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Sendent current s pute 100 mH Sketch the current waveform b. At what instant of time is the current maximum e Express the voltage across the terminals of the 100 mit inductor, v, as a function of time d. Sketch the voltage waveform Is the voltage maximum when the current is maximu? At what isnt of time does the voltage change polarity? Is there ever an instantaneous change in the voltage across the inductor? If so, at what time?

Sendent current s pute 100 mH Sketch the current waveform b. At what instant of time is the current maximum e Express the voltage across the terminals of the 100 mit inductor, v, as a function of time d. Sketch the voltage waveform Is the voltage maximum when the current is maximu? At what isnt of time does the voltage change polarity? Is there ever an instantaneous change in the voltage across the inductor? If so, at what time?

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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continued on f and g.

### Educational Content on Inductance and Current Decay #### Problem Statement A circuit with an independent current source generates a current \( I(t) \) for \( t \ge 0 \) and a pulse "10e\(^s\)" for \( t > 0 \). #### Tasks: 1. Sketch the current waveform. 2. At what instant of time is the current maximum? 3. Express the voltage across the terminals of the 100 mH inductor, \( V \), as a function of time. 4. Sketch the voltage waveform. 5. At what instant of time is the current maximum? 6. Is there ever an instantaneous change in the voltage across the inductor? If so, at what time? ### Solutions #### a) Current Waveform Sketch Upon \( t \) increasing, the current decreases. **Graph Description:** - The y-axis represents current \( i(t) \) in Amperes. - The x-axis represents time \( t \) in seconds. - The maximum value of current occurs at around 0.16 A at \( t = 0.2 \) seconds. - The graph shows an exponential decay. #### b) Determining the Instant of Maximum Current Given: \[ \text{Current} = 10te^{-5t} \] Differentiate with respect to \( t \): \[ \frac{di}{dt} = \frac{d}{dt}[10te^{-5t}] = 10[(e^{-5t} - 5te^{-5t})] = [10e^{-5t} (1-5t)] \] Set \(\frac{di}{dt} = 0 \) to find when the current is maximum: \[ 10 e^{-5t} (1-5t) = 0 \] \[ (1-5t) = 0 \] \[ 5t = 1 \] \[ t = 0.2s \] Current is maximum at \( t = 0.2 \) seconds. #### c) Voltage Across the Inductor Given: \[ V_L = L \frac{di}{dt} \] \[ L = 0.1 H \] Then: \[ V_L = 0.1 \frac{d}{dt}[10te^{-5t}] = 0.1 \times 10 (e^{-5t} - 5
Transcribed Image Text:### Educational Content on Inductance and Current Decay #### Problem Statement A circuit with an independent current source generates a current \( I(t) \) for \( t \ge 0 \) and a pulse "10e\(^s\)" for \( t > 0 \). #### Tasks: 1. Sketch the current waveform. 2. At what instant of time is the current maximum? 3. Express the voltage across the terminals of the 100 mH inductor, \( V \), as a function of time. 4. Sketch the voltage waveform. 5. At what instant of time is the current maximum? 6. Is there ever an instantaneous change in the voltage across the inductor? If so, at what time? ### Solutions #### a) Current Waveform Sketch Upon \( t \) increasing, the current decreases. **Graph Description:** - The y-axis represents current \( i(t) \) in Amperes. - The x-axis represents time \( t \) in seconds. - The maximum value of current occurs at around 0.16 A at \( t = 0.2 \) seconds. - The graph shows an exponential decay. #### b) Determining the Instant of Maximum Current Given: \[ \text{Current} = 10te^{-5t} \] Differentiate with respect to \( t \): \[ \frac{di}{dt} = \frac{d}{dt}[10te^{-5t}] = 10[(e^{-5t} - 5te^{-5t})] = [10e^{-5t} (1-5t)] \] Set \(\frac{di}{dt} = 0 \) to find when the current is maximum: \[ 10 e^{-5t} (1-5t) = 0 \] \[ (1-5t) = 0 \] \[ 5t = 1 \] \[ t = 0.2s \] Current is maximum at \( t = 0.2 \) seconds. #### c) Voltage Across the Inductor Given: \[ V_L = L \frac{di}{dt} \] \[ L = 0.1 H \] Then: \[ V_L = 0.1 \frac{d}{dt}[10te^{-5t}] = 0.1 \times 10 (e^{-5t} - 5
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