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What is the initial energy stored in the inductor in Figure 7 at t=0? Ow=84.4 m.J Ow=37.5mJ Ow-75.0 m.J O Correct Answer Not Given Ow-21.6 m.J

What is the initial energy stored in the inductor in Figure 7 at t=0? Ow=84.4 m.J Ow=37.5mJ Ow-75.0 m.J O Correct Answer Not Given Ow-21.6 m.J

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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### Initial Energy Stored in an Inductor **Question:** What is the initial energy stored in the inductor in Figure 7 at \( t=0 \)? **Options:** - \( w_i = 84.4 \, \text{mJ} \) - \( w_i = 37.5 \, \text{mJ} \) - \( w_i = 75.0 \, \text{mJ} \) - Correct Answer Not Given - \( w_i = 21.6 \, \text{mJ} \) **Explanation:** The question prompts the calculation or identification of the initial energy stored in an inductor at a given moment ( \( t=0 \) ). Each option provides a different value for \( w_i \), the initial energy in millijoules (mJ). If none of the given values are correct, one should choose "Correct Answer Not Given." The user must refer to Figure 7 to make the appropriate determination.
Transcribed Image Text:### Initial Energy Stored in an Inductor **Question:** What is the initial energy stored in the inductor in Figure 7 at \( t=0 \)? **Options:** - \( w_i = 84.4 \, \text{mJ} \) - \( w_i = 37.5 \, \text{mJ} \) - \( w_i = 75.0 \, \text{mJ} \) - Correct Answer Not Given - \( w_i = 21.6 \, \text{mJ} \) **Explanation:** The question prompts the calculation or identification of the initial energy stored in an inductor at a given moment ( \( t=0 \) ). Each option provides a different value for \( w_i \), the initial energy in millijoules (mJ). If none of the given values are correct, one should choose "Correct Answer Not Given." The user must refer to Figure 7 to make the appropriate determination.
### Inductor Current Analysis in RL Circuit **Problem Statement:** The switch in the circuit shown in Figure 7 has been closed for a long time before opening at \( t = 0 \). Select the correct expression for the inductor current valid for \( t \geq 0 \). **Circuit Description:** - **Voltage Source:** 12 V - **Resistors:** - \( 16 \Omega \) (in series) - \( 8 \Omega \) (in parallel with inductor and \( 6 \Omega \) resistor) - \( 6 \Omega \) (in parallel with inductor and \( 8 \Omega \) resistor) - **Inductor:** 300 mH (in parallel with the \( 6 \Omega \) and \( 8 \Omega \) resistors) **Figure 7:** Figure 7 illustrates the described RL circuit. The circuit includes a 12 V power supply connected in series with a \( 16 \Omega \) resistor, followed by a branch where an inductor of 300 mH is connected in parallel with two resistors (\( 8 \Omega \) and \( 6 \Omega \)), as depicted when the switch is open at \( t = 0 \). **Explanation:** 1. **Initial Condition (Steady State before \( t = 0 \)):** - The switch has been closed for a long time. - In steady state, the inductor will act as a short circuit (ideal inductor behavior). - Currents and voltages in the circuit will have stabilized. 2. **At \( t = 0 \):** - The switch opens. - The inductor will oppose changes in current, creating a decaying exponential current through it. 3. **Finding \( i_L(t) \):** - Use Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL). - Determine the time constant \( \tau = \frac{L}{R_t} \), where \( R_t \) is the Thevenin equivalent resistance seen by the inductor once the switch is open. - Evaluate \( i_L(t) \) using the exponential decay formula for inductor current \( i_L(t) = i_{L0} e^{-\frac{t}{\tau}} \), where \( i_{L0}
Transcribed Image Text:### Inductor Current Analysis in RL Circuit **Problem Statement:** The switch in the circuit shown in Figure 7 has been closed for a long time before opening at \( t = 0 \). Select the correct expression for the inductor current valid for \( t \geq 0 \). **Circuit Description:** - **Voltage Source:** 12 V - **Resistors:** - \( 16 \Omega \) (in series) - \( 8 \Omega \) (in parallel with inductor and \( 6 \Omega \) resistor) - \( 6 \Omega \) (in parallel with inductor and \( 8 \Omega \) resistor) - **Inductor:** 300 mH (in parallel with the \( 6 \Omega \) and \( 8 \Omega \) resistors) **Figure 7:** Figure 7 illustrates the described RL circuit. The circuit includes a 12 V power supply connected in series with a \( 16 \Omega \) resistor, followed by a branch where an inductor of 300 mH is connected in parallel with two resistors (\( 8 \Omega \) and \( 6 \Omega \)), as depicted when the switch is open at \( t = 0 \). **Explanation:** 1. **Initial Condition (Steady State before \( t = 0 \)):** - The switch has been closed for a long time. - In steady state, the inductor will act as a short circuit (ideal inductor behavior). - Currents and voltages in the circuit will have stabilized. 2. **At \( t = 0 \):** - The switch opens. - The inductor will oppose changes in current, creating a decaying exponential current through it. 3. **Finding \( i_L(t) \):** - Use Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL). - Determine the time constant \( \tau = \frac{L}{R_t} \), where \( R_t \) is the Thevenin equivalent resistance seen by the inductor once the switch is open. - Evaluate \( i_L(t) \) using the exponential decay formula for inductor current \( i_L(t) = i_{L0} e^{-\frac{t}{\tau}} \), where \( i_{L0}
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