Recitations_4_chapter_18_Ghamlouche

Recitations_4_chapter_18_Ghamlouche - 13.5 (a) The...

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Unformatted text preview: 13.5 (a) The equivalent resistance of the two parallel 7.00 E) resistors is Thus. Rub = R4 +Rp +129 =(4.00+4.12+9100) Q: 111;: (m 3. (13) Int: "5: 40V=].99A,s0 14:19:1qu R 17.19 ab Also, (AV)? = labRF = {1.99 A](4.i2 :2): 3.13 v (Ml 313v Then. L=' “"2 ' = LUA ’ R 10.09 :i 0.818A 18.9 (3} Turn the circuit given in Figure P189 90° counterclockwise to observe that it is equivalent to that shown in Figure 1 below. This reduces. in stages. as shown in the following figures. b i- b )- 120 _ 00 I20 2%0: 0Q Ill-"GillT 500 10.0 5.00 C 10.0 Q 25.0 (2 Q Q S). 10.0 g 20.0 10.0 g o o o E I: Figure 1 Figure 2 i1 1’ 1" 25.U__ 25.0__ V —— V ‘— 2.94 12.9 Q Q 10.0 g o .0 Figure 3 Figure 4 From Figure 4. i=fl— 25DV=L93A R _ 12.9 o (b) From Figure 3. (swim = mm = (1.93 A)(2.94 o}: 5.53 V (a) From Figures 1 and 2, the current through the 20.0 9 resistor is (2.1!) 5.68 V l. = b” = = 0.22TA ‘0 Rm 253 Q 18.13 The resistors in the circuit can be combined in the stages shown below to yield an equivalent resistance of Rmi = [63]] I) Q. 3.0 n L? fl IIIII II'IIIVIIII _nJI".¥.'I.II,—-C an— 309 3.051 W I. t J 3.0 o *I 6-0 Q a firm—b, 1 4.0 £2 vf 'l'v'l‘l "RN 2 ‘ 2.0 fl - Wt 12 £2 " 18V Figure1 Figurez 6.051 30 as —Q —Q a 3!! b H d a 1 a, I 1' 18V 18V 18V Figure 3 Figure 4 Figure 5 {Av} 13 v rom iguie _ Rad (63:51]) 9 Then. from Figure 4. {AVde = i'RM = {3.14 A][30/1] Q] = 8.5? V Now. look at Figure 2 and observe that (AWN _ 3.5? V =—— =l.?lA 3.0 Q+2.U Q 5.09 1 so {Al/lb: = QR,” = {131 A]{3.{] Q): 5.14 V || || || .0 .p. Lu 3" Finally, from Figure I. In RIB 12 fl 13.13 Observe that the center branch ofthis circuit. that is the branch containing points a and b. is not a continuous conducting path, so no current can flow in this branch. The only current in the circuit flows counterclockwise around the perimeter of this circuit. Going counterclockwise around the this outer loop and applying Kirchhoff’s loop rule gives —s.ov—{2.o (in—{3.0 n)r+12V—[10 cur—{5.0 oji=o _ 12 v—sov 209 Now, we start at point D and go around the upper panel of the circuit to point a. keeping track of changes in potential as they occur. This gives or I = 0.20 A avg = v; — v; = —4.0 V+{6.oo)(o]—(3.0 o)(o.2o A)+12 V— (10 n)[0.2o A): +5.4 v Since AVab >~ 0, point a is 5.4 V higher in potential than point if). 18.23 (a) We name the currents I, . 13, and I3 as shown. 4 00 kg 5 C W d Applying Kji'chhoff’s loop rule to loop abcfa. T gives +£,—52— Rgrg— R11] = o It 91 El 82 R3 93 it; 0i“ 31’: + 21'] =10-0 m 70.0 v “60.0 v son v__ and I.=5.00ntA—1.5012 [1] R2 g 3.00 m Applying the loop rule to loop edcfe yields +83—RJIS—82—R313 =0 2.00 kfl or 3L+4l =20.0mA will : _ 3 R R1 f L and I3 = 5.00 mA— 0.7"50l’2 [2] Finally, applying Kirchhoff ‘s junction rule at junc— tion 6 gives 12 = llr1 + llr3 [3] Substituting Equations [1] and [2] into [3] yields 12 = 5.00 mA— 1.501"2 + 5.00 mA— 0.7"50l'2 or 3.25!2 =10.0 mA and I3 = 3.08 mA .Then [1] gives 1'] = 0.380 mA , and from [2] I; = 2.69 mA . (b) Start at point c and go to pointf, recording changes in potential to obtain VJ. VC— 8; R213— 60.0v (3.00X103Q][3.08><10‘3A}=—69.2V (I = 69.2 V and point c is at the higher potential or |AV 13.39 From {P = { civil/R. the resistance of the element is (at!)2 _ (240 it)2 re _ 3000w R: =19.2 Q When the element is connected to a 120—3! source. we find that av 120v r=—= = sass . d (a) R .929 a" (s) e=tavjr=uzo Wises a): 750w 18.60 The total resistance in the circuit is —l —| R=[l+] = 1 + 1 =1.2ko R. a. 2.0m 3.0 to and the total capacitance is C = C, + C1 = 2.0 ttF + 3.011]: = 5.0 MP. Thus. Q... = CE: (5.0 ammo V] = 600 pc 6.0 and I: RC: (1.2x103 o](5.0x10—5 F]: 6.0x10‘3 5: 5 - 1000 The total stored charge at any time I is then Q = Q. + Q. = Q... (1— at”) or Qt+Qz=(600t1C]i1- it [1] Since the capacitors are in parallel with each other, the same potential difference exists across both at any time. Therefore. (Al/if = % = 01' Q; = [%]Q. = 1.5 Q. [2] l | C1 Substituting Equation [2] into [1] gives ZSQI = (600 HC)(1_ e—Inoou'sns] and _ (240HC)(1_ e_I000=,-'s.ot] Then. Equation [2] yields Q2 = l.5(240 yCl(l— 9—1000""6'°‘)= (360 Inc)“ _ e—l Wort-6.05) 18.58 Consider a battery of emf 8 connected between points a and b as shown. Applying Kirchhoff '3 loop rule to loop acben gives —(l.0)[1—(1.0][1|—I3]+8= 0 or 13=2l,—8 [1] Applying the loop rule to loop adbea gives —[3.0]I3—(5.D)[l2+l3]+8=0 or 812+513=8 [2] For loop adcn, the loop rule yields 11H“; 43.0)12 +0.0):3 +(].0}l, = 0 or 13 = [3] Substituting Equation [1] into [3] gives [3 = I. — 8/3. [4] . . . . 26 , 13 Now, substitute Equations [1] and [4] into [2] to obtain 181’] = g8, which reduces to II = ES. Then E uation [4] ives I — E— i 8— i8 and [l] ields l ——i&‘ ' q g 2 2? 27 27 ' y 3 27 ' Then. applying Kirchhoff’sjunction rule atjunction (I gives I = {I ‘3": =£5+i5 = £8.Therefore._ Rab =§=LZ E Q _ 27 27 27 l' (178.327) 1 ...
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Recitations_4_chapter_18_Ghamlouche - 13.5 (a) The...

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