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title: "A circuit consists of an ideal battery of electromotive force \\(\\mathcal{E}\\), a resistor of resistance \\(R\\), two initially uncharged capacitors of capacitances \\(C_1\\) and \\(C_2\\), and an open switch, all connected in series. At time \\(t = 0\\), the switch is closed. Which of the following expressions represents the potential difference \\(V_1(t)\\) across capacitor \\(C_1\\) as a function of time \\(t\\)?"
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url: "https://nerd-notes.com/ubq/118438/"
date_modified: "2026-08-04T08:10:04+00:00"
---

# A circuit consists of an ideal battery of electromotive force \(\mathcal{E}\), a resistor of resistance \(R\), two initially uncharged capacitors of capacitances \(C_1\) and \(C_2\), and an open switch, all connected in series. At time \(t = 0\), the switch is closed. Which of the following expressions represents the potential difference \(V_1(t)\) across capacitor \(C_1\) as a function of time \(t\)?

A circuit consists of an ideal battery of electromotive force \(\mathcal{E}\), a resistor of resistance \(R\), two initially uncharged capacitors of capacitances \(C_1\) and \(C_2\), and an open switch, all connected in series. At time \(t = 0\), the switch is closed. Which of the following expressions represents the potential difference \(V_1(t)\) across capacitor \(C_1\) as a function of time \(t\)?

![A single closed rectangular circuit loop. On the left vertical branch is an ideal DC battery labeled \mathcal{E} with its long positive terminal line on top. On the top horizontal branch is an open switch labeled S and a resistor labeled R. On the right vertical branch are two capacitors connected in series, the upper one labeled C_1 and the lower one labeled C_2. Straight wires connect all components into a single series loop. No other labels, text, or components appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-circuit-1-1785831004-dTaMK9.jpg)

- **A.** \(V_1(t) = \mathcal{E} \left(1 - e^{-\frac{t(C_1 + C_2)}{R C_1 C_2}}\right)\)
- **B.** \(V_1(t) = \dfrac{C_2 \mathcal{E}}{C_1 + C_2} \left(1 - e^{-\frac{t}{R (C_1 + C_2)}}\right)\)
- **C.** \(V_1(t) = \dfrac{C_1 \mathcal{E}}{C_1 + C_2} \left(1 - e^{-\frac{t(C_1 + C_2)}{R C_1 C_2}}\right)\)
- **D.** \(V_1(t) = \dfrac{C_2 \mathcal{E}}{C_1 + C_2} \left(1 - e^{-\frac{t(C_1 + C_2)}{R C_1 C_2}}\right)\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118438/*