hw07solution - P72 [a] t<0 15 k9||15ks2 = 7.5m . _ 9...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P72 [a] t<0 15 k9||15ks2 = 7.5m . _ 9 ¢1(0-) = M0“) 2 (0.4) x 104% z 0.2 mA [b] i1(0+) = 210—) = 0.2 mA i2(0+) = ~z'1(0+) = —O.2mA (when switch is open) L 30x10-3 *6 1 6 [C]T"E_30><103“10’ $"10 i1(t) == i1(0+)€at/T i1(t) = 0.28406‘t mA, t > 0 [(1] i205) = ——i1(t) when t 2 0+ 12(23) = "0.25% mA, t 2 0+ {e} The current in a resistor can change instantaneously. The switching operation forces 25(0“) to equal 0.2 mA and ig(0+) 2 ~02 mA. P 7.10 120mH [a] t<0 4k?! 2.5m 10m ilgfl') *2.’ 16 _ 40 x 10—3 “ 103 v0 = x 10—3)e—25,000t : 2e—25,000t V, = 40 x 104‘; 1/7» = 25,000 ,7. 1 [b] wdel = 5(40 >< 10*3)(4 x 10—6) 2 80 nJ [C] 0.9510016] = 76 DJ to 48—50,000t 1*“92/ ___.__ 76x 0 0 1000 dt t20+ t 76 X 10_9 = 80 x 10—98'50’00‘” °= 80 x 10—9(1 — 6'50’000t0) 0 6"50’0‘00“ #— 0.05 50,000t0 m In 20 so to x 59.9 gs to 59.9 —; -—- 2-6—- —-- SO to ~ 1.5T P 7.21 [a] m0") = v1(0+) = 75V 1)2(0+) = 0 Ceq = 2 x 8/10 = 1.6 HF T = (5)(l.6) x 10—3 = 8ms; = 125 ~1|H 1 = 3- x 10-35123“ = 15e_125th, t 2 0+ 51:11 + a]; + 295E '“ v1 v2 “L EMF t 1 5t 01 = 2 f 15 x 10413-125“ d2: + 75 = 6012” 2 + 15V, t 2 0 0 1 6 t 112 = % f 15 x 104512“ dm + 0 : —15e"125t + 15V, t 2 0 0 l 2 [c] Wrapped = $(2 >< 10*6)(225) + —;—(8 x 1043225 = 1125,11. [b] «11(0) = (2 x 10‘6)(5625) = 5625 #J wdiss z $0.6 >< 1045x5625) = 4500 ,uJ. Check: wtmpped + 111th = 1125 + 4500 2 5625 ,uJ; 111(0) = 5625 ,uJ. P 7.33 After making a Thévenin equivalent we have t=El 141;) 15kg! 51:9 + 180V vét) ZSDmH L, = 180/15 = 12 mA ’7’ = (0.25/20) x 10—3 = 0.125 x 10—4; % 2 80,000 V 180 10 3 9 + — 9)8_80’000t = 9 + 33‘803000t mA '12,, = [180 —— 1290158030“ = ——6Oe“8°>°0°t V P 7.42 t < 0; 10:2 15E? 2min. 5 MO“) 2 5 + 15 (0.002) = 0.5 mA 040+) : 240—) = 0.5 mA 75 > 0; 10;: 210A v0 “Iva 15 +5><10_4+%°—4z’A—0.001=0 00 M = Z — 4iA ~— 0.001 Solving, vo(0+) 2 2 111V We also know that vo(oo) == 0 Find the Thévenin resistance seen by the 2 mH inductor: ZT=~2—6+~Zf—4ZA W I “LT :11" 2A Z—4ZA 52A“ 4, 2A 20 En :2 109 _ 2 X 10’3 T 10 = 0.2 ms; 1/7' = 5000 '00 = 0 + (2 —— mes-500‘” = 25500” mV, t > 0+ ...
View Full Document

This note was uploaded on 07/29/2008 for the course ECE 376 taught by Professor Milenkovic during the Summer '06 term at University of Wisconsin Colleges Online.

Page1 / 9

hw07solution - P72 [a] t<0 15 k9||15ks2 = 7.5m . _ 9...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online