Consider the RL circuit below. In that circuit, R1 = 10 k2, R2 = 14 k2, L1 = 39 mH, and the battery outputs 34 V. The switch has been closed for a long time. At t = 0, the switch is %3D opened. Calculate the current through the inductor after 1.6 us, in mA. S2 R2 R1 L1 V1+
Consider the RL circuit below. In that circuit, R1 = 10 k2, R2 = 14 k2, L1 = 39 mH, and the battery outputs 34 V. The switch has been closed for a long time. At t = 0, the switch is %3D opened. Calculate the current through the inductor after 1.6 us, in mA. S2 R2 R1 L1 V1+
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(Please answer to the fourth decimal place - i.e 14.3225)
![**RL Circuit Analysis**
Consider the RL circuit below. In this circuit, the values are as follows:
- \( R1 = 10 \, k\Omega \)
- \( R2 = 14 \, k\Omega \)
- \( L1 = 39 \, mH \)
- The battery voltage \( V1 = 34 \, V \)
The switch \( S2 \) has been *closed* for a long time. At \( t = 0 \), the switch is opened. Calculate the current through the inductor after 1.6 µs, in mA.
### RL Circuit Diagram:
The diagram provided shows a circuit where:
- \( R1 \) is connected in parallel with the inductor \( L1 \).
- A switch \( S2 \) is in series with this parallel combination.
- \( R2 \) is connected in series with the battery \( V1 \) and the switch \( S2 \).
When \( S2 \) is closed, the battery provides a voltage of 34 V, and the circuit reaches a steady state after a long time. The current through the inductor will be calculated after the switch \( S2 \) is opened at \( t = 0 \).
**Steps to Solve:**
1. **Steady State Current:**
In the steady state, the inductor \( L1 \) behaves like a short circuit, so the voltage across \( R1 \) is the same as across the parallel combination.
2. **Current Through R2:**
In steady state:
\[
I_{\text{total}} = \frac{V1}{R2}
\]
\[
I_{\text{total}} = \frac{34 \, V}{14 \, k\Omega} = 2.4286 \, mA
\]
3. **Current Through the Inductor:**
The current that was flowing through the inductor \( L1 \) just before \( t = 0 \) will continue to flow through \( R1 \) immediately after the switch opens.
4. **Current Decay in RL Circuit:**
The time constant \(\tau\) for the RL circuit is given by:
\[
\tau = \frac{L1}{R1} = \frac{39\, mH}{10 \, k\Omega} =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F021aca7b-3c6d-4d3c-bfc7-736350e47cc8%2F0597f5ff-77fa-4e0c-89a8-70af8f947a3f%2F4ykxrs_processed.png&w=3840&q=75)
Transcribed Image Text:**RL Circuit Analysis**
Consider the RL circuit below. In this circuit, the values are as follows:
- \( R1 = 10 \, k\Omega \)
- \( R2 = 14 \, k\Omega \)
- \( L1 = 39 \, mH \)
- The battery voltage \( V1 = 34 \, V \)
The switch \( S2 \) has been *closed* for a long time. At \( t = 0 \), the switch is opened. Calculate the current through the inductor after 1.6 µs, in mA.
### RL Circuit Diagram:
The diagram provided shows a circuit where:
- \( R1 \) is connected in parallel with the inductor \( L1 \).
- A switch \( S2 \) is in series with this parallel combination.
- \( R2 \) is connected in series with the battery \( V1 \) and the switch \( S2 \).
When \( S2 \) is closed, the battery provides a voltage of 34 V, and the circuit reaches a steady state after a long time. The current through the inductor will be calculated after the switch \( S2 \) is opened at \( t = 0 \).
**Steps to Solve:**
1. **Steady State Current:**
In the steady state, the inductor \( L1 \) behaves like a short circuit, so the voltage across \( R1 \) is the same as across the parallel combination.
2. **Current Through R2:**
In steady state:
\[
I_{\text{total}} = \frac{V1}{R2}
\]
\[
I_{\text{total}} = \frac{34 \, V}{14 \, k\Omega} = 2.4286 \, mA
\]
3. **Current Through the Inductor:**
The current that was flowing through the inductor \( L1 \) just before \( t = 0 \) will continue to flow through \( R1 \) immediately after the switch opens.
4. **Current Decay in RL Circuit:**
The time constant \(\tau\) for the RL circuit is given by:
\[
\tau = \frac{L1}{R1} = \frac{39\, mH}{10 \, k\Omega} =
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