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Q5 - 1 A wire of resistance 5.0 Q is connected to a battery...

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Unformatted text preview: 1. A wire of resistance 5.0 Q is connected to a battery wire whose emf e is 2.0 v ( Sack“ AB and whose internal resistance is 1.0 Q. In 2.0 min, (a) how'much energy is transferred from chemical to electrical form? (b) How much energy appears in the wire as thermal energy? (c) Account for the difference between (a) and (b). Also 5% W) “Wm B} AX—7 Trba+ Hi; Wm 05:9.DJ1 015 qgabb, hegfl'kbr‘ K. This 9’”th Nap/AFC} U)‘\I) &{I.F Lufivte (A) ‘21:? ; F 3 EL IL J‘Z'SU‘XU "Fla JbJ-Ll 1Lrov3'fer of @ W CL?“"‘“‘ 4v gust] W97. (CM 01130 L) 56 P: IV H, CAICJ1A'H/ 1 Q2 ESQ/MW) (D \J 9 I i ‘ U: P9- : —Lt : a + : (Q'OVXDOS) :‘I‘K/Oj CD (3.3;, ({«H S.0JL+J.0J1 L) 0w ' TL"; egval‘é-a 1's UiC'R’l ‘Rr {\‘J’\'*‘ @14— -—— 2 (2+ :1?7 z P: R Q/Q/ ‘L I tov) l (ox-£0- \\—, 0i $\h5{€ ,hcgi'l'kV‘) WJ - @ Wt” M17 (“if”) a? “4 wine. the («VLMHWW : vafl ultra =>I=£E® ‘ we. :3 3 of HWLI’ @037 MA“ (>va l\'\ P) WLw'LLtX {‘Lp/ 'OL‘W‘L LLW MML) » @ > 1. (a) In the figure below. what value must R have if the current in the circuit is to be 1.0 mA? Take 8, =30 V. e, = 3.0 V. and I; = r2 = 3.0 Q. (b) What is the rate at which thermal energy appears in R? (Sadnwu B) 4'50. 5411. H-NSDLAWS ‘For $346 a) OM, W07 '1» CLoago, Lib/s 0‘11 What": If} I m3 1%, L»? +mw‘ Cow‘LJ‘cZothw'y, [(xVLLoija Loaf NL: (Huvwfl 9m We” M3q4‘kt of +1‘"SI'S“'309D°-\1)0‘\-LH @r 8‘ 7' Ir. "Fife 4’er ~29 _’ O at $WYLL-‘VD'LLQ53'10fE H e - , ref; :r‘ (ah/eh) 10119:) A7181 +I M’NM‘L :K '7' ga‘g‘ ’1 (F-t-r') ‘ . \ -— . V‘ @ R: £1 é, ’Dr : 30/ )0 ‘— 3(30J1) I "ONOKBA MHJL @ D QMg‘x-l‘w}, KBW (307;, ’IooL/fl ) “LL“ ‘10” “"2" 3° {794% 04 H» H» Sign a} l ;v- LL, ewe‘m‘m 1» M‘F ‘th corhec'rww-h P : IAR (/Vlvs'} wt +11): 0.4, '1» {VJ ‘Hv. JTLt/MJ 242/97 ® mot Spur-Fa. £93,“) “WV New - ® cted across a 6.0 pF capacitor and a 0 the switch is closed, calculate: (SGJWN A‘MB) 2. A battery of emf E = 20 Volts is conne resistor in series. Assuming that at t = (a) the initial current in the circuit. (b) the final charge on the capacitor. L3) A—+ 43,00 )‘HV' 01,,va LasLeen {WE/(7 LLWjLQ- "Th (Wat Hwy.» .1 Hum O) M :4 (44)} Inch} °\ ¢Lioxflo¢ On My LJl'tt'LL‘ 21 5;an {’7 Cl/ 3 CV Or. 3/=C€“ : (6.0/4F)(20\/) : ESL/:3 C9 = ‘AXIOJ‘C G) {or 5'54““5 1:0) RWAQtVCH' W‘Un’7L 7”)», wtva ‘MtW V: i—IK fwk’dv V _ 3. (a) In the figure below what' IS the equivalent resistance of the network shown? (b) What are the currents in each resistor? PutR =100S'2, R =R =SOQ, R =7SQ, 5: 6.0V. ( Sate fir) , let; . [76 6V NM MJo. Not (A) Au... WA mo +1. + No. NW "z %& W47 WY; 03L {'14, +WLraAelt; TLV; .> ' -6 ‘f 1,? #- Ix'érE 0-) 1";("3- 8—HR1 ’:..~~ [V " 1, I [ — fl :: __‘,_\,/_,_ a (0+LU MLHVJL; NC Li ‘ R4 75‘ 5 O 0'3/4’ = Y-Q) PUFkVLIV WM”): 3. (a) In the figure below, what power appears as thermal energy in R.? In R2? In R,? (b) What power is supplied by e, ? By 52 ? (c) Discuss the energy balance in this circuit. Assume that =3.0 V, 52 = 1.0 V, RI =5.0 Q, R, = 2.0 Q, and R3 = 4.0 Q. ( Sum B) (A) TLUMgi 2mm” POW-Zr I; 53$ in, 6); I? K .1 So wfi- Mfl‘LV +71} +)‘¢ CUHJ’W'L ‘LVV'OJAK 9,0914 reji'g-‘i‘o;l QASCQL/ l’kffie C/Jr/J/HLS o‘rfll [47°95 : Tie 355*?“ mile .j-WS (mi alimnglw //L0J£.).- . ._ .. _ ”Mi-Lit Up.” WJov-hi} Ln? 3 '- ’53 +1327 i’j—i R: = O (D “M’s'sniwokl TALK; J‘Ze Loop 1 0,71%” 0&9 (ii/3 ‘Hv, j‘mqlln lob; ee/oL-LMM. -_§’ if; K: #173: +73“: .317 1 . *1 RV 'Rl'f‘g} V Pivf) +1“! 93.1 i‘kjvnuibq (vise/ML i'lvv (g)? 3 21,44qu .42; +IQK3+I,K, 4. (my :0 .Kgé—RJ I OppoxLe &'ree{~‘sn HAP £2 {Harv} "Ls muffle, C) TL, 03%: La L) $11;ka Link E. K supply‘m'owan. ‘nl .2 4/ 7L: lVLL Orw1)WL-We 2: rs ’34") Qnel‘l)~7 —FNH +Le Li‘ffiu‘g‘,l Tie Gav-Jr; L» R) 0% He row»; envy Lewj Ju‘FSAQ ‘\n'\r~a LL, I‘Q-Sx'ljbr: 0x3 LLM. ”FMS-’4) 1“; W MJ‘O L‘QVL Q'\Qf‘j\11;a)w¢e/)+l€/ 9.9ij qus MVSfiL an GM '4’ "71> 21am . (TA-s Q vJv‘fl Wld’. +LL‘C/Ufr2/1V’WS AD’XI’ q ¥wu\r~‘u’x of; LN”) (11,004 " f 7 Pi" {'ng " (R +Pa +P5> : O In *Dlé ”WWW” *0' 7’) ’ O \/ . (MHLLM (1)an V3 QTIOrj) 33f Q COW}, mm» WWW (Nuwm alwmw “A. "Lug”? ) Lgfi‘ Jq S'LijJz/y wauw‘l‘ .) ...
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