What is a circuit?
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Textbook Solutions for Physics for Scientists and Engineers: A Strategic Approach, Standard Edition (Chs 1-36)
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 \(Q_{\text {max }}=C \varepsilon\).
Solution
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)
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