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Unformatted text preview: P 2.20 [a] Use KVL for the right loop to calculate the voltage drop across the
right—hand branch 110. This is also the voltage drop across the middle
branch, so once we is known, use Ohm’s law to calculate in: v0 = 1000ia + 4000ia + 3000ia = 8000iat : 8000(0.002) = 16 V 16 2: 20002},
. 16
1° * M‘smA [b] KCL at the top node: 719 = ta + 710 = 0.002 + 0008 = 0.010 A = 10 mA. [0] The voltage drop across the source is 110, seen by writing a KVL equation
for the left loop. Thus,
pg 2 ~voz'g = —(16)(0.01) = —0.160 W = —160 mW.
Thus the SOIII’CQ delivers 160 mW. P 2.22 [a] .. ‘ lﬂﬂ 11a: (9+6)(3) : 45V —125+va+'vb =0 so vb = 125~va = 125—45 2 80V
iezvb/(10+6) =80/16=5A id=ie—3=5—3=2A '06: 5id+vb =1 ic = vc/30 = 90/30 = 3A vd=125——vc=125—90==35V ia=id+ic=2+3=ﬁ5A Rn vd/z'a = 35/5: 79 pg (supplied) = (125)(8) :: 1000 W P 2.24 [a] ted = SOD/15,000 = 33.33 mA ibd + icd = 0.1 so ibd : 0.1 —— 0.033 = 66.67mA 40002},c + 500 ~— 75002‘bd :2 0 so the = (500 ~— 500) / 4000 2 0
2'“ :3 ted — the 2 33.33 — 0 = 33.33 mA 0.1 = iab + 7133 so tab 2 0.1  33.33 = 66.67 mA Calculate the power dissipated by the resistors using the equation
2
pR 2 RZR: 1951“) 2 (5000) (0.0667)2 = 22.22W 197.51“) = (7500) (0.0667)2 :2 33.33W
10ka = (10,000)(0.03333)2 = 11.11W pm = (15,000)(0.0333)2 2: 16.67W
pm = (4000)(0)2 2 0w [b] Calculate the voltage drop across the current source: cad = 50002;”, + 750mm = 5000(0.0667) + 7500(0.0667) = 833.33V N ow that we have both the voltage and the current for the source, we can
calculate the power supplied by the source: pg 2 ~~833.33(0.1) = —83.33W thus pg (supplied) 2 83.33W [c] 2 Pdis = 22.22 + 33.33 + 1111 + 16.67 + 0 = 83.33W
Therefore, lesupp:zpdis P 2.29 [a] i0 = 0 because no current can exist in a single conductor connecting two
parts of a circuit. 60 2 600% 2'9 = 10 mA m = 500% 2 50V 6 x 10*?“ = 300 mA 20002} = 5002'2, so '51 + 47:1 2 ~300 mA; therefore, i1 2 —60 mA
[c] 300 —~ 60 + 2'2 2 0, so 2‘2 2 —240 mA. P 2.31 [a] ~50 w 202;, + 182'A = 0
—182'A + 5x}, + 402;, = 0 so 187$A = 452;, Therefore, — 50 ~ 20% + 452}, = 0, so 2}, == 2 A
181}; = 452} = 90; SO iA = 5A
v0 2 402", = 80V b i 2 current out of the ositive terminal of the 50 V source
9 P
11d = voltage drop across the SM source ig=iA+iU+8iA=9iA+ia=47A
m=w~m=mv 213m = 5% + 20iaz‘g = 50(47) + 20(2)(47) = 4230 W
ZPdiSS = 182'2A + 5%(2‘9 —~ Q) + 4023 + 82mg; + 8mm)
:= (18)(25)4~10(47—5)4—4(40)440(60)+—40(20)
z 4230 W; Therefore,
2%ﬂzykﬂmw ...
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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.
 Summer '06
 Milenkovic

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