Edits to “Statement”, “Solution”, “Answer”
en/11.4.5.md
+8 −7
| @@ -1,23 +1,24 @@ | |||
| ### Statement | |||
| − | $11.4.5.$ [Insert the problem statement] | ||
| + | $11.4.5.$ In the circuit shown in the figure, the diode $D$ and the inductor $L$ are connected to an alternating voltage source $V =V_0 cosωt$ using a key $K$. At time $t = 0$, the key $K$ is closed. Determine the current in the coil as a function of time.Plot the graph of this function. The diode and coil are considered ideal. Ignore the internal resistance of the source. | ||
| + |  | ||
| ### Solution | |||
| − | Запишем правило Кирхгофа: | ||
| + | Let's write down Kirchhoff's rule: | ||
| $$L\frac{dI}{dt} = V_0 cos(\omega t)$$ | |||
| − | Проинтегрируем и получим: | ||
| + | Integrate and get: | ||
| $$I = \frac{V_0}{\omega L} sin(\omega t)$$ | |||
| − | Этому уравнению подчиняется сила тока пока ток не идет в обратную сторону и диод не закрывается, то есть до времени $\frac{\pi}{\omega}$. Затем , пока напряжение на источнике снова не поменяет знак на положительный, это момент $\frac{3\pi}{2\omega}$. Затем ток снова начинает расти, но посмотрим теперь на уравнение. | ||
| + | This equation obeys the current strength until the current goes in the opposite direction and the diode closes, that is, until the time $\frac{\pi}{\omega}$. Then, until the voltage at the source changes sign to positive again, this is the moment $\frac{3\pi}{2\omega}$. Then the current starts to rise again, but now let's look at the equation. | ||
| − | I = \int^{t}_{\frac{3\pi}{2\omega}} \frac{V_0}{\omega L} cos(\omega t) = \frac{V_0}{\omega L} (sin(\omega t)+1) | ||
| + | $$I = \int^{t}_{\frac{3\pi}{2\omega}} \frac{V_0}{\omega L} cos(\omega t) = \frac{V_0}{\omega L} (sin(\omega t)+1)$$ | ||
| − | Период тот же, а амплитуда в два раза выше | ||
| + | The period is the same, but the amplitude is twice as high, and this type of dependence persists. | ||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + |  | ||
| @@ -1,23 +1,24 @@ | |||
| ### Statement | ### Statement | ||
| $11.4.5.$ [Insert the problem statement] | $11.4.5.$ In the circuit shown in the figure, the diode $D$ and the inductor $L$ are connected to an alternating voltage source $V =V_0 cosωt$ using a key $K$. At time $t = 0$, the key $K$ is closed. Determine the current in the coil as a function of time.Plot the graph of this function. The diode and coil are considered ideal. Ignore the internal resistance of the source. | ||
|  | |||
| ### Solution | ### Solution | ||
| Запишем правило Кирхгофа: | Let's write down Kirchhoff's rule: | ||
| $$L\frac{dI}{dt} = V_0 cos(\omega t)$$ | $$L\frac{dI}{dt} = V_0 cos(\omega t)$$ | ||
| Проинтегрируем и получим: | Integrate and get: | ||
| $$I = \frac{V_0}{\omega L} sin(\omega t)$$ | $$I = \frac{V_0}{\omega L} sin(\omega t)$$ | ||
| Этому уравнению подчиняется сила тока пока ток не идет в обратную сторону и диод не закрывается, то есть до времени $\frac{\pi}{\omega}$. Затем , пока напряжение на источнике снова не поменяет знак на положительный, это момент $\frac{3\pi}{2\omega}$. Затем ток снова начинает расти, но посмотрим теперь на уравнение. | This equation obeys the current strength until the current goes in the opposite direction and the diode closes, that is, until the time $\frac{\pi}{\omega}$. Then, until the voltage at the source changes sign to positive again, this is the moment $\frac{3\pi}{2\omega}$. Then the current starts to rise again, but now let's look at the equation. | ||
| I = \int^{t}_{\frac{3\pi}{2\omega}} \frac{V_0}{\omega L} cos(\omega t) = \frac{V_0}{\omega L} (sin(\omega t)+1) | $$I = \int^{t}_{\frac{3\pi}{2\omega}} \frac{V_0}{\omega L} cos(\omega t) = \frac{V_0}{\omega L} (sin(\omega t)+1)$$ | ||
| Период тот же, а амплитуда в два раза выше | The period is the same, but the amplitude is twice as high, and this type of dependence persists. | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] |  | ||