---
title: "A single closed circuit consists of an ideal battery with electromotive force \\(\\mathcal{E}\\), a linear resistor of resistance \\(R\\), and a non-ohmic circuit component connected in series. The potential difference across the non-ohmic component depends quadratically on the current \\(I\\) flowing through it according to \\(V_N = \\beta I^2\\), where \\(\\beta\\) is a known positive constant. Which of the following expressions correctly represents the steady-state current \\(I\\) in the circuit?"
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url: "https://nerd-notes.com/ubq/118402/"
date_modified: "2026-08-04T08:09:55+00:00"
---

# A single closed circuit consists of an ideal battery with electromotive force \(\mathcal{E}\), a linear resistor of resistance \(R\), and a non-ohmic circuit component connected in series. The potential difference across the non-ohmic component depends quadratically on the current \(I\) flowing through it according to \(V_N = \beta I^2\), where \(\beta\) is a known positive constant. Which of the following expressions correctly represents the steady-state current \(I\) in the circuit?

A single closed circuit consists of an ideal battery with electromotive force \(\mathcal{E}\), a linear resistor of resistance \(R\), and a non-ohmic circuit component connected in series. The potential difference across the non-ohmic component depends quadratically on the current \(I\) flowing through it according to \(V_N = \beta I^2\), where \(\beta\) is a known positive constant. Which of the following expressions correctly represents the steady-state current \(I\) in the circuit?

![A single rectangular circuit loop. The left vertical wire contains an ideal DC voltage source labeled with EMF symbol \(\mathcal{E}\), oriented with the longer positive plate pointing upward. The top horizontal wire contains a standard zig-zag resistor labeled R. The right vertical wire contains a rectangular box representing a non-ohmic component labeled V_N = \beta I^2. A clockwise directed arrow labeled I indicates the current in the loop. No other components, text, or symbols appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-circuit-1-1785830995-FYUuXq.jpg)

- **A.** \(\dfrac{\sqrt{R^2 + 2\beta\mathcal{E}} - R}{\beta}\)
- **B.** \(\dfrac{\sqrt{R^2 + 4\beta\mathcal{E}} - R}{2\beta}\)
- **C.** \(\dfrac{\sqrt{R^2 + 4\beta\mathcal{E}} + R}{2\beta}\)
- **D.** \(\dfrac{\sqrt{R^2 - 4\beta\mathcal{E}} - R}{2\beta}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118402/*