The current in the RL circuit shown below reaches half its maximum value in 1.75 ms after the switch S1 is thrown. Determine (a) the time constant of the circuit and (b) the resistance of the circuit if ?=250mH.
The current in the RL circuit shown below reaches half its maximum value in 1.75 ms after the switch S1 is thrown. Determine (a) the time constant of the circuit and (b) the resistance of the circuit if ?=250mH.
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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57) The current in the RL circuit shown below reaches half its maximum value in 1.75 ms after the switch S1 is thrown. Determine (a) the time constant of the circuit and (b) the resistance of the circuit if ?=250mH.
![### Transcription and Explanation of RL Circuit Analysis
#### Given:
- Equation: \( I(t = 1.75 \times 10^{-3}s) = \frac{I_{\text{max}}}{2} \)
- Inductance: \( L = 250 \times 10^{-3} \, \text{H} \)
#### Questions:
- a) Find \(\tau_{\text{RL}}\).
- b) Determine \(R\).
#### Circuit Diagram:
The circuit consists of a resistor (\(R\)) and an inductor (\(L\)) in series with a voltage source (\(\mathcal{E}\)).
#### Solution Steps:
1. **Time Constant \(\tau_{\text{RL}}\):**
The time constant for an RL circuit is given by:
\[
\tau_{\text{RL}} = \frac{L}{R}
\]
2. **Current Equation:**
\[
I(t) = I_{\text{max}} \left(1 - e^{-\frac{R}{L}t} \right)
\]
Rearranging for \(\frac{I(t)}{I_{\text{max}}}\):
\[
\frac{I(t)}{I_{\text{max}}} = 1 - e^{-\frac{t}{\tau_{\text{RL}}}}
\]
Solving for \(e^{-\frac{t}{\tau_{\text{RL}}}}\):
\[
1 - \frac{I(t)}{I_{\text{max}}} = e^{-\frac{t}{\tau_{\text{RL}}}}
\]
3. **Taking the Natural Logarithm:**
\[
\ln \left(1 - \frac{I(t)}{I_{\text{max}}}\right) = -\frac{t}{\tau_{\text{RL}}}
\]
Rearrange to find \(\tau_{\text{RL}}\):
\[
\tau_{\text{RL}} = \frac{-t}{\ln \left(1 - \frac{I(t)}{I_{\text{max}}}\right)}
\]
Substitute \(t = 1.75 \times 10^{-3}\) and \(\frac{I(t)}{I_{\text{max}}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F183a4de2-60b8-4988-9fb9-0a70f8d6fd02%2F7e9a7a6e-0da6-422b-81ae-49fc135add8d%2Fd2nhlhq_processed.png&w=3840&q=75)
Transcribed Image Text:### Transcription and Explanation of RL Circuit Analysis
#### Given:
- Equation: \( I(t = 1.75 \times 10^{-3}s) = \frac{I_{\text{max}}}{2} \)
- Inductance: \( L = 250 \times 10^{-3} \, \text{H} \)
#### Questions:
- a) Find \(\tau_{\text{RL}}\).
- b) Determine \(R\).
#### Circuit Diagram:
The circuit consists of a resistor (\(R\)) and an inductor (\(L\)) in series with a voltage source (\(\mathcal{E}\)).
#### Solution Steps:
1. **Time Constant \(\tau_{\text{RL}}\):**
The time constant for an RL circuit is given by:
\[
\tau_{\text{RL}} = \frac{L}{R}
\]
2. **Current Equation:**
\[
I(t) = I_{\text{max}} \left(1 - e^{-\frac{R}{L}t} \right)
\]
Rearranging for \(\frac{I(t)}{I_{\text{max}}}\):
\[
\frac{I(t)}{I_{\text{max}}} = 1 - e^{-\frac{t}{\tau_{\text{RL}}}}
\]
Solving for \(e^{-\frac{t}{\tau_{\text{RL}}}}\):
\[
1 - \frac{I(t)}{I_{\text{max}}} = e^{-\frac{t}{\tau_{\text{RL}}}}
\]
3. **Taking the Natural Logarithm:**
\[
\ln \left(1 - \frac{I(t)}{I_{\text{max}}}\right) = -\frac{t}{\tau_{\text{RL}}}
\]
Rearrange to find \(\tau_{\text{RL}}\):
\[
\tau_{\text{RL}} = \frac{-t}{\ln \left(1 - \frac{I(t)}{I_{\text{max}}}\right)}
\]
Substitute \(t = 1.75 \times 10^{-3}\) and \(\frac{I(t)}{I_{\text{max}}
Transcribed Image Text:The image depicts an electrical circuit diagram featuring an inductor (L), a resistor (R), a switch (S1), and a battery (denoted by
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