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The question is as follows: There are two equations: IL(t) = e-0.599t [0.678 sin (1.3227 ?) + cos (1.3227 ?)] ILnew(t) = -0.774e-5.277t + 1.0744e-0.379t Orange Curve is ILnew Blue curve is IL 2-D) Which equation shows better performance in reaching within 5% of the maximum value? Explain clearly

The question is as follows: There are two equations: IL(t) = e-0.599t [0.678 sin (1.3227 ?) + cos (1.3227 ?)] ILnew(t) = -0.774e-5.277t + 1.0744e-0.379t Orange Curve is ILnew Blue curve is IL 2-D) Which equation shows better performance in reaching within 5% of the maximum value? Explain clearly

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

There are two equations:

IL(t) = e-0.599t [0.678 sin (1.3227 ?) + cos (1.3227 ?)]

ILnew(t) = -0.774e-5.277t + 1.0744e-0.379t

Orange Curve is ILnew

Blue curve is IL

2-D) Which equation shows better performance in reaching within 5% of the maximum value? Explain clearly

2-D) Which equation shows better performance in reaching within 5% of the maximum value? Explain clearly. *11*
Transcribed Image Text:2-D) Which equation shows better performance in reaching within 5% of the maximum value? Explain clearly. *11*
The electric circuit in the figure below contains a current source driving a series inductive and resistive load with a shunt capacitor across the load. The circuit is representative of motor drive systems and induction heating systems used in manufacturing processes. Excessive peak currents during transients in the input could damage the inductor. Therefore, the equation below is used to compute response of the current through the inductor to a step in the input current to ensure that the manufacturers stated maximum current is not exceeded during start up as per the underdamped system. (1) i(t) = Load = e-0.599 t [0.678 sin (1.3227 t) + cos (1.3227 t)] . (1)
Transcribed Image Text:The electric circuit in the figure below contains a current source driving a series inductive and resistive load with a shunt capacitor across the load. The circuit is representative of motor drive systems and induction heating systems used in manufacturing processes. Excessive peak currents during transients in the input could damage the inductor. Therefore, the equation below is used to compute response of the current through the inductor to a step in the input current to ensure that the manufacturers stated maximum current is not exceeded during start up as per the underdamped system. (1) i(t) = Load = e-0.599 t [0.678 sin (1.3227 t) + cos (1.3227 t)] . (1)
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