Правка разделов «Statement», «Answer»
en/11.3.18.md
+8 −2
| @@ -1,7 +1,13 @@ | |||
| ### Statement | |||
| − | $11.3.18.$ | ||
| + | $11.3.18.$ | ||
| + | Show that in an ideal transformer with a short-circuited secondary winding, | ||
| + | the ratioI1=NN12 | ||
| + | holds, where I1and I2are the currents, and N1and N2are | ||
| + | I2 | ||
| + | the number of turns in the windings. | ||
| + | |||
| ### Solution | |||
| In an ideal transformer (ideal means no losses, no leakage, and infinite magnetic permeability) | |||
| the voltage induced in each winding is proportional to the number of turns | |||
| $ V_1 = N_1 \frac{d\Phi}{dt}, V_2 = N_2 \frac{d\Phi}{dt}$ | |||
| With the secondary winding short‑circuited, the voltage $ V_2 = 0$ | |||
| Since$ V_2 = N_2 \frac{d\Phi}{dt}$ | |||
| this implies that the magnetic flux $\Phi $in the core must be constant (or zero, if starting from zero initial conditions). For the net flux to be zero, the total magnetomotive force must be zero | |||
| $N_1 I_1 + N_2 I_2 = 0$ | |||
| From this, the ratio of magnitudes is : | |||
| $\boxed{\frac{I_1}{I_2} = \frac{N_2}{N_1}}$ | |||
| The negative sign indicates that the currents are in opposite phase, but the ratio of amplitudes is as given. | |||
| @@ -24,4 +30,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{\frac{I_1}{I_2} = \frac{N_2}{N_1}}$ | ||
| @@ -1,7 +1,13 @@ | |||
| ### Statement | ### Statement | ||
| $11.3.18.$ |
$11.3.18.$ | ||
| Show that in an ideal transformer with a short-circuited secondary winding, | |||
| the ratioI1=NN12 | |||
| holds, where I1and I2are the currents, and N1and N2are | |||
| I2 | |||
| the number of turns in the windings. | |||
| ### Solution | ### Solution | ||
| In an ideal transformer (ideal means no losses, no leakage, and infinite magnetic permeability) | In an ideal transformer (ideal means no losses, no leakage, and infinite magnetic permeability) | ||
| the voltage induced in each winding is proportional to the number of turns | the voltage induced in each winding is proportional to the number of turns | ||
| $ V_1 = N_1 \frac{d\Phi}{dt}, V_2 = N_2 \frac{d\Phi}{dt}$ | $ V_1 = N_1 \frac{d\Phi}{dt}, V_2 = N_2 \frac{d\Phi}{dt}$ | ||
| With the secondary winding short‑circuited, the voltage $ V_2 = 0$ | With the secondary winding short‑circuited, the voltage $ V_2 = 0$ | ||
| Since$ V_2 = N_2 \frac{d\Phi}{dt}$ | Since$ V_2 = N_2 \frac{d\Phi}{dt}$ | ||
| this implies that the magnetic flux $\Phi $in the core must be constant (or zero, if starting from zero initial conditions). For the net flux to be zero, the total magnetomotive force must be zero | this implies that the magnetic flux $\Phi $in the core must be constant (or zero, if starting from zero initial conditions). For the net flux to be zero, the total magnetomotive force must be zero | ||
| $N_1 I_1 + N_2 I_2 = 0$ | $N_1 I_1 + N_2 I_2 = 0$ | ||
| From this, the ratio of magnitudes is : | From this, the ratio of magnitudes is : | ||
| $\boxed{\frac{I_1}{I_2} = \frac{N_2}{N_1}}$ | $\boxed{\frac{I_1}{I_2} = \frac{N_2}{N_1}}$ | ||
| The negative sign indicates that the currents are in opposite phase, but the ratio of amplitudes is as given. | The negative sign indicates that the currents are in opposite phase, but the ratio of amplitudes is as given. | ||
| @@ -24,4 +30,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{\frac{I_1}{I_2} = \frac{N_2}{N_1}}$ | ||