Question

Apps Super Mario 63 Blessed Session 6 S Sketchpad 5.1 - Dra... MG 5 Calculus Volume 3 myPay Web Site Guhaul truck - Reche... -OEC compLWT <HW 9 Chapter 6 Inductor Problem 6.9 The current in a 25 mH inductor is known to be --10A fort < 0 and (-10 cos 4004 - 5 sin 400€) 200 A fort > 0. Assume the passive sign convention Part A At what instant of time is the voltage across the inductor maximum Express your answer with the appropriate units NA ? Value Units Submit Request Answer Part B What is the maximum voltage? Express your answer with the appropriate units. μΑ ? Value Units Submit Request Answer Type here to search O

Answer 1

I- - -lo tso -lo cos (uoot) - 5 Sin (4007) je 300t tro 2= 25mH V- I di at 0 too U= (25x103 300 ne-3001 [rlo cos ( 400+) - 5 sů 400t ] te 300€ +4000 si (400t) 2000 cos 400t tzo 2. 253103 1ooo cos (400 t) +5500 sur 400+] -3000 W tzo (a) fos ze to be maximum: dre -0 =D at [- - 400 x 1000 sin ( 400t) + 5500 - 400 Eos (uoot) 5500 a 400 €os (uoot) ] e Boot Зоо хе-Зоо - 1000 cos (uoot) 5500 si (4oot) 17 0. का -2056109 Sin (yoot) +190 x109 Cos(400 t) = 0 =D tan (400t) = 0.99 68 t=1.8685 msec

(b) 20 al ( 400t)= 402.89° and to 1.8685 msec. re= 25x10-3 [ 1000 cos (42-82°) + 5500 « sín (42.82') ке- 300x 1.86852103 V= 111.795 x 0.57 20= 63.82 V