. | ||
| For an industrial mains frequency of $f_{\text{grid}} = 50\ \text{Hz}$ | For an industrial mains frequency of $f_{\text{grid}} = 50\ \text{Hz}$ | ||
| $\boxed{\nu = 2 \times 50\ \text{Hz} = 100\ \text{Hz}}$ | $\boxed{\nu = 2 \times 50\ \text{Hz} = 100\ \text{Hz}}$ | ||
| @@ -17,4 +20,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{\nu = 2 \times 50\ \text{Hz} = 100\ \text{Hz}}$ | ||