Новое решение
en/11.3.24.md
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| + | ### Statement | ||
| + | |||
| + | $11.3.24.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
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| + | When an alternating voltage is applied to a coil, the total flux generated is split into two identical paths, so that the other coil, placed in one of the branches, is only linked by half of the total flux. | ||
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| + | In the first case, with$ V_1 = 40\ \text{V} $on coil 1, a voltage$ V_2 = 10\ \text{V} $appears on coil 2. The voltage ratio is | ||
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| + | $\frac{V_2}{V_1} = \frac{N_2}{2N_1} = \frac{1}{4}$ | ||
| + | |||
| + | from which it follows that $N_2 = N_1/2$ (coil 2 has half the number of turns as coil 1). | ||
| + | |||
| + | When the connection is reversed and $V_2' = 10\ \text{V}$ is applied to coil 2, the generated flux divides just as before. Coil 1 is in one branch and receives half of the total flux. The voltage induced in it is | ||
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| + | $V_1' = N_1 \frac{d}{dt}\!\left(\frac{\Phi_2}{2}\right) = \frac{N_1}{2N_2}\,V_2'$ | ||
| + | |||
| + | Substituting $N_2 = N_1/2 $ gives | ||
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| + | $V_1' = V_2' = 10\ \text{V}$ | ||
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| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $11.3.24.$ [Insert the problem statement] | |||
| ### Solution | |||
| When an alternating voltage is applied to a coil, the total flux generated is split into two identical paths, so that the other coil, placed in one of the branches, is only linked by half of the total flux. | |||
| In the first case, with$ V_1 = 40\ \text{V} $on coil 1, a voltage$ V_2 = 10\ \text{V} $appears on coil 2. The voltage ratio is | |||
| $\frac{V_2}{V_1} = \frac{N_2}{2N_1} = \frac{1}{4}$ | |||
| from which it follows that $N_2 = N_1/2$ (coil 2 has half the number of turns as coil 1). | |||
| When the connection is reversed and $V_2' = 10\ \text{V}$ is applied to coil 2, the generated flux divides just as before. Coil 1 is in one branch and receives half of the total flux. The voltage induced in it is | |||
| $V_1' = N_1 \frac{d}{dt}\!\left(\frac{\Phi_2}{2}\right) = \frac{N_1}{2N_2}\,V_2'$ | |||
| Substituting $N_2 = N_1/2 $ gives | |||
| $V_1' = V_2' = 10\ \text{V}$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||