---
title: "A space heater and a television are both designed to operate when connected across a standard household potential difference of \\(120\\text{ V}\\). Under normal operating conditions, the space heater dissipates \\(1200\\text{ W}\\) of electric power, while the television dissipates \\(300\\text{ W}\\) of electric power. Assuming both appliances behave as ohmic resistors, what is the ratio of the internal resistance of the space heater to the internal resistance of the television, \\(\\dfrac{R_{\\text{heater}}}{R_{\\text{TV}}}\\)?"
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url: "https://nerd-notes.com/ubq/120377/"
date_modified: "2026-08-23T04:34:19+00:00"
---

# A space heater and a television are both designed to operate when connected across a standard household potential difference of \(120\text{ V}\). Under normal operating conditions, the space heater dissipates \(1200\text{ W}\) of electric power, while the television dissipates \(300\text{ W}\) of electric power. Assuming both appliances behave as ohmic resistors, what is the ratio of the internal resistance of the space heater to the internal resistance of the television, \(\dfrac{R_{\text{heater}}}{R_{\text{TV}}}\)?

A space heater and a television are both designed to operate when connected across a standard household potential difference of \(120\text{ V}\). Under normal operating conditions, the space heater dissipates \(1200\text{ W}\) of electric power, while the television dissipates \(300\text{ W}\) of electric power. Assuming both appliances behave as ohmic resistors, what is the ratio of the internal resistance of the space heater to the internal resistance of the television, \(\dfrac{R_{\text{heater}}}{R_{\text{TV}}}\)?

- **A.** \(\dfrac{1}{4}\)
- **B.** \(\dfrac{1}{2}\)
- **C.** \(2\)
- **D.** \(4\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120377/*