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For the RL circuit as shown in the following: R = 150 Ohms, L= 0.38 mH and E = 12.00 V a Leele R 23. find the time constant.

For the RL circuit as shown in the following: R = 150 Ohms, L= 0.38 mH and E = 12.00 V a Leele R 23. find the time constant.

College Physics
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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### Inductor Problems and Solutions **25. Find the magnetic energy stored in an inductor when it reaches the steady state (time = infinity).** - a. \(5.8 \times 10^{-9} \, \text{J}\) - b. \(5.8 \times 10^{-3} \, \text{J}\) - c. \(0.015 \, \text{J}\) - d. none of these **26. If the switch is thrown from a to b suddenly, how long will it take for the current to drop to 0.02 A?** - a. \(1.82 \, \mu s\) - b. \(2.73 \, \mu s\) - c. \(3.51 \, \mu s\) - d. \(4.82 \, \mu s\) - e. none of these **27. What is the voltage across the inductor when the current drops down to 0.02 A? (Hint: use \(V_L = L \frac{di}{dt}\) by taking the derivative.)** - a. \(2.00 \, \text{V}\) - b. \(3.00 \, \text{V}\) - c. \(4.00 \, \text{V}\) - d. \(5.00 \, \text{V}\) - e. none of these **28. Two inductors, \(L_1 = 600 \, \text{mH}\) and \(L_2 = 400 \, \text{mH}\), are connected in parallel. What is the equivalent inductance?** - a. \(100 \, \text{mH}\) - b. \(292 \, \text{mH}\) - c. \(200 \, \text{mH}\) - d. \(240 \, \text{mH}\)
Transcribed Image Text:### Inductor Problems and Solutions **25. Find the magnetic energy stored in an inductor when it reaches the steady state (time = infinity).** - a. \(5.8 \times 10^{-9} \, \text{J}\) - b. \(5.8 \times 10^{-3} \, \text{J}\) - c. \(0.015 \, \text{J}\) - d. none of these **26. If the switch is thrown from a to b suddenly, how long will it take for the current to drop to 0.02 A?** - a. \(1.82 \, \mu s\) - b. \(2.73 \, \mu s\) - c. \(3.51 \, \mu s\) - d. \(4.82 \, \mu s\) - e. none of these **27. What is the voltage across the inductor when the current drops down to 0.02 A? (Hint: use \(V_L = L \frac{di}{dt}\) by taking the derivative.)** - a. \(2.00 \, \text{V}\) - b. \(3.00 \, \text{V}\) - c. \(4.00 \, \text{V}\) - d. \(5.00 \, \text{V}\) - e. none of these **28. Two inductors, \(L_1 = 600 \, \text{mH}\) and \(L_2 = 400 \, \text{mH}\), are connected in parallel. What is the equivalent inductance?** - a. \(100 \, \text{mH}\) - b. \(292 \, \text{mH}\) - c. \(200 \, \text{mH}\) - d. \(240 \, \text{mH}\)
For the RL circuit as shown in the following: \(R = 150 \, \text{Ohms}\), \(L = 0.38 \, \text{mH}\), and \(E = 12.00 \, \text{V}\). **Diagram Explanation:** The diagram depicts a basic RL circuit, consisting of: - A voltage source \(E\) - A switch with two positions, a and b - A resistor \(R\) labeled within the circuit - An inductor \(L\) also labeled The circuit is connected in series with the resistor and the inductor. **Questions:** 23. Find the time constant. - a. \(2.53 \, \text{ms}\) - b. \(2.53 \, \mu\text{s}\) - c. \(2.53 \, \text{sec}\) - d. None of these 24. When the switch is thrown to point a, what is the current at \(t = 5.0 \, \mu\text{s}\)? - a. \(28.7 \, \text{mA}\) - b. \(34.7 \, \text{mA}\) - c. \(68.9 \, \text{mA}\) - d. \(11.0 \, \text{mA}\)
Transcribed Image Text:For the RL circuit as shown in the following: \(R = 150 \, \text{Ohms}\), \(L = 0.38 \, \text{mH}\), and \(E = 12.00 \, \text{V}\). **Diagram Explanation:** The diagram depicts a basic RL circuit, consisting of: - A voltage source \(E\) - A switch with two positions, a and b - A resistor \(R\) labeled within the circuit - An inductor \(L\) also labeled The circuit is connected in series with the resistor and the inductor. **Questions:** 23. Find the time constant. - a. \(2.53 \, \text{ms}\) - b. \(2.53 \, \mu\text{s}\) - c. \(2.53 \, \text{sec}\) - d. None of these 24. When the switch is thrown to point a, what is the current at \(t = 5.0 \, \mu\text{s}\)? - a. \(28.7 \, \text{mA}\) - b. \(34.7 \, \text{mA}\) - c. \(68.9 \, \text{mA}\) - d. \(11.0 \, \text{mA}\)
Expert Solution
Step 1

27)I max = 12/150  = 0.08 ANow, current = 0.02 Achange in current = |0.02 -0.08| = 0.06 AUsing relation I = Imax(1 - e-Rt/L)0.02 = 0.08 ( 1 - e-150t/0.00038)0.75 = e-394736.8 t-0.28768 = -394736.8 t t = 7.288 * 10-7 sec

 

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