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hw02solution

hw02solution - P 2.20[a Use KVL for the right loop to...

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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)4-40(60)+—40(20) z 4230 W; Therefore, 2%ﬂzykﬂmw ...
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