Edits to “Statement”, “Answer”
en/11.3.24.md
+9 −2
| @@ -1,7 +1,14 @@ | |||
| ### Statement | |||
| − | $11.3.24.$ | ||
| + | $11.3.24.$ | ||
| + | Two coils are wound on the iron core. The magnetic flux generated by each | ||
| + | coil does not leave the core and is divided equally in its branches. When coil | ||
| + | 1 is connected to an AC circuit with a voltage of 40 V, the voltage across coil 2 | ||
| + | is 10 V. What is the voltage at the open terminals of coil 1 if coil 2 is connected | ||
| + | to an AC circuit with a voltage of 10 V? | ||
| + | |||
| + | |||
| ### 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 | |||
| $V_1' = V_2' = 10\ \text{V}$ | |||
| @@ -22,4 +29,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{V_1' = V_2' = 10\ \text{V}}$ | ||
| @@ -1,7 +1,14 @@ | |||
| ### Statement | ### Statement | ||
| $11.3.24.$ |
$11.3.24.$ | ||
| Two coils are wound on the iron core. The magnetic flux generated by each | |||
| coil does not leave the core and is divided equally in its branches. When coil | |||
| 1 is connected to an AC circuit with a voltage of 40 V, the voltage across coil 2 | |||
| is 10 V. What is the voltage at the open terminals of coil 1 if coil 2 is connected | |||
| to an AC circuit with a voltage of 10 V? | |||
| ### Solution | ### 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. | 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} $appears on coil 2. The voltage ratio is | In the first case, with$ V_1 = 40\ \text{V} $on coil 1, a voltage$ V_2 = 10\ \text{V} $appears on coil 2. The voltage ratio is | ||
| $\frac{V_2}{V_1} = \frac{N_2}{2N_1} = \frac{1}{4}$ | $\frac{V_2}{V_1} = \frac{N_2}{2N_1} = \frac{1}{4}$ | ||
| from which it follows that $N_2 = N_1/2$ (coil 2 has half the number of turns as coil 1). | 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 | 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 | ||
| $V_1' = N_1 \frac{d}{dt}\!\left(\frac{\Phi_2}{2}\right) = \frac{N_1}{2N_2}\,V_2'$ | $V_1' = N_1 \frac{d}{dt}\!\left(\frac{\Phi_2}{2}\right) = \frac{N_1}{2N_2}\,V_2'$ | ||
| Substituting $N_2 = N_1/2 $ gives | Substituting $N_2 = N_1/2 $ gives | ||
| $V_1' = V_2' = 10\ \text{V}$ | $V_1' = V_2' = 10\ \text{V}$ | ||
| @@ -22,4 +29,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{V_1' = V_2' = 10\ \text{V}}$ | ||