3. Consider the circuit shown below. The switch S is closed for a very long time so that the current through the inductor I, is constant. At time t=0 the switch is opened, what is the ratio of the current I through the inductor 75 msec after the switch was opened to the initial current: 1/1₂? A. 0.1 B. 0.2 C. 0.3 D. 0.4 E. 0.5 10.0V- 4,000 W L 8.00 22 8.000 W 1.00 H
3. Consider the circuit shown below. The switch S is closed for a very long time so that the current through the inductor I, is constant. At time t=0 the switch is opened, what is the ratio of the current I through the inductor 75 msec after the switch was opened to the initial current: 1/1₂? A. 0.1 B. 0.2 C. 0.3 D. 0.4 E. 0.5 10.0V- 4,000 W L 8.00 22 8.000 W 1.00 H
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 3: Analysis of Inductor Circuit**
**Problem Statement:**
Consider the circuit shown below. The switch \( S \) is closed for a very long time so that the current through the inductor \( I_0 \) is constant. At time \( t = 0 \) the switch is opened, what is the ratio of the current \( I \) through the inductor 75 msec after the switch was opened to the initial current \( I_0 \)?
**Multiple Choice Options:**
- A. 0.1
- B. 0.2
- C. 0.3
- D. 0.4
- E. 0.5
**Circuit Diagram Explanation:**
The circuit diagram includes:
- A 10.0 V voltage source.
- A switch \( S \).
- Three resistors with values of 4.00 Ω, 8.00 Ω, and 8.00 Ω.
- An inductor with inductance of 1.00 H.
The 4.00 Ω and 8.00 Ω resistors are in series with each other and parallel to another branch containing the 8.00 Ω resistor and 1.00 H inductor.
**Understanding the Circuit Behavior:**
1. **Initial Condition:** When the switch \( S \) is closed for a long time, the inductor acts like a short circuit (since inductors oppose changes in current and eventually act as a short for DC current after long periods). The current through the circuit reaches a steady-state value \( I_0 \).
2. **Switch Opened at \( t = 0 \):** Upon opening the switch, the circuit now relies on the inductor to oppose the sudden change in current. The inductor discharges and the current decays exponentially.
**Calculation Process:**
The decay of current through an inductor \( L \) in a series RL circuit follows the exponential form:
\[ I(t) = I_0 e^{-\frac{R_{\text{eq}}}{L} t} \]
Where:
- \( I_0 \) is the initial current through the inductor.
- \( R_{\text{eq}} \) is the equivalent resistance seen by the inductor when the switch is open.
- \( L \) is the inductance.
- \( t \) is the time.
Given:
- \( L = 1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa90f48aa-fe1f-4446-8e54-ba82906f66bc%2F981781c8-0dd6-4347-94a5-e28ef7386d49%2Fd0coaaq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 3: Analysis of Inductor Circuit**
**Problem Statement:**
Consider the circuit shown below. The switch \( S \) is closed for a very long time so that the current through the inductor \( I_0 \) is constant. At time \( t = 0 \) the switch is opened, what is the ratio of the current \( I \) through the inductor 75 msec after the switch was opened to the initial current \( I_0 \)?
**Multiple Choice Options:**
- A. 0.1
- B. 0.2
- C. 0.3
- D. 0.4
- E. 0.5
**Circuit Diagram Explanation:**
The circuit diagram includes:
- A 10.0 V voltage source.
- A switch \( S \).
- Three resistors with values of 4.00 Ω, 8.00 Ω, and 8.00 Ω.
- An inductor with inductance of 1.00 H.
The 4.00 Ω and 8.00 Ω resistors are in series with each other and parallel to another branch containing the 8.00 Ω resistor and 1.00 H inductor.
**Understanding the Circuit Behavior:**
1. **Initial Condition:** When the switch \( S \) is closed for a long time, the inductor acts like a short circuit (since inductors oppose changes in current and eventually act as a short for DC current after long periods). The current through the circuit reaches a steady-state value \( I_0 \).
2. **Switch Opened at \( t = 0 \):** Upon opening the switch, the circuit now relies on the inductor to oppose the sudden change in current. The inductor discharges and the current decays exponentially.
**Calculation Process:**
The decay of current through an inductor \( L \) in a series RL circuit follows the exponential form:
\[ I(t) = I_0 e^{-\frac{R_{\text{eq}}}{L} t} \]
Where:
- \( I_0 \) is the initial current through the inductor.
- \( R_{\text{eq}} \) is the equivalent resistance seen by the inductor when the switch is open.
- \( L \) is the inductance.
- \( t \) is the time.
Given:
- \( L = 1.
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