A 500 N resistor and a 300 mH inductor are connected in series with an ac generator with an rms voltage of 17.0 V and a frequency of 65.0 Hz. You may want to review (Pages 860 - 863) Part A What is the rms current in this circuit? | να ΑΣφ ? Ims = Submit Request Answer Provide Feedback 國
A 500 N resistor and a 300 mH inductor are connected in series with an ac generator with an rms voltage of 17.0 V and a frequency of 65.0 Hz. You may want to review (Pages 860 - 863) Part A What is the rms current in this circuit? | να ΑΣφ ? Ims = Submit Request Answer Provide Feedback 國
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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![**Problem Description:**
A 500 Ω resistor and a 300 mH inductor are connected in series with an AC generator. The generator provides an RMS voltage of 17.0 V and operates at a frequency of 65.0 Hz. To review the relevant concepts, you may want to consult pages 860 to 863.
---
**Part A:**
**Question:** What is the RMS current in this circuit?
**Answer Input:**
\[ I_{\text{rms}} = \, \_\_\_ \, \text{mA} \]
**Submit Answer** | [Request Answer]
[Provide Feedback]
---
**Explanation of the Circuit Concepts:**
When a resistor and inductor are connected in series in an AC circuit, the total impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + (X_L)^2} \]
where \( R \) is the resistance and \( X_L \) is the inductive reactance, calculated by:
\[ X_L = 2\pi f L \]
In this scenario:
- \( R = 500 \, \Omega \)
- \( L = 300 \, \text{mH} = 0.3 \, \text{H} \)
- \( f = 65.0 \, \text{Hz} \)
The RMS current \( I_{\text{rms}} \) can be determined using Ohm’s Law for AC circuits:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \]
where \( V_{\text{rms}} = 17.0 \, \text{V} \).
Calculate \( X_L \), then \( Z \), and finally \( I_{\text{rms}} \) using the above formulas.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81f7b2d1-b7ba-44aa-b963-9939c9f44db6%2F74a6de74-3e5a-4f99-a38f-1649b163d815%2Ftn1gako_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
A 500 Ω resistor and a 300 mH inductor are connected in series with an AC generator. The generator provides an RMS voltage of 17.0 V and operates at a frequency of 65.0 Hz. To review the relevant concepts, you may want to consult pages 860 to 863.
---
**Part A:**
**Question:** What is the RMS current in this circuit?
**Answer Input:**
\[ I_{\text{rms}} = \, \_\_\_ \, \text{mA} \]
**Submit Answer** | [Request Answer]
[Provide Feedback]
---
**Explanation of the Circuit Concepts:**
When a resistor and inductor are connected in series in an AC circuit, the total impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + (X_L)^2} \]
where \( R \) is the resistance and \( X_L \) is the inductive reactance, calculated by:
\[ X_L = 2\pi f L \]
In this scenario:
- \( R = 500 \, \Omega \)
- \( L = 300 \, \text{mH} = 0.3 \, \text{H} \)
- \( f = 65.0 \, \text{Hz} \)
The RMS current \( I_{\text{rms}} \) can be determined using Ohm’s Law for AC circuits:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \]
where \( V_{\text{rms}} = 17.0 \, \text{V} \).
Calculate \( X_L \), then \( Z \), and finally \( I_{\text{rms}} \) using the above formulas.
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