When an inductor is connected to a 60.0 Hz source it has an inductive reactance of 50.2 0. Determine the maximum current in the inductor (in A) if it is connected to a 47.0 Hz source that produces a 110 V rms voltage.
When an inductor is connected to a 60.0 Hz source it has an inductive reactance of 50.2 0. Determine the maximum current in the inductor (in A) if it is connected to a 47.0 Hz source that produces a 110 V rms voltage.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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![**Problem Statement: Inductor in AC Circuit**
When an inductor is connected to a 60.0 Hz source, it has an inductive reactance of 50.2 Ω. Determine the maximum current in the inductor (in Amperes) if it is connected to a 47.0 Hz source that produces a 110 V rms voltage.
**Answer:**
\[ \text{Maximum Current} (I_{\text{max}}) = \_\_\_ \text{ A} \]
*Note: There is a text box provided for the answer.*
**Explanation:**
To solve this problem, use the formula for inductive reactance (\(X_L\)):
\[ X_L = 2 \pi f L \]
Where:
- \( f \) is the frequency of the AC source
- \( L \) is the inductance of the coil
Given the existing reactance at 60.0 Hz, you can find \( L \) and then determine the new reactance at 47.0 Hz. Use Ohm's Law for AC circuits (\( V = I X_L \)) to find the current. Remember \( V \) here refers to the rms voltage, and \( I \) is the rms current related to the maximum current by \( I_{\text{max}} = \sqrt{2} \times I_{\text{rms}} \).
**Equations to Consider:**
1. \( X_L = 2 \pi f L \)
2. \( I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L} \)
3. \( I_{\text{max}} = \sqrt{2} \times I_{\text{rms}} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21f50632-2fc7-432c-bfea-564ad2558809%2F91056782-747b-4027-8094-5ab4d8a9b45f%2F9cxlymi_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement: Inductor in AC Circuit**
When an inductor is connected to a 60.0 Hz source, it has an inductive reactance of 50.2 Ω. Determine the maximum current in the inductor (in Amperes) if it is connected to a 47.0 Hz source that produces a 110 V rms voltage.
**Answer:**
\[ \text{Maximum Current} (I_{\text{max}}) = \_\_\_ \text{ A} \]
*Note: There is a text box provided for the answer.*
**Explanation:**
To solve this problem, use the formula for inductive reactance (\(X_L\)):
\[ X_L = 2 \pi f L \]
Where:
- \( f \) is the frequency of the AC source
- \( L \) is the inductance of the coil
Given the existing reactance at 60.0 Hz, you can find \( L \) and then determine the new reactance at 47.0 Hz. Use Ohm's Law for AC circuits (\( V = I X_L \)) to find the current. Remember \( V \) here refers to the rms voltage, and \( I \) is the rms current related to the maximum current by \( I_{\text{max}} = \sqrt{2} \times I_{\text{rms}} \).
**Equations to Consider:**
1. \( X_L = 2 \pi f L \)
2. \( I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L} \)
3. \( I_{\text{max}} = \sqrt{2} \times I_{\text{rms}} \)
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