Edits to “Statement”, “Solution”, “Answer”
en/8.4.8.md
+4 −4
| @@ -1,18 +1,18 @@ | |||
| ### Statement | |||
| − | $8.4.8.$ [Insert the problem statement] | ||
| + | $8.4.8.$ The diode has the current-voltage characteristic shown in the figure. At $V_0$, the diode opens. The capacitor is not charged at first. How much heat will be released on the resistance after the key is closed? | ||
| ### Solution | |||
| − |  | ||
| Clearly, if $\mathcal{E}\le V_0$, then there is no current in the circuit and hence no heat generated in the resistor. If $\mathcal{E}>V_0$, then the charge $q$ collected on the capacitor satisfies the differential equation | |||
| \[\dot qR+\frac{q}{C}=\mathcal{E}-V_0,\] | |||
| where the terms on the left are voltage drops across the resistor and the capacitor, respectively. Solving this equation subject to the condition that $q$ is initially zero, we get | |||
| − | \[q=C(\mathcal{E}-V_0) | ||
| + | \[q=C(\mathcal{E}-V_0)[1-e^{-t/(RC)}],\] | ||
| so the current in the circuit is | |||
| \[I=\dot q=\frac{\mathcal{E}-V_0}{R}\cdot e^{-t/(RC)}.\] | |||
| Thus, the heat generated in the resistor is | |||
| \[\int_0^\infty I^2R\,dt=\frac{(\mathcal{E}-V_0)^2}{R}\int_0^\infty e^{-2t/(RC)}=\frac{C(\mathcal{E}-V_0)^2}{2}.\] | |||
| @@ -24,4 +24,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\frac{C(\mathcal{E}-V_0)^2}{2}$ if $\mathcal{E}>V_0$ and zero otherwise. | ||
| @@ -1,18 +1,18 @@ | |||
| ### Statement | ### Statement | ||
| $8.4.8.$ [Insert the problem statement] | $8.4.8.$ The diode has the current-voltage characteristic shown in the figure. At $V_0$, the diode opens. The capacitor is not charged at first. How much heat will be released on the resistance after the key is closed? | ||
| ### Solution | ### Solution | ||
|  | ||
| Clearly, if $\mathcal{E}\le V_0$, then there is no current in the circuit and hence no heat generated in the resistor. If $\mathcal{E}>V_0$, then the charge $q$ collected on the capacitor satisfies the differential equation | Clearly, if $\mathcal{E}\le V_0$, then there is no current in the circuit and hence no heat generated in the resistor. If $\mathcal{E}>V_0$, then the charge $q$ collected on the capacitor satisfies the differential equation | ||
| \[\dot qR+\frac{q}{C}=\mathcal{E}-V_0,\] | \[\dot qR+\frac{q}{C}=\mathcal{E}-V_0,\] | ||
| where the terms on the left are voltage drops across the resistor and the capacitor, respectively. Solving this equation subject to the condition that $q$ is initially zero, we get | where the terms on the left are voltage drops across the resistor and the capacitor, respectively. Solving this equation subject to the condition that $q$ is initially zero, we get | ||
| \[q=C(\mathcal{E}-V_0) |
\[q=C(\mathcal{E}-V_0)[1-e^{-t/(RC)}],\] | ||
| so the current in the circuit is | so the current in the circuit is | ||
| \[I=\dot q=\frac{\mathcal{E}-V_0}{R}\cdot e^{-t/(RC)}.\] | \[I=\dot q=\frac{\mathcal{E}-V_0}{R}\cdot e^{-t/(RC)}.\] | ||
| Thus, the heat generated in the resistor is | Thus, the heat generated in the resistor is | ||
| \[\int_0^\infty I^2R\,dt=\frac{(\mathcal{E}-V_0)^2}{R}\int_0^\infty e^{-2t/(RC)}=\frac{C(\mathcal{E}-V_0)^2}{2}.\] | \[\int_0^\infty I^2R\,dt=\frac{(\mathcal{E}-V_0)^2}{R}\int_0^\infty e^{-2t/(RC)}=\frac{C(\mathcal{E}-V_0)^2}{2}.\] | ||
| @@ -24,4 +24,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\frac{C(\mathcal{E}-V_0)^2}{2}$ if $\mathcal{E}>V_0$ and zero otherwise. | ||