Edits to “Statement”, “Solution”, “Answer”
en/11.4.2.md
+43 −3
| @@ -1,11 +1,51 @@ | |||
| ### Statement | |||
| − | $11.4.2.$ [Insert the problem statement] | ||
| + | $11.4.2.$ How should the voltage change in an electrical circuit consisting of a series-connected inductor $L$ and resistance $R$, so that the current in it: (a) increases linearly: $I = \alpha t$? (b) changed sinusoidally: $I = I_0\sin\omega t$? | ||
| + | |||
| ### Solution | |||
| − | 123 | ||
| + | From the Kirchoff's voltage law: | ||
| + | $$ | ||
| + | iR+L\frac{di}{dt}=u,\tag{1} | ||
| + | $$ | ||
| + | |||
| + | In the case (а): | ||
| + | |||
| + | \begin{gather} | ||
| + | \alpha tR+L\alpha=u, | ||
| + | \end{gather} | ||
| + | |||
| + | or equivalently | ||
| + | |||
| + | \begin{gather} | ||
| + | u=\alpha(Rt+L). | ||
| + | \end{gather} | ||
| + | |||
| + | In the case (b): | ||
| + | |||
| + | \begin{gather} | ||
| + | RI_0\sin\omega t + L\omega I_0\cos\omega t=u, | ||
| + | \end{gather} | ||
| + | |||
| + | or equivalently | ||
| + | |||
| + | \begin{gather} | ||
| + | u=I_0(R\sin\omega t + L\omega\cos\omega t). | ||
| + | \end{gather} | ||
| + | |||
| + | |||
| #### Answer | |||
| − | |||
| + | (а) | ||
| + | |||
| + | $$ | ||
| + | u=\alpha(Rt+L); | ||
| + | $$ | ||
| + | |||
| + | (b) | ||
| + | |||
| + | $$ | ||
| + | u=I_0(R\sin\omega t + L\omega\cos\omega t). | ||
| + | $$ | ||
| @@ -1,11 +1,51 @@ | |||
| ### Statement | ### Statement | ||
| $11.4.2.$ [Insert the problem statement] | $11.4.2.$ How should the voltage change in an electrical circuit consisting of a series-connected inductor $L$ and resistance $R$, so that the current in it: (a) increases linearly: $I = \alpha t$? (b) changed sinusoidally: $I = I_0\sin\omega t$? | ||
| ### Solution | ### Solution | ||
| 123 | From the Kirchoff's voltage law: | ||
| $$ | |||
| iR+L\frac{di}{dt}=u,\tag{1} | |||
| $$ | |||
| In the case (а): | |||
| \begin{gather} | |||
| \alpha tR+L\alpha=u, | |||
| \end{gather} | |||
| or equivalently | |||
| \begin{gather} | |||
| u=\alpha(Rt+L). | |||
| \end{gather} | |||
| In the case (b): | |||
| \begin{gather} | |||
| RI_0\sin\omega t + L\omega I_0\cos\omega t=u, | |||
| \end{gather} | |||
| or equivalently | |||
| \begin{gather} | |||
| u=I_0(R\sin\omega t + L\omega\cos\omega t). | |||
| \end{gather} | |||
| #### Answer | #### Answer | ||
| (а) | |||
| $$ | |||
| u=\alpha(Rt+L); | |||
| $$ | |||
| (b) | |||
| $$ | |||
| u=I_0(R\sin\omega t + L\omega\cos\omega t). | |||
| $$ | |||