### Problem 7.8: Analysis of an RL Circuit**Given:**In the circuit shown in Figure P7.8, the switch has been open for a long time. The voltage and current expressions are:\[ v = 400e^{-5t} \, \text{V,} \quad t \geq 0^{+} \]\[ i = 10e^{-5t} \, \text{A,} \quad t \geq 0^{+} \]**Figure P7.8:**- A simple RL circuit with a resistor \( R \) and an inductor \( L \) in series with a voltage source. The current \( i \) flows clockwise.**Tasks:**1. **Find \( R \):** Determine the resistance in the circuit. 2. **Find \( \tau \) (in milliseconds):** Calculate the time constant \( \tau \) of the circuit, where \( \tau = \frac{L}{R} \).3. **Find \( L \):** Determine the inductance of the inductor.4. **Calculate the initial energy stored in the inductor:** Use the initial current to find the energy, given by the formula: \[ W = \frac{1}{2} L i^2 \]5. **Determine the time (in milliseconds) it takes to dissipate 80% of the initial stored energy:** Calculate the time required for the energy in the inductor to decrease to 20% of its initial value.### Explanation of the Diagram- **Diagram Setup:** - The circuit layout is a simple series connection. - The inductor \( L \) and resistor \( R \) are connected, with the direction of current marked. - The circuit is exposed to a decaying exponential voltage function over time.### Concepts Involved- **RL Circuit Dynamics:** Understanding how voltage and current decrease over time in an RL circuit.- **Time Constant \( \tau \):** Governs the rate at which current and voltage change.- **Energy Storage in Inductors:** Calculating how much energy is stored and how it dissipates over time.This exercise helps in understanding exponential responses in circuits and the energy dynamics associated with inductors.

V= 400 e-5t V t>,o+, i=10e-5t A t>,o R=vi =400e-5t10 e-5t (a) R=40Ω (b) e=5see…

In the circuit in Fig. P7.8, the voltage and current expressions are v = 400e-51 V, t2 0; i = 10e A, t2 0. Find a) R. b) 7 (in milliseconds). c) L. d) the initial energy stored in the inductor. e) the time (in milliseconds) it takes to dissipate 80% of the initial stored energy. Figure P7.8 L. R.

In the circuit in Fig. P7.8, the voltage and current expressions are v = 400e-51 V, t2 0; i = 10e A, t2 0. Find a) R. b) 7 (in milliseconds). c) L. d) the initial energy stored in the inductor. e) the time (in milliseconds) it takes to dissipate 80% of the initial stored energy. Figure P7.8 L. R.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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