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Unformatted text preview: Engineering Circuit Analysis, 7th Edition Chapter Ten Solutions 10 March 2006 Lu . {—3 1 —B W . .
(a) Note that A cos x + 8 5111 x = A" + B" cos i x + tan‘1 . FOIL/(U). The angle is in the
x K ' ,
second quadrant; inost calculators will return —30.96°, which is off by 180°.
f0) = 750 cos (of i 30 sin (of = 58.31cos(cof+149.04°) g(r) 2 55cosmf7153incor = 57.01cos((o{ +15.255°) anipl. ofo) = anipl. of g(r) 2 (b) f0) leads g(1‘)byl49.04°—15.255° 133.8o PROPRIETARY MATERLLXL. '1C 200? The MeGrawHill Companies. Inc. Limited distribution permitted only to
teachers and educators for course preparation. If you are a student using this Manual. you are using it Without permission. Engineering Circuit Analysis, 7th Edition Chapter Ten Solutions 10 March 2006 7. (a) 6 cos (21:60r— 9°) —; 649"
6 COS(212'601+ 9°) —> 64189“ 6 cos (2360! + 9°) lags 6 cos (2 7:60( — 9°) by 360— 9 — 189 =162°. 64189O 619O (b) cos(r100°) —> 14100"
43030100") —> 14100°=1480° 14800 cos (I  100°) lags cos (I  100°) by 180°. 111000 (c) si111r —> 44900 I 14900 sin r a 11 90° sin( lags sinr by 180°.
1190B
1 A 900
(d) 7000 cos (I7 7r) A 7000 A v”! = 7000 A 1800 9:208 (1—3.140) —> 913140
7000 cos [I — Jr) lags 9 cos (I— 3.140)
bV 180 — 3.14 =176.9D. 7000 A 1800 PROPRIETARY MATERLLXL. '1C 200? The McGrawHill Companies. Inc. Limited distribution permitted 0111}! to
teachers and educators for course preparation. If you are a student using this Manual. you are using it Without permission. Engineering Circuit Analysis, 7th Edition 14.
Atx—x : Rm : 80H20 :169
var :—0.4(1585):—3c035001 var = 4.8 cos SOOIV (a) .1sz03 SOOI—tan'l—
16+10 \ 15. : 0.2544COS(5001—32.010)A Chapter Ten Solutions 10 March 2006 sin (5003‘ — 32.010) 2—2.5«44sin (soar—32.01 0) V vi : 2.544 cos (500! + 57.99") V, ix
: 31.80cos(500r+57.99°) 111A PROPRIETARY MATERLLXL. '1C 200? The McGrawHill Companies. Inc. Limited distribution permitted only to
teachers and educators for course preparation. If you are a student using this Manual. you are using it Without permission. Engineering Circuit Analysis, 7th Edition Chapter Ten Solutions 10 March 2006 16. i=5 2 3c05105.’ v? is zones105m
v in series with 309 —> 0.1colejrAH309 5 Add, getting 0.2 cos 105m H309
change to 6 cos 105 I V in series with 309; 309 + 209 = 509 6 f a _110\  ..
.‘.f z—eos lO'f—tan —J:0.l1767'cos(10:'r—11.310”)A
L J502+102 [K 50 Atrzloiis,105r21.‘.iL20.1167cos(lmd—11.310°) 31.7mm vi = 0.11767x10cos(1m —11.300+900)= —0.8462V’ PROPRIETARY MATERLLXL. '1C 200? The MeGrawHill Companies. Inc. Limited distribution permitted only to
teachers and educators for course preparation. If you are a student using this Manual. you are using it Without permission. ...
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This homework help was uploaded on 04/14/2008 for the course EECE 112 taught by Professor Fitzpatrick during the Spring '06 term at Vanderbilt.
 Spring '06
 FITZPATRICK

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