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Figure below shows a series RL circuit with an applied voltage, Obtain a differential equation giving the current i at time t. Solve the differential equation for the initial condition when t=0 and i=0

Figure below shows a series RL circuit with an applied voltage, Obtain a differential equation giving the current i at time t. Solve the differential equation for the initial condition when t=0 and i=0

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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Figure below shows a series RL circuit with an applied voltage,

  • Obtain a differential equation giving the current i at time t.
  • Solve the differential equation for the initial condition when t=0 and i=0
E = 220(1-e¹/3) R = 3kn L = 5mH E R
Transcribed Image Text:E = 220(1-e¹/3) R = 3kn L = 5mH E R
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in the total solution the part  that says what follows the 0

but i(0)=0 ?

75 --Total solution is i(t) = iclt) + ip(t) -6X105E i(t) = Ae But i = 0 AMP :2 i(t) + 150 [1-e +13] A for too A (1) + !! A (1) + 11/12 (1-1) 150 A=0 = 11 [₁-== "¹3] ₁ A for too t/3² / 150
Transcribed Image Text:75 --Total solution is i(t) = iclt) + ip(t) -6X105E i(t) = Ae But i = 0 AMP :2 i(t) + 150 [1-e +13] A for too A (1) + !! A (1) + 11/12 (1-1) 150 A=0 = 11 [₁-== "¹3] ₁ A for too t/3² / 150
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Follow-up Question

is the (6x105) correct in the differential equation part 

 

di(t) + (6x105) i(t) = 200 E (t) dt Differential equation is dilt) + (6×105) i(t) = 200 (220) (1-e¯-t1₂) dt (6x105) i(t) = 44000 [l-e-²/3] (7)! P - equations D, dr +
Transcribed Image Text:di(t) + (6x105) i(t) = 200 E (t) dt Differential equation is dilt) + (6×105) i(t) = 200 (220) (1-e¯-t1₂) dt (6x105) i(t) = 44000 [l-e-²/3] (7)! P - equations D, dr +
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