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The capacitor in Figure 28.38a begins to charge after the
Chapter 28, Problem 28.119(choose chapter or problem)
The capacitor in Figure 28.38a begins to charge after the switch closes at \(t = 0 s\). Analyze this circuit and show that \(Q=Q_{\max}\left(1-e^{-t/t}\right)\), where \(Q_{\text {max }}=C \varepsilon\).
Questions & Answers
QUESTION:
The capacitor in Figure 28.38a begins to charge after the switch closes at \(t = 0 s\). Analyze this circuit and show that \(Q=Q_{\max}\left(1-e^{-t/t}\right)\), where \(Q_{\text {max }}=C \varepsilon\).
ANSWER:Step 1 of 2
The value of charge is \(Q_{\max }=C \varepsilon\).
The equation of charge is \(Q=Q_{\max }\left(1-e^{-t / \tau}\right)\).
In order to equation of charge, we have to:
Apply the loop rule in which electrostatic force is conservative.
\(\begin{array}{r} \sum V=0 \\ \varepsilon-\Delta V-I R=0 \end{array}\)
\(\text { For } \varepsilon=\frac{Q_{\max }}{C} \quad I=\frac{d Q}{d t} \text { and } \Delta V=\frac{Q}{C}\)
Therefore,
\(\begin{aligned} \left(\frac{Q_{\max }}{C}\right)-\left(\frac{Q}{C}\right)-\left(\frac{d Q}{d t}\right) R & =0 \\ \frac{Q_{\max }-Q}{C} & =\frac{d Q}{d t} R \\ \frac{d Q}{Q_{\max }-Q} & =\left(\frac{1}{R C}\right) d t \end{aligned}\)
(1)