Question

The triangular current pulse shown in (Figure 1) is applied to a 20 mH inductor. Use the passive sign convention. O AEQ If vec ? y = Submit Request Answer Part F Derive the expression for the inductor voltage in the interval 0 <t<5 ms. Suppose that t is in seconds. Express your answer in volts in terms of t. TIVI AS01 1. vec ? Submit Request Answer Figure < 1 of 1 > Part G the interval ms <t<10 ms. Suppose that t is in seconds. Derive the expression for the inductor voltage Express your answer in volts in terms of t. i (mA) 250 TVI AST J1 vec 0 5 10 1 (ms) y = V Submit Request Answer

Answer 1

It ma + VLLA) voltage VLH) = L OILLA) dt given wave Form ear of line-i; CCMA) 250 inert (5,250) line-a. A You (-34) yoy, 39-34 (0,01 5 (10914(ms) 250-0 (-0) lc4-0 ELLA) 0-250 (7-5) 5-0 507 0 in Ampere. not wha. ear of line-2; Ć (H)-250 = (10-5) Ca = 250- 50 (t-5) 250-500+250 500- 5004 o in Ampere not 114. PART(F). Weltage from 0 <t <ms. ult L (504) = pom 50 = ivelt dt Answer. bom<t<loms. PART CG)! Veltage from VLGH = LC500-50+) = 20ř* (0-50) = -10 & (H) = & CH) Answers

&GH=0 det ܝܬ PARTCH)! for tzlomku VICH) = idlich 2 at S. VLHOV Anus PART(I)for t<o, ÉCH=, V6H=0 Power P=OW An PART(J): For ost< 5mfll &H= 50t and VL lt) = IV Power, P= Vilts alt I tot e P= sot at Plost) = owatt And at Plts5m) Sotom = 0.250 W Avi PART(K): For t7.com sec. &170, VLE=0 & P=ow And PARTCM). for to: E=oj Ans. PAKT (P). For t>lom seer ESOJ ANA - OO 00