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title: "A circuit consists of an ideal battery of emf \\(\\varepsilon\\), an open switch \\(S\\), three resistors with resistances \\(R_1 = R\\), \\(R_2 = 2R\\), and \\(R_3 = R\\), and an initially uncharged capacitor of capacitance \\(C\\). Branch 1 contains resistors \\(R_1\\) and \\(R_2\\) in series with node \\(X\\) located between them. Branch 2 contains capacitor \\(C\\) and resistor \\(R_3\\) in series with node \\(Y\\) located between them. At time \\(t = 0\\), switch \\(S\\) is closed. What are the potential difference \\(V_X – V_Y\\) immediately after the switch is closed (\\(t = 0^+\\)) and the initial rate of change of this potential difference, \\(\\left.\\dfrac{d(V_X – V_Y)}{dt}\\right|_{t=0^+}\\)?"
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url: "https://nerd-notes.com/ubq/118461/"
date_modified: "2026-08-04T08:10:17+00:00"
---

# A circuit consists of an ideal battery of emf \(\varepsilon\), an open switch \(S\), three resistors with resistances \(R_1 = R\), \(R_2 = 2R\), and \(R_3 = R\), and an initially uncharged capacitor of capacitance \(C\). Branch 1 contains resistors \(R_1\) and \(R_2\) in series with node \(X\) located between them. Branch 2 contains capacitor \(C\) and resistor \(R_3\) in series with node \(Y\) located between them. At time \(t = 0\), switch \(S\) is closed. What are the potential difference \(V_X – V_Y\) immediately after the switch is closed (\(t = 0^+\)) and the initial rate of change of this potential difference, \(\left.\dfrac{d(V_X – V_Y)}{dt}\right|_{t=0^+}\)?

A circuit consists of an ideal battery of emf \(\varepsilon\), an open switch \(S\), three resistors with resistances \(R_1 = R\), \(R_2 = 2R\), and \(R_3 = R\), and an initially uncharged capacitor of capacitance \(C\). Branch 1 contains resistors \(R_1\) and \(R_2\) in series with node \(X\) located between them. Branch 2 contains capacitor \(C\) and resistor \(R_3\) in series with node \(Y\) located between them. At time \(t = 0\), switch \(S\) is closed. What are the potential difference \(V_X - V_Y\) immediately after the switch is closed (\(t = 0^+\)) and the initial rate of change of this potential difference, \(\left.\dfrac{d(V_X - V_Y)}{dt}\right|_{t=0^+}\)?

![A schematic diagram of a direct-current circuit. An ideal battery labeled \(\varepsilon\) is on the left side, oriented vertically with its long positive terminal at the top and short negative terminal at the bottom. A switch labeled \(S\) is located on the top horizontal wire leading from the positive battery terminal. Beyond the switch, the circuit splits into two parallel vertical branches before returning to the negative battery terminal at the bottom rail. The left branch contains a top resistor labeled \(R_1 = R\) and a bottom resistor labeled \(R_2 = 2R\), with a junction node labeled \(X\) located between them. The right branch contains a top capacitor labeled \(C\) drawn as two parallel plates and a bottom resistor labeled \(R_3 = R\), with a junction node labeled \(Y\) located between them. The bottom wire connects the lower terminals of \(R_2\) and \(R_3\) back to the negative terminal of the battery. No other labels, components, or lines appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785831017-NaQhIa.jpg)

- **A.** \((V_X - V_Y)_{t=0^+} = -\dfrac{1}{3}\varepsilon\), \quad \left.\dfrac{d(V_X - V_Y)}{dt}\right|_{t=0^+} = +\dfrac{\varepsilon}{RC}\)
- **B.** \((V_X - V_Y)_{t=0^+} = +\dfrac{1}{3}\varepsilon\), \quad \left.\dfrac{d(V_X - V_Y)}{dt}\right|_{t=0^+} = -\dfrac{\varepsilon}{RC}\)
- **C.** \((V_X - V_Y)_{t=0^+} = -\dfrac{1}{3}\varepsilon\), \quad \left.\dfrac{d(V_X - V_Y)}{dt}\right|_{t=0^+} = -\dfrac{\varepsilon}{RC}\)
- **D.** \((V_X - V_Y)_{t=0^+} = +\dfrac{2}{3}\varepsilon\), \quad \left.\dfrac{d(V_X - V_Y)}{dt}\right|_{t=0^+} = +\dfrac{\varepsilon}{RC}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118461/*