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# HW2 - P 2.8 P 2.15 The interconnection is invalid In the...

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Unformatted text preview: P 2.8 P 2.15 The interconnection is invalid. In the middle branch, the value of the current iA must be —25 A7 since the 25 A current source supplies current in this branch in the direction Opposite the direction of the current z'A. Therefore, the voltage supplied by the dependent voltage source in the left hand branch is 6(—25) = *150 V. This gives a voltage drop from the top terminal to the bottom terminal in the left hand branch of 50 —~ (~150) : 200 V. But the voltage drop between these same terminals in the right hand branch is 250 V, due to the voltage source in that branch. Therefore, the interconnection is invalid. [a] Plot the v —— i characteristic 100 BO 50 V: (V) 4a 20 O u 2 4 . B e 1:: 1t. {A} From the plot: A11 (90 —- 50) R = .— = ._..___ 2 Ai (10 —~ 0) 4 Q When it = 0, at = 50 V; therefore the ideal voltage source has a voltage of 50 V. 49 ‘ 4_lt + 50V v, b 49 - [ ] +41 + 50v vi = n —5 When ct = 0, it z —4—0 2 ~12.5A Note that this result can also be obtained by extrapolating the 'u —— z' characteristic to “(It = 0. P 2.18 [a] 5'] V BE] 9 202'a = 80¢b z'g 2 71a + ib = 521, éa = 42}, 50 2 42'!) + 802'}3 = 2071b + 80% 2 10074, db 2 0.5 A, therefore, ia = 2 A and i9 2: 2.5 A [b] ib = 0.5 A [c] 7.10 = 802‘b = 40 V [(1] p49 2 £344) 2 625(4) = 25 W 20209 = £320) 2 (4X20) = 80 W p309 :2 i§(80) : 0.25(80) = 20 W [e] p50V (delivered) = 50% =2 125 W Check: 2 Pdel = 125W P 2.23 189 id = 60/12 = 5A; therefore, 00d = 60 + 18(5) = 150V ——240 + 2),,C + Dad = 0; therefore, vac = 240 —— 150 = 90V ib = vac/45 : 90/45 = 2A; therefore, ic = id ~— ib = 5 — 2 = 3A 7de = 102'C + vcd = 10(3) + 150 = 180V; therefore, ia 2 mod/180 = 180/180 2 1 A iezia+iC=1+3=4A ~240 + ﬂab + vbd = 0 therefore, val) = 240 — 180 2: 60V R = «Jab/i, = 60/4 2 159 CHECK: ig=ib+ie=2+426A Pdev = (240)(6) = 1440 W ZPdis = 12(180) + 42(15) + 32(10) + 52(12) + 52(18) + 22(45) = 1440 W (CHECKS) P 2.25 [a] BC! 80V 12Q 109 v2 = 80+4(12) : 128V; '01 = 128— (8+ 12+4)(2) = 80v _ 711 __§(_)_ "6+10‘16— u, 2 v1 + 2423, = 80 + 24(3) = 152V Z'4=2+4::6A igz—i4~i3=—6—3=—9A [b] Calmilate power using the formula p = Rig: P80 1: (8X32 = 32 W; P129 = (12)(2)2 = 48W P49 2 (4X2)2 = 16W; 1249 = (4)(6)2 = 144W 22249 = (24x3)2 = 216W; Pen = (6)(5)2 = new plan 2 (10)(5)2 : 250W; 19129 2 (12)(4)2 = 192W [0] 09 = 152V [(1] Sum the power dissipated by the resistors: Zpdiss == 32 + 48 + 16 + 144 + 216 + 150 + 250 + 192 = 1048W The power associated with the sources is pvolt—source = (80) (4) : 320W pcurr—source = _'Ug’l:g 2 “‘(152)(9) = *1368W Thus the total power dissipated is 1048 + 320 z 1368 W and the total power developed is 1368 W, so the power balances. P 2.26 [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. Finally, KCL at each of the middle three nodes yields the currents in the two sources and the current in the middle 109 resistor. These calculations are summarized in the ﬁgure below: P130 = “(130)05) = ~1950 W P460 = —(460)(30) = ~13,800 W P 2.28 [b] ZPdis = (15)2(2) + (15)2(10) + (550)2(2) + (10)2(25) + (25)2(10) + (5)2(100) = 450 + 2250 + 1800 + 2500 + 6250 + 2500 = 15,750 W ZPSHP = 1950 + 13,800 a 15,750 W Therefore, :Pdis z ZPSUP = 15,750 W First note that we know the current through all elements in the circuit except the 200 S2 resistor (the current in the three elements to the left of the 2009 resistor is ﬁg; the current in the three elements to the right of the 200 Q resistor is 290g). To ﬁnd the current in the 2009 resistor, write a KCL equation at the top node: iﬁ + 2923 = @009 = 3023 We can then use Ohm’s law to ﬁnd the voltages across each resistor in terms of 1,3. The results are shown in the ﬁgure below: + 10,00015— 10kg: 0.8V 2953 -14,50015+ 500\$} [a] To ﬁnd iﬁ, write a KVL equation around the left—hand loop, summing voltages in a clockwise direction starting below the 15.2V source: —15.2V + 10,0002‘1 — 0.8V + 6000756 = 0 Solving for ig 10,000725 + 600003 = 16V so 16,0002}; 3: 16V Thus, 7: _ 16 ’3 _ 16,000 Now that we have the value of i5, we can calculate the voltage for each component except the dependent source. Then we can write a KVL equation for the right—hand loop to ﬁnd the voltage vy of the dependent source. Sum the voltages in the clockwise direction, starting to the left of the dependent source: —1}y - 14,5002'g + 25V — 6000735 = 0 Thus, 113, = 25V — 20,500t'5 = 25V — 20,500(10‘3) = 25V w 20.5V : 4.5V zlmA [b] We now know the values of voltage and current for every circuit element. Let’s construct a power table: Power Equation 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 :— —741mW Thus, the total power generated in the circuit 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 InW = 741 mW Thus, the total power absorbed in the circuit is 741 111W and the power in the circuit balances. P 2.31 [a] ~50 — 20730 + 181); = 0 —182'A + 52}, + 402}, = 0 so 182'A = 45%}, Therefore, —- 50 — 202}, + 452}, = 0, so 7;, z 2 A 182'A = 452}, = 90; so iA :2 5A 110 = 402} = 80V [b] 2'9 2 current out of the positive terminal of the 50 V source vd = voltage drop across the 8% source ig=iA+ig+8iA=9iA+ia=47A vd=80~20=60V ZPgen :Pdiss H 502}, + 2011,29 = 50(47) + 20(2)(47) = 4230 W 182'2A + 5mg, ._ m) + 40¢; + 82pm + 8iA(20) (18)(25) + 10(47 — 5) + 4(40) + 40(60) + 40(20) : 4230 W; Therefore, Zpgen = ZPdiss = 4230 W H ...
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