**Inductance and Current Decay in an Inductor****Equation Overview:**The current through inductance \( L \) is given by the formula:\[ I = I_0 e^{-t/\tau} \]where:- \( I \) is the current at time \( t \),- \( I_0 \) is the initial current,- \( \tau \) is the time constant,- \( t \) is the time.**Part B:**Evaluate \( \Delta V_L \) at \( t = 0 \, \text{ms} \), given:- \( L = 20 \, \text{mH} \),- \( I_0 = 50 \, \text{mA} \),- \( \tau = 1.0 \, \text{ms} \).**Answer:**\[ \Delta V_L = 1.0 \, \text{V} \]- This part confirms the answer is correct.**Part C:**Evaluate \( \Delta V_L \) at \( t = 1.0 \, \text{ms} \), given:- \( L = 20 \, \text{mH} \),- \( I_0 = 50 \, \text{mA} \),- \( \tau = 1.0 \, \text{ms} \).**Answer:**\[ \Delta V_L = 0.37 \, \text{V} \]- This part confirms the answer is correct.**Part D:**Evaluate \( \Delta V_L \) at \( t = 3.0 \, \text{ms} \), given:- \( L = 20 \, \text{mH} \),- \( I_0 = 50 \, \text{mA} \),- \( \tau = 1.0 \, \text{ms} \).**Answer:**- Input box for \( \Delta V_L \) is provided with placeholders for value and units. A "Submit" button is available to confirm the answer.**Note:** Each part requires the expression of the answer using two significant figures and the appropriate units.

Answered: Image /qna-images/answer/797471ad-d544-40be-8291-52e07a2d8a42.jpg

The current through inductance L is given by I = Io e-t/T. ♥ ▼ Part B Evaluate AVL at t = 0 ms, if L = 20 mH, Io = 50 mA, and T = 1.0 ms. Express your answer using two significant figures and include the appropriate units. AVL = 1.0 V Submit ✓ Correct Part C Evaluate AVL at t = 1.0 ms, if L = 20 mH, Io = 50 mA, and T = 1.0 ms. Express your answer using two significant figures and include the appropriate units. AVL = 0.37 V Submit ✓ Correct Part D Previous Answers AVL = Previous Answers Evaluate AVL at t = 3.0 ms, if L = 20 mH, Io = 50 mA, and T = 1.0 ms. Express your answer using two significant figures and include the appropriate units. Submit μA Value Request Answer Units ?

The current through inductance L is given by I = Io e-t/T. ♥ ▼ Part B Evaluate AVL at t = 0 ms, if L = 20 mH, Io = 50 mA, and T = 1.0 ms. Express your answer using two significant figures and include the appropriate units. AVL = 1.0 V Submit ✓ Correct Part C Evaluate AVL at t = 1.0 ms, if L = 20 mH, Io = 50 mA, and T = 1.0 ms. Express your answer using two significant figures and include the appropriate units. AVL = 0.37 V Submit ✓ Correct Part D Previous Answers AVL = Previous Answers Evaluate AVL at t = 3.0 ms, if L = 20 mH, Io = 50 mA, and T = 1.0 ms. Express your answer using two significant figures and include the appropriate units. Submit μA Value Request Answer Units ?

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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Concept explainers

Voltage Across Inductor

An actor is a passive electrical component that consists of a coil bent in the form of wire which uses electromagnetism to produce electric current through the coil. When the electric current flows through a conductor that is the coil, a magnetic flux is developed around the conductor. This flow of current through the coil creates a magnetic field based on Fleming's Right-hand thumb rule. If there is any decrease in the current passing through the inductor, then the magnetic field will also decrease and energy is released through the generation of electric current. There is also a secondary voltage inducing in the same coil due to the magnetic flux, since it opposes any changes in the electric current flowing through the circuit. An inductor is also called choke, coil, or solenoid. A choke is designed to block certain frequencies when the DC current passes. A winding coil is coated with insulation to give extra insulation and protect the circuit. They are usually tightly wrapped around a central core which is cylindrical in nature so that the magnetic flux can be induced. The solenoid is a three-dimension coil that is used to exhibit electromagnetism when the magnetic field is developed due to the current flowing through the coil. The sign convention of the inductor is the same as the power dissipating device. When the current flows into the positive side of the voltage across the inductor, the value is positive and the inductor dissipates power whereas when the inductor releases the power back to the circuit, the current has a negative sign convention. 

Question

**Inductance and Current Decay in an Inductor**

**Equation Overview:**
The current through inductance \( L \) is given by the formula:
\[ I = I_0 e^{-t/\tau} \]
where:
- \( I \) is the current at time \( t \),
- \( I_0 \) is the initial current,
- \( \tau \) is the time constant,
- \( t \) is the time.

**Part B:**
Evaluate \( \Delta V_L \) at \( t = 0 \, \text{ms} \), given:
- \( L = 20 \, \text{mH} \),
- \( I_0 = 50 \, \text{mA} \),
- \( \tau = 1.0 \, \text{ms} \).

**Answer:**
\[ \Delta V_L = 1.0 \, \text{V} \]
- This part confirms the answer is correct.

**Part C:**
Evaluate \( \Delta V_L \) at \( t = 1.0 \, \text{ms} \), given:
- \( L = 20 \, \text{mH} \),
- \( I_0 = 50 \, \text{mA} \),
- \( \tau = 1.0 \, \text{ms} \).

**Answer:**
\[ \Delta V_L = 0.37 \, \text{V} \]
- This part confirms the answer is correct.

**Part D:**
Evaluate \( \Delta V_L \) at \( t = 3.0 \, \text{ms} \), given:
- \( L = 20 \, \text{mH} \),
- \( I_0 = 50 \, \text{mA} \),
- \( \tau = 1.0 \, \text{ms} \).

**Answer:**
- Input box for \( \Delta V_L \) is provided with placeholders for value and units. A "Submit" button is available to confirm the answer.

**Note:** Each part requires the expression of the answer using two significant figures and the appropriate units.

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