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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} =
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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