For the circuit shown below, ε = 24 V, L = 6 mH, and R = 4.50. After steady state is reached with S₁ closed and S₂ open, at t = 0, S₂ is closed and S₁ is opened. Determine (a) the current through L at t = 0, (b) the current through L at t = 5.5 x 10-4 s, (c) the voltage across R on the right at t = 5.5 × 10-4 s, and (d) the rate at which current through I changes at t = 5.5 × 10-4 L S. E Hint R S₁ eeee a. The current through L at t = 0 is 5.3 for (a) b. The current through Lat t = A. Additional Hints 5.5 × 10-4 s is A. Additional Hints for (b). c. The voltage across R at t = 5.5 x 10-4 s is V. Additional Hints for (c). d. The rate at which current through L changes at t = 5.5 × 10-4 s is A/s. Additional Hints for (d).
For the circuit shown below, ε = 24 V, L = 6 mH, and R = 4.50. After steady state is reached with S₁ closed and S₂ open, at t = 0, S₂ is closed and S₁ is opened. Determine (a) the current through L at t = 0, (b) the current through L at t = 5.5 x 10-4 s, (c) the voltage across R on the right at t = 5.5 × 10-4 s, and (d) the rate at which current through I changes at t = 5.5 × 10-4 L S. E Hint R S₁ eeee a. The current through L at t = 0 is 5.3 for (a) b. The current through Lat t = A. Additional Hints 5.5 × 10-4 s is A. Additional Hints for (b). c. The voltage across R at t = 5.5 x 10-4 s is V. Additional Hints for (c). d. The rate at which current through L changes at t = 5.5 × 10-4 s is A/s. Additional Hints for (d).
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![### Analyzing an RL Circuit: A Case Study
**Introduction to the Circuit**
For the circuit shown below, the parameters provided are:
- Voltage (\(\epsilon\)): 24 V
- Inductance (L): 6 mH
- Resistance (R): 4.5 Ω
After the steady state is reached with switch \(S_1\) closed and switch \(S_2\) open, at t = 0, switch \(S_2\) is closed and \(S_1\) is opened. The task is to determine the following:
1. The current through the inductor (\(L\)) at \(t = 0\).
2. The current through the inductor (\(L\)) at \(t = 5.5 \times 10^{-4} \) s.
3. The voltage across the resistor (\(R\)) on the right at \(t = 5.5 \times 10^{-4} \) s.
4. The rate at which the current through the inductor (\(L\)) changes at \(t = 5.5 \times 10^{-4} \) s.
**Circuit Diagram**
[Insert Circuit Diagram]
\[ \begin{equation} \epsilon - L - R - S_2 \end{equation} \]
**Problem Analysis**
1. **Current through \(L\) at \(t = 0\):**
- The current through \(L\) at \(t = 0\) is \(\boxed{5.3}\) A.
2. **Current through \(L\) at \(t = 5.5 \times 10^{-4} \) s:**
- The current through \(L\) at \(t = 5.5 \times 10^{-4} \) s is \(\boxed{}\) A.
3. **Voltage across \(R\) at \(t = 5.5 \times 10^{-4} \) s:**
- The voltage across \(R\) at \(t = 5.5 \times 10^{-4} \) s is \(\boxed{}\) V.
4. **Rate of change of current through \(L\) at \(t = 5.5 \times 10^{-4} \) s:**
- The rate at which current through \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9bd1217a-91ba-44de-b552-38ea0e6629c6%2Faf3487f2-0ef1-4d9d-ba96-c00b8eb85ee8%2F3zia8qa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Analyzing an RL Circuit: A Case Study
**Introduction to the Circuit**
For the circuit shown below, the parameters provided are:
- Voltage (\(\epsilon\)): 24 V
- Inductance (L): 6 mH
- Resistance (R): 4.5 Ω
After the steady state is reached with switch \(S_1\) closed and switch \(S_2\) open, at t = 0, switch \(S_2\) is closed and \(S_1\) is opened. The task is to determine the following:
1. The current through the inductor (\(L\)) at \(t = 0\).
2. The current through the inductor (\(L\)) at \(t = 5.5 \times 10^{-4} \) s.
3. The voltage across the resistor (\(R\)) on the right at \(t = 5.5 \times 10^{-4} \) s.
4. The rate at which the current through the inductor (\(L\)) changes at \(t = 5.5 \times 10^{-4} \) s.
**Circuit Diagram**
[Insert Circuit Diagram]
\[ \begin{equation} \epsilon - L - R - S_2 \end{equation} \]
**Problem Analysis**
1. **Current through \(L\) at \(t = 0\):**
- The current through \(L\) at \(t = 0\) is \(\boxed{5.3}\) A.
2. **Current through \(L\) at \(t = 5.5 \times 10^{-4} \) s:**
- The current through \(L\) at \(t = 5.5 \times 10^{-4} \) s is \(\boxed{}\) A.
3. **Voltage across \(R\) at \(t = 5.5 \times 10^{-4} \) s:**
- The voltage across \(R\) at \(t = 5.5 \times 10^{-4} \) s is \(\boxed{}\) V.
4. **Rate of change of current through \(L\) at \(t = 5.5 \times 10^{-4} \) s:**
- The rate at which current through \(
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