ECE109 HW 2 SOLUTION

# ECE109 HW 2 SOLUTION - 2.21 ’11 current i in lhc circuil...

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Unformatted text preview: 2.21 ’11): current i, in lhc circuil in ﬁg P721 IS A A HP“! 3) Fund 1,. 1)) Fund the power diszipalcd in cach ICSL'JHI c) Verify that the total powcr dissipated m Ihr. «.ir- cuit equals the power dcvcmpcd by lhc 180 V SOUICC. [a] 1'3 'f‘ 4 = 2‘2, “gm. PI“ v1 = 1053 + 8i2 = 10(6) + 8(10) 7—. MW Note also that i4 xil+i3= 2+6=~8A ig=14+la=8+4=§12A “1] Pan Pun P700 P109 Pan _. .— -.— .— — 89(5) ='320 w (4)2(25) = 400 w 22(70) = 280 w 62(10) = 360 w 102(8) = 800 w [c] ZR». = 320 + 400 + 280 + 360 + 800 = 2160W PM = 1802', = 180(12) = 2160W 2.22 For the circuit shown in ﬁg. P222. find (5‘) R and 5"“ (b) the powcr supplied by the 125 V source. Figure P222 125V ”a = (9+6)(3)=45V ‘125+va+vb= 0 SO ‘05 = 125 2': =vb/(10+6) = 80/16 = 5A id=ie—3=5—3=2A vc=5id+vb=5(2)+80=90V ic =vc/30= 90/30: 3A vd=125—vc=125~90=35V ia=id+ic=2+3=5A R=Udﬁo =35/5=7Q [b1 1;, =ia+3='5+3=3A Pg (supplied) = (1%)(8) = 1000 W —va=125f45=sov ("1" . ~——--....‘ ......‘.....>..__.. v_.-vu.-.rr V 2.23 The variable resistor R in the circuit in Fig. P223 is '5'“! adjusted until v4 equals 60 V. Find the value of R. Figure 92.23 180 id = 60/12 = 5A; therefore, vod = 60 + 18(5) = 150V —240 + v.3 + vod = 0; therefore, v“ = 240 — 150 = 90V ib =vu/45=90/45=2A; therefore, ic=id—ib=5—2=‘3A vbd = 101‘c + val = 10(3) + 150 = 180V; therefore, i, = vbd/180 = 180/180 = 1A i3=ir+ic= 1+3=4A —240 +0.“, +vbd = 0 therefore, v35 = 240 ‘— 180 = 60V R = U‘b/‘ie = 60/4 = 150 CHECK: ig=ib+ie=2+4=6A Pdcv = (240)(6) = 1440 W ' dei. = 19(180) + 42(15) + 3.2(10) + 5’02) + 5’88) + 22(45) L... i=1f140W(CHECKS) ‘ ea... ® :3: 'lhevoltage across the 15 k0 resistor in the circuit m Fig. P224 ts 500V. positive at the upper terminal. 3) Find the pawn dissipated in each resistor. b) Find the power supplied by the 100 mA ideal current source. (1) Verify that the powar supplied equals the total power dissipated. ﬁgure P2.2t. la] iod = SOD/15,000 = 33.33mA ibd + icd = 0.1 SO ibd = 0.1 —- 0.033 = 66.67mA 4000i},c + 500 ~ 75001521 = 0 so the = (500 -— 500)/4000 = 0 in: = icd — ibc = 33.33 — 0 = 33.33 mA 0.1 = m, + i" so 12.1, = 0.1 — 33.33 = 66.67mA Calculate the power dissipated by the resistors using the equation -2 pg = R1 R: ”51‘“ = (5000)(°‘°667)2 = 2222‘” mm = (7500)(0.0667)2 = 33.33w pmkn = (10,000)(0.03333)2 = 11.11 w pm = (4000)(0)a = ow pwkn = (15,000)(0.0333)2 = 16.67W [b] Calculate the voltage drop across the current source: 1}“, = 5000111; + 7500ibd = 5000(0.0667) + 7500(0.0667) = 833.33V Now that we have both the voltage and the current for the source, we can calculate the power supplied by the source: p9 = —833.33(O.1)= —83.33W thus ‘11,7 (supplied) = 83.33W [c] 2 Pdi, = 22.22 + 33.33 + 11.11 + 16.67 + 0 = 83.33W Therefore, ZP-upp =2Pdi' ‘ ' ‘ Fi ure P215 em: 4 A and 2 A, rresr>ecti{'t:ly.hc “mm m Eg' P225 are g a) Fmd ix- b) Find the power dissi c) Find 1):. d) Show that the power delivered by the current source is equal to the 1" power absorbed by all the 80 V o o 6 L other elements. pated in each resistor. 8 n 4 n 80V 12C) 10C} 1); = 80 + 40.2) =-l28 V; U1 80 . . =-—= ' = 2_— -—2= A 6 10 16 5A, 1,3 7.1 5 3 v1 = 128 — (8 + 12 + 4)(2) ‘= 80V 2:1: . vg 2 v1 + 242, = 80 + 24(3) = 152V i4 = 2 + 4 = 6 A 2", = 44—23 = —6—3=—9A [b] Calculate power using the formula p 2 12?: I Pan = (8)(2)2 = 32W; Pm = (‘99)2 = 16 W; p249 = (24;)(3)2 = 216W; P100 = (10)(5)2 = 250 W; [c] '09 = 152V [d] Sum the power dissipated by the resistors: 10120.: (1“?!)(2)2 = 48W Pm‘v: (4) (5)2 = 144W ‘ Pen = (6)06)” =‘150W P129 = (12)(4)2 = 192W 2%, =32+48+16+144+216+150+250+192=1048W The power associated with the sources is pVOR-aouxca 2 (80) (4) = 320W pmw z: "ugig = —(152) (9) = -1368W Thus the total power dissipated is 1048 + 320 = 1368 W and the total power developed is 1368 W, so the power bal 8.1168. 2.26 The currents i, and i; 'm the circuit in Fig. P2 26 are 10 A and 25 A, respectively. a) Fmd the power supplied by each voltage source. b) Show that the total power supphed equals the total power dissipated in the resistors Figure P226 ”(M/0250 1000 [a] Start by calculating the voltage drops due to the currents i1 and i2. Then ‘ use KVL to calculate the voltage drop across and 100 Q resistor, and Ohm's law to ﬁnd the current in the 100 Q resistor. FtnellgﬁlﬁCL at each of the middle three nodes yields the currents in the two sources and the . current to the middle 10 Q resistor. These calculations are summarized in the ﬁgure below: Pm = —(130)(15)=—1950w page = —(460)(30)=—13,800w . [bl M :3’ u (15)’(2) + (15)‘(10) + (30)’(2) + (10)’(25) + (25)’(10) + (5)2000) ~ ‘450 + 2250+ 1800 + 2500 + 6250 + 2500 = 15,750 w 2P”, .. 1950 + 13,800 = 15,750 W Therefore, 2P4, = 2PM, = 15,750 W I! I"- _,. [I . . . Figure P2.28 ’ I . a) ﬁn d the voltage '0, 1n the cucuit in F1g. P128. 0 s v ”:5 b) Show that the total power generated in the cir- 10 m I 500 n cuit equals the total power absorbed. , It. . . I, . 15.2 v o a 25 v I/‘ 5: F' tisgtﬁlg Phettwe know the Current through all elements in the circuit accept resisto - \$0313 or (the Current in the three elements to.the left of the 200 Q r 15 "ﬁ: the Current in the three elements to the right of the 200 Q resistor is 29735). To ﬁnd the current in the 2009 resistor, write a. KCL equatlon at the top node: iﬁ + 292/3 = 2.2009 : 301% We can then use Ohm’s law to ﬁnd the voltages across each resistor in terms of 1,9. The results are shown in the ﬁgure below: + 10,00013— 101:0 0.8V ”is 5009 [a] To ﬁnd z'p, write a KVL equation around the left-hand loop, summing voltages in a clockwise.direction starting below the 152V source: —15.2 V + 10,000;1 — 0.8V + 60005,; = 0 Solving for £5 10,00in + 6000215 = 16V so 16,00in = 16V Thus, i _ 42. __ ﬂ "‘ 16,000 ‘ ow that we have the value of 115, we can calculate the voltage for each . ponent except the dependent source. Then we can write a KVL 7n ‘ if for the right-hand loop to ﬁnd the voltage '03, of the dependent . nun-the voltages in the clockwise direction, starting to the left of 1111A 25v— 20,500(10-3) = 25V — 20.5V = 4.5V [a] 1a = 0 because no current can exist in a. single conductor connecting two Parts of a circuit. [13} 60:6000ig ig=IOmA vb = 50001;: 50v 6 x 10-30A = 300 mA 20001} = 5002}, so i1 + 4131 = -300 mA; therefore, i1 = —60 mA [c] 300— 60+i2 = 0, so i2 = —240 mA. Figure P2.3D 2.30 ﬁnd 0, and u, in the circuit shown 1h ﬁg. P230 100 mm When v0 equals 250 mV. (Hint Start at the right end ' 't d work back toward 1) . of the cucm an x) v‘ o 0 25 n 100 n o 12.5 n 20 i, 0.250 0.250 . ' —-— --—— = ' = —-O.5 mA 5013 + 50 + 125 0, 12 '1 _..-... ‘01 = 100i2 = —50 HIV 20’i1 + Egg—0'2 + (—0.0005) =- 0; i1 = 125 [1A ‘ ‘09 = 102.1 + 401:1 = 501:1 4;; Therefore, 09 = 6.25 mV. [b] We. now know the “dues of voltage and current for every circuit element. Let‘s construct a ower table: Element Current Voltage Power ' (mA) (V) Equation (mW) m-m The total power generated in the circuit is the sum of the negative power values in the power table: —15.2 mW + —-0.8 mW + —725 mW = —741 mW Thus, the total power generated in the encuit is 741 mW. The total power absorbed in the circuit is the sum of the positive power values in the power table: 10 mW + 180 mW + 130.5 mW + 420.5 mW = 741 111W the circuit ham. ...
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