---
title: "A circuit contains an ideal battery of voltage \\(V_0\\) connected in series with resistor \\(R_A\\). Downstream of \\(R_A\\), the circuit splits into two parallel branches: Branch 1 contains resistor \\(R_B\\), and Branch 2 contains resistors \\(R_C\\) and \\(R_D\\) in series. Resistor \\(R_D\\) suddenly burns out and becomes an open circuit. Which of the following claims correctly describes the change in power dissipated by resistor \\(R_B\\) and provides the correct physical justification?"
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url: "https://nerd-notes.com/ubq/118416/"
date_modified: "2026-08-04T08:09:58+00:00"
---

# A circuit contains an ideal battery of voltage \(V_0\) connected in series with resistor \(R_A\). Downstream of \(R_A\), the circuit splits into two parallel branches: Branch 1 contains resistor \(R_B\), and Branch 2 contains resistors \(R_C\) and \(R_D\) in series. Resistor \(R_D\) suddenly burns out and becomes an open circuit. Which of the following claims correctly describes the change in power dissipated by resistor \(R_B\) and provides the correct physical justification?

A circuit contains an ideal battery of voltage \(V_0\) connected in series with resistor \(R_A\). Downstream of \(R_A\), the circuit splits into two parallel branches: Branch 1 contains resistor \(R_B\), and Branch 2 contains resistors \(R_C\) and \(R_D\) in series. Resistor \(R_D\) suddenly burns out and becomes an open circuit. Which of the following claims correctly describes the change in power dissipated by resistor \(R_B\) and provides the correct physical justification?

![A rectangular circuit schematic with an ideal DC battery labeled V_0 on the left vertical wire, positive terminal facing up. The top horizontal wire contains a resistor labeled R_A. To the right of R_A, the wire reaches a junction node that splits into two parallel vertical branches. The middle vertical branch contains a single resistor labeled R_B. The right vertical branch contains two resistors in series, top one labeled R_C and bottom one labeled R_D. Both branches reconnect at a bottom junction node on the bottom horizontal wire, which leads back to the negative terminal of the battery. No other labels, lines, text, or components appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830997-S0GnOb.jpg)

- **A.** The power dissipated by \(R_B\) decreases because the total current supplied by the battery decreases, reducing the current through all remaining components in the circuit.
- **B.** The power dissipated by \(R_B\) remains unchanged because \(R_B\) is in a parallel branch and continues to receive the same fraction of the battery potential difference.
- **C.** The power dissipated by \(R_B\) increases because all current that previously flowed through Branch 2 is redirected into \(R_B\) while the total current from the battery remains constant.
- **D.** The power dissipated by \(R_B\) increases because the reduction in total current decreases the potential difference across \(R_A\), which increases the potential difference across \(R_B\) by Kirchhoff's loop rule.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118416/*