ECE109 HW 3 SOLUTIONS

# ECE109 HW 3 SOLUTIONS - 1 EC:09 Hw#123 9...

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Unformatted text preview: 1 EC? :09 Hw #123 9% , @ P313 [a]uo=w=ssv ____—-t- (4700 + 3300) [b] 2‘ = 160/8000 :20 mA P3, = (400 x 10-6)(4.7 x 103) = 1.88 W PR; = (400 x 10-5)(3.3 x 103) = 1.32 w [0] Since R1 and R2 carry the same current and R1 > R2 to satisfy the voltage requirement, ﬁrst pick R1 to meet the 0.5 W speciﬁcation _ 160 — 66 94 2 r 13, — T, Therefore, (E) R1 3 0.5 942 Thus, R1 2 E 01' R1 2 17,6720 Now use the voltage speciﬁcation: 32 __ 5 = R2 + 17,672 (1 O) 66 _Thus, R2 = 12,4089 20R2 @ 4 R2+40 so R2 109 20RE 120 3 = = — 40+Re so Re 17 a 120 10R Thus, —- — L so RL 2 249 17 "10+RL / f \ /—@ 133.15 [a] '00: 100R? ~20 so 1121:4122 . @‘3 R1+R2 — RzRL Let R: = RzllRL = R2 + RL 100R: 0: =16 so R =5.2 v _.Ri+Re ‘ 1 5R, 5.2501832) 48 + R2 Thus, R2=15kQ and R1=4(15k)=60k§2 Then, m2 = 5.25125 = [b] The resistor that must dissipate the most power is R1, as it has the largest resistance and carries the same current as the parallel combination of R2 and the load resistor. The power dissipated in R1 will be maximum when the voltage across R1 is maximum. This will occur when the voltage divider has a resistive load. Thus, ' v31=100—16=84 V 842 =' 60 k = 117.6 In W PR, E Thus the minimum power rating for all resistors should be 1/8 W. (11> P 3.16 Refer to the solution to- Problem 3.15.._The voltage divider will reach the -‘...—- maximum power it can safely dissipate when the power dissipated in R1 equals .«....-_. . ....._ . 0.15 w. Thus, ‘1 "’2?! — 015 94 87 v ‘1 , 60 k — . SO DR; — . v0 = 100 — 94.87 = 5.13 V 1001-?e -—-—-—— : . ' d = . SO'GOk+Re 513 an R: 325k!) (15 1012;, '” ——-——= 2 d = ‘ Thus, 15 k+RL 3 50 an RL 414 kg The minimum value for Ry, is thus 4.14 k9. .' (111331! e: ‘2 2‘2- .~ 4.“ 1““: v ~, ~- ,\I :w -'-' 3"?" _ \.\'-,'x::'_.‘= 7%.: ...-~-._J: 180 m + 60 m = 240 k9 so knu240 m = 60 m 60 000 = ____#_:_____ 480 = 360 v 1101 (20,000 + 50,000)( ) 180,000 ”° = mm = 270 V M 20m 60 k0 480 i = = 4. 100,000 8 80,000'i = 384 V _ 180,000 " 240,000 0., (384) = 288 V [c] It removes loading eﬁect of second voltage divider on the ﬁrst voltage divider. Observe that the open circuit voltage of the ﬁrst divider is 80,000 _ WHEN) —— 384 V Now note this is the input voltage to the second voltage divider when the current controlled voltage source is used. I ”01 P 3.19 [a] At unload: v9 = 1:11. = Ug— Thereforc I: II and a R]: R1+Rg l—a R2Ro]__(1_k) Thus (a )[Ro-f-Rz _ ’3 R2 81' ‘r n 'Ids R—(k_“)12., ~ ovmg or 2}"? 2_a(1—k) _, i:— A150, 131:0 HR: Rl=( 0)]{0 k a}: 0.05 b = __ 1 =2. .0 H R‘ 0.68)z° 51' 0.05 R; _ (m)Ro—l4.167kﬂ [CI Maximum dissipation in R2 occurs at no load, therefore, [(60) (0135)]2 14,167 Maximum dissipation in R1 occurs at full load. 60 — 0.80 60 2 P3103“) = L-ﬂ—EéJ—H = 57.60 m\‘»’ 103““) = = 183.6 111W (m) P 3.25 sonao = 200 _ 25 75 ”°“"'( 112(5 )=15A v2 = (15)(20) = 300 V v; + 30850 = 750 \I 111 — 12(25) = 750 U; = 1050 V @ P 3.26 1.10" = W = 6.75 IDA v15]. = —(6.75 m)(15 1:): —101.25 V £33. =._18 m — 6.75 m = 11.25 mA um = —(12 k)(11.25 m) = —135 V 1)., = —101.25 — (—135): 33.75 v 8,13‘ . '_ , - _ .. 1I”«(qﬁgﬁiei'f‘z‘ygﬁ' .. .~‘- .. . ﬂit-“£32155: ‘».am‘;£~ra...=;u-.;a a- .’ P 3.52 Begin by transforming the A-connected resistors (10 Q, 30 (1,60 0) to (Q Y-connected resistors- Both the Y—connected and A—connected resistors are ' shown below to assist in using Eqs._ 3.44 — 3.46: Now use Eqs. 3.44 —— 3-46 to calculate the values of the Y—connected resistors: (30)(60) (60)(10) ' (30)(10) u=——= 89- R =§——=6' R.=—4———_= 13- 10+30+50 1 ’ 2 10+30+60 0’ 3 10+30+60 3“ The transformed circuit is shown below: 23 Q The equivalent resistance seen by the 80 V source can be calculated by making series and parallel combinations of the resistors to the right of the 24 V source: RE: = (28+6)||(16+ 18) + 3 = 34||34+3 =_1_7+3 = 209 Therefore, the current i in the 80 V source is given by ._mv_ _————._A z 2004 Use current division to calculate the currents i1 and i2. Note that the current I 1'; ﬂow in the brandi containing the 289 and 6 9 series connected resistors, 1° - while the current i; ﬂows in the parallel branch that contains the series connection of the 16 Q and 18 Q resistors: 16+18 - 34 z =~x - = = .= _ =2A 1 16+18+28+6m 68(4 A) 2A, and 12 4A 2A NOW use KVL and Ohm’s law to calculate v1. Note that 111 is the sum of the voltage drop across the 18 Q resistor, 18112, and the voltage drop across the 3 O resistor, 31?: v1=18i2+3i=18(2 A)+3(4A)#36-l—'12=48"V Finally, use KVL and Ohm’s law to calculate 112. Note that 112 is the sum of the voltage drop across the 6 Q resistOr, 6731, and the voltage drop across the 39 resistor, 31': P 3.53 [a] Calculate the values of the Y—connected resistors that are equivalent to the 10 Q, 30 Q, and 609 A—connected resistors: _ (10) (30) _ . _ (30X60) _ . RX_10~+30+60—3Q’ R"‘10+30+60“180’ _ (10)(60) 2 69 Z_lO+3O+60 Replacing the 122—123—1124 delta with its equivalent Y gives 2] 0 Now calculate the equivalent resistance Rab by making series and parallel combinations of the resistors: Rab = 20+3+[(30+6)H(18+18)]+9= 509 <3) @ U66 = n 41:51-31: ”“3 1309 = ——°9—z——— =“Hf1501 = g; =1‘Hius‘n = - (001)(9z1) (dOIst) (ssz) = ._=oe= f=m=‘=os___ 3“ Ema ”‘3 ”’3 (91mm ”‘3 “9 (0mg) m3 mmmmlnaqlmpmpmpuenddaamm ig‘gd app 3mm! 0: 93—33—33 mum 35.9. sq; 3.1351103 ‘95:; 311312.13an 51; o: BEL—93% W 1:15;? sq: ammo {3] uos=6+ta+oz=ﬂz ma = (91391 —:— EYE-“9i" 69.1.39! = 1591mm: 18927:: = amines :Q‘Hmlmjnbe ampugmmmampmmmmmm or ‘ 1?:— L0.» m ~=£________ um m 3mm mmn-x 809 m menu-m m :mjaha 3331;: mm 93:9me my: gag-m 3;: m :33 /\ (9/ circuit reduces to A __ _. delta, the P 3 5 ]After the 200 100 Q 50 Q. wye is replaced b its equivalent @ 01 ____~,.9..[a y ' 00 =240IDA 1.: T906000 ) 400 1000 80 [1)] i1— _ @940): 48 mA [c] Now that 2,, and 21 are known return to the original circuit —(240)_ — 96 mA 1at: . [d] 03 ‘= 1);; + 20(O.6 + 0.048) = 60 + 12.96 = 72.96 V pg = —(vg)(l) = —72.96 W Thus the current source delivers 72.96 W. {b} (600)? = 4(31 + 32ml 9x 104=R§+R1R2 R2 2R1 + R2 + 600 ‘Uo —- = 0.6 ‘Ui 1.2111 + 0.6R2 + 360 = R2 0.4122 = 1.2111 + 360 R2 = 3111 + 900 ' 9 x 104 = R;- + 111(3121 + 900) = 412% + 900111 R? + 225111 —- 22,500 = 0 'R1 = —112.5 d: W/(112.5)2 + 22,500.? —112.5 :I: 187.5 R1 = 759. 122 = 3(75) + 900 =1125§2 193.64 [a] After maldn th - . g ‘3 Y tO-A transformation, the circuit reduces m b Combining the parallel rwistors reduces the circuit to DJSR 33.3% b 3RRL _ 2.25122 + 3.75RRL 3R + RL _ 312 + 12L 2.251%2 + 3.75RRL 3R ( 3R + RL ) 312(312 + SRL) . Th f R“: = _.____...___ = ____._ are m . (2.25122 + 3.75RRL) 15R + 912;, Now note: 0.751% + 3 R + 3R + RL When R. = 12L, we have 1511212L + QRE = 9R2 + 151C212L Therefore R2 =‘R2 or RL = R I [b] When R = RL, the circuit reduces to E 0.75:?L 3. —; c -—P 0.753:L d ,_i,—(3RL)'__}.._l_vi __ ,_1 ’°" 4512,. ‘ 1.5“" 1.5 RL’ ”°_0'75R”°” 2”" Therefore 3)3—05 pm ,9 = 550 x 1&4ﬁzzmo) = 6 W pam=i250xm3§4mg=2sw pam=(53xm“‘)1(m)=1 W Pa “420% mﬂﬂm) = 15 W @ P - . 3‘33; Curry-11' m Zn .3 ; .'. have -4c~6"-l¢. =0 @ So It. =g-01u) a) 5: DP 3-0 i' 5A with. swifch. 0pm or closcd. for {he pawcr +0 vary as a fundfon. of Mg swifch we. and CL chrdrollcd. source Inside. +hc box. (-5‘ PM .- lsow (as before.) so 'fhc ckpwduaf ﬂour“. suppln'cs ‘25 w=%& 9' bit-"1.44 R = 3.:sn. up 3-7 at am , mm m. 5:111 50 ("a I 1/“ Hum ML. current into Terminal «3. IS ii = ' =&—- +1 snafu“, : '1 =z+ b 3 (in... 52-3” ‘——5°" 2> vac—km) => I: =leZ ...
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