### Problem StatementA light bulb filament has an RMS current of 0.880 A flowing through it. The voltage in the bulb depends on the time $ t $ according to the relation:\[ V = (300 \text{ V}) \sin(345 \text{ Hz} \cdot t) \]**Graph Explanation:**The provided graph is a sinusoidal waveform, which depicts the variation of voltage (amplitude) with respect to time. The oscillating nature of the sine wave indicates alternating current (AC) with a peak voltage of 300 V.### QuestionWhat is the **resistance** of the light bulb filament?#### Answer Choices:- 375 Ω- 110 Ω- 241 Ω- 187 Ω- 552 Ω- 126 Ω### Detailed Explanation:To determine the resistance of the light bulb filament, apply Ohm's Law in the context of RMS values for AC circuits:\[ V_{\text{rms}} = I_{\text{rms}} \times R \]Given:- $ I_{\text{rms}} = 0.880 \text{ A} $- $ V_{\text{peak}} = 300 \text{ V} $The RMS voltage ( $ V_{\text{rms}} $ ) for a sinusoidal AC voltage is given by:\[ V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \]\[ V_{\text{rms}} = \frac{300 \text{ V}}{\sqrt{2}} \approx 212.1 \text{ V} \]Using Ohm's Law:\[ 212.1 \text{ V} = 0.880 \text{ A} \times R \]\[ R = \frac{212.1 \text{ V}}{0.880 \text{ A}} \approx 241 \, \Omega \]The resistance of the light bulb filament is **241 Ω**.

Given rms current is irms=0.880 A Voltage in the bulb is V=(300 V)sin345 Hzt We have to find the…

A light bulb filament has a rms current of 0.880 A flowing through it. The voltage in the bulb depends on the time t according to the relation: V = (300 V) sin[(345 Hz)t] MM What is the resistance of the lightbulb filament? 375 Ω 110 Ω 241 02 187 Ω • 552 Ω Amplitude 126 Ω

A light bulb filament has a rms current of 0.880 A flowing through it. The voltage in the bulb depends on the time t according to the relation: V = (300 V) sin[(345 Hz)t] MM What is the resistance of the lightbulb filament? 375 Ω 110 Ω 241 02 187 Ω • 552 Ω Amplitude 126 Ω

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

### Problem Statement
A light bulb filament has an RMS current of 0.880 A flowing through it. The voltage in the bulb depends on the time $ t $ according to the relation:
\[ V = (300 \text{ V}) \sin(345 \text{ Hz} \cdot t) \]

**Graph Explanation:**
The provided graph is a sinusoidal waveform, which depicts the variation of voltage (amplitude) with respect to time. The oscillating nature of the sine wave indicates alternating current (AC) with a peak voltage of 300 V.

### Question
What is the **resistance** of the light bulb filament?

#### Answer Choices:
- 375 Ω
- 110 Ω
- 241 Ω
- 187 Ω
- 552 Ω
- 126 Ω

### Detailed Explanation:
To determine the resistance of the light bulb filament, apply Ohm's Law in the context of RMS values for AC circuits:
\[ V_{\text{rms}} = I_{\text{rms}} \times R \]

Given:
- $ I_{\text{rms}} = 0.880 \text{ A} $
- $ V_{\text{peak}} = 300 \text{ V} $

The RMS voltage ( $ V_{\text{rms}} $ ) for a sinusoidal AC voltage is given by:
\[ V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \]
\[ V_{\text{rms}} = \frac{300 \text{ V}}{\sqrt{2}} \approx 212.1 \text{ V} \]

Using Ohm's Law:
\[ 212.1 \text{ V} = 0.880 \text{ A} \times R \]
\[ R = \frac{212.1 \text{ V}}{0.880 \text{ A}} \approx 241 \, \Omega \]

The resistance of the light bulb filament is **241 Ω**.

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