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title: "A circuit contains an ideal battery of emf \\(\\mathcal{E}\\), a switch \\(S\\), three identical resistors of resistance \\(R_1 = R_2 = R_3 = R\\), and an initially uncharged capacitor of capacitance \\(C\\). Resistor \\(R_1\\) is connected in series with the battery, while resistors \\(R_2\\) and \\(R_3\\) are in two parallel branches, with \\(R_3\\) in series with the capacitor \\(C\\) in its branch. The switch \\(S\\) is closed at time \\(t = 0\\). Which of the following claims correctly describes how the current \\(I_2\\) through resistor \\(R_2\\) changes for \\(t > 0\\), and provides the correct physical justification?"
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url: "https://nerd-notes.com/ubq/118390/"
date_modified: "2026-08-04T08:09:52+00:00"
---

# A circuit contains an ideal battery of emf \(\mathcal{E}\), a switch \(S\), three identical resistors of resistance \(R_1 = R_2 = R_3 = R\), and an initially uncharged capacitor of capacitance \(C\). Resistor \(R_1\) is connected in series with the battery, while resistors \(R_2\) and \(R_3\) are in two parallel branches, with \(R_3\) in series with the capacitor \(C\) in its branch. The switch \(S\) is closed at time \(t = 0\). Which of the following claims correctly describes how the current \(I_2\) through resistor \(R_2\) changes for \(t > 0\), and provides the correct physical justification?

A circuit contains an ideal battery of emf \(\mathcal{E}\), a switch \(S\), three identical resistors of resistance \(R_1 = R_2 = R_3 = R\), and an initially uncharged capacitor of capacitance \(C\). Resistor \(R_1\) is connected in series with the battery, while resistors \(R_2\) and \(R_3\) are in two parallel branches, with \(R_3\) in series with the capacitor \(C\) in its branch. The switch \(S\) is closed at time \(t = 0\). Which of the following claims correctly describes how the current \(I_2\) through resistor \(R_2\) changes for \(t > 0\), and provides the correct physical justification?

![A schematic diagram of a direct current circuit. An ideal battery with emf \mathcal{E} has its positive terminal oriented upward on the left vertical branch. A switch S is in series with the battery on the upper horizontal wire, followed by a resistor labeled R_1 = R. The circuit then splits at a three-way junction into two vertical parallel branches. The left vertical branch contains a resistor labeled R_2 = R. The right vertical branch contains a resistor labeled R_3 = R in series with an initially uncharged capacitor labeled C. Both parallel branches reconnect at a bottom junction on the lower horizontal wire, returning to the negative terminal of the battery. Arrows indicate current I_1 through R_1, current I_2 through R_2, and current I_3 through R_3 and C. No other labels, lines, text, or components appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830992-DtIsst.jpg)

- **A.** \(I_2\) decreases over time because as the capacitor charges, the net equivalent resistance of the circuit increases, which reduces all branch currents in the circuit.
- **B.** \(I_2\) decreases over time because charge accumulating on the capacitor plates generates a back emf that repels incoming charge from both parallel branches.
- **C.** \(I_2\) increases over time because as the capacitor charges, total circuit current decreases, which reduces the potential drop across \(R_1\) and increases the potential difference across \(R_2\).
- **D.** \(I_2\) remains constant over time because \(R_2\) is in a parallel branch independent of the capacitor branch, keeping the potential difference across \(R_2\) fixed.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118390/*