### Analysis of the Given WaveformIn this educational exercise, we will analyze the given waveform to determine several key characteristics. The waveform graph displays the voltage \( v \) in millivolts (mV) over time \( t \) in milliseconds (ms).**Tasks:**1. **Peak Value**2. **Peak to Peak Value**3. **Period**4. **Frequency**5. **Number of Cycles Shown**#### Description of the Waveform- **Graph Axes** - The vertical axis represents the voltage \( v \) in millivolts (mV), ranging from \( -100 \) mV to \( 100 \) mV. - The horizontal axis represents time \( t \) in milliseconds (ms), ranging from 0 ms to 40 ms.- **Waveform Characteristics** - The waveform appears sinusoidal with peaks at \( 100 \) mV and troughs at \( -100 \) mV. - Points of interest along the time axis include: 0 ms, 10 ms, 20 ms, 30 ms, and 40 ms.#### Details to Find:- **Peak Value:** The maximum value reached by the waveform (100 mV).- **Peak to Peak Value:** The total distance from the highest peak to the lowest trough (200 mV).- **Period:** The time duration over which one complete cycle of the waveform occurs, which can be calculated by observing the repetitive nature of the wave.- **Frequency:** The number of cycles the waveform completes in one second, calculated as the reciprocal of the period.- **Number of Cycles Shown:** The count of complete waves displayed within the 40 ms time frame on the graph.Use the graphical data provided to determine each of these characteristics, which are essential in understanding the behavior of sinusoidal waveforms in various applications, including electrical engineering and signal processing. **Title: Solving Inductor Circuits: Time Constant and Waveform Analysis****Question 7: For the Circuit Below**Given the circuit diagram:- **Resistor (R1)**: 1 kΩ- **Inductor (L)**: 1 mH- **Voltage Source (V)**: 10V- **Switch**: Initially open, then closed.**Tasks:**a) Find the time constant of the circuit.b) Sketch the waveforms for the current through the inductor (\(i_L\)) and the voltage across the inductor (\(v_L\)) when you close the switch.### Solution:#### Part (a): Finding the Time ConstantThe time constant (\(\tau\)) for an RL circuit is given by the formula:\[ \tau = \frac{L}{R} \]Where:- \( \tau \) is the time constant,- \( L \) is the inductance,- \( R \) is the resistance.Using the given values:\[L = 1 \text{ mH} = 1 \times 10^{-3} \text{ H}\]\[R = 1 \text{ kΩ} = 1 \times 10^{3} \text{ Ω}\]Therefore:\[\tau = \frac{1 \times 10^{-3}}{1 \times 10^3} = 1 \times 10^{-6} \text{ seconds} = 1 \mu\text{s}\]#### Part (b): Sketching the WaveformsWhen the switch is closed at \( t = 0 \):1. **Current through the inductor (\(i_L\))**: Initially, the current (\(i_L\)) through the inductor is zero. As time progresses, the current increases exponentially towards a steady-state value determined by the voltage source and the resistor. The exponential increase can be represented by the equation: \[ i_L(t) = \frac{V}{R} (1 - e^{-t/\tau}) \] Given \( V = 10 \text{V} \) and \( R = 1 \text{kΩ} \): \[ i_L(t) = \frac{10}{1000} (1 - e^{-t/ \tau})

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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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