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The figure shows an LR circuit with L= 0.15 H, R = 25 2, and Vo = 36 V. The switch is initially open. Eight milliseconds (t = 8 ms) after the switch is closed, what is the current in the circuit and the potential difference between points a and b, Vab ? a L Vo

The figure shows an LR circuit with L= 0.15 H, R = 25 2, and Vo = 36 V. The switch is initially open. Eight milliseconds (t = 8 ms) after the switch is closed, what is the current in the circuit and the potential difference between points a and b, Vab ? a L Vo

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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### Analysis of an LR Circuit with Switch Closure The figure below depicts an LR circuit consisting of an inductor (L) and a resistor (R) connected in series with a voltage source (V₀). The given parameters for the circuit are: - Inductance (L) = 0.15 H - Resistance (R) = 25 Ω - Voltage (V₀) = 36 V The switch in the circuit is initially open. The objective is to determine the current in the circuit and the potential difference between points a and b (V_ab) eight milliseconds (t = 8 ms) after the switch is closed. #### Circuit Diagram: ``` R _____ | | | R V₀ | | | | ___ | | | L | |___| | | |_____| ``` ### Explanation: When the switch is closed, the current in the circuit starts to increase according to the relation governed by the LR circuit's differential equations. The time constant for an LR circuit is given by: \[ \tau = \frac{L}{R} \] Calculating the time constant: \[ \tau = \frac{0.15 \text{ H}}{25 \, \Omega} = 0.006 \text{ s} = 6 \text{ ms} \] The current at any time \( t \) after the switch is closed in an LR circuit is given by: \[ I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \] Given \( t = 8 \text{ ms} \): \[ I(8 \text{ ms}) = \frac{36 \text{ V}}{25 \, \Omega} \left(1 - e^{-\frac{8 \, \text{ms}}{6 \, \text{ms}}}\right) \] Simplifying the exponential term: \[ I(8 \text{ ms}) = 1.44 \left(1 - e^{-\frac{4}{3}}\right) \text{ A} \] \[ I(8 \text{ ms}) \approx 1.44 \left(1 - e^{-1.333}\right) \text{ A}
Transcribed Image Text:### Analysis of an LR Circuit with Switch Closure The figure below depicts an LR circuit consisting of an inductor (L) and a resistor (R) connected in series with a voltage source (V₀). The given parameters for the circuit are: - Inductance (L) = 0.15 H - Resistance (R) = 25 Ω - Voltage (V₀) = 36 V The switch in the circuit is initially open. The objective is to determine the current in the circuit and the potential difference between points a and b (V_ab) eight milliseconds (t = 8 ms) after the switch is closed. #### Circuit Diagram: ``` R _____ | | | R V₀ | | | | ___ | | | L | |___| | | |_____| ``` ### Explanation: When the switch is closed, the current in the circuit starts to increase according to the relation governed by the LR circuit's differential equations. The time constant for an LR circuit is given by: \[ \tau = \frac{L}{R} \] Calculating the time constant: \[ \tau = \frac{0.15 \text{ H}}{25 \, \Omega} = 0.006 \text{ s} = 6 \text{ ms} \] The current at any time \( t \) after the switch is closed in an LR circuit is given by: \[ I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \] Given \( t = 8 \text{ ms} \): \[ I(8 \text{ ms}) = \frac{36 \text{ V}}{25 \, \Omega} \left(1 - e^{-\frac{8 \, \text{ms}}{6 \, \text{ms}}}\right) \] Simplifying the exponential term: \[ I(8 \text{ ms}) = 1.44 \left(1 - e^{-\frac{4}{3}}\right) \text{ A} \] \[ I(8 \text{ ms}) \approx 1.44 \left(1 - e^{-1.333}\right) \text{ A}
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