Assignment8 - 26.1. IDENIIFY: The newly—formed 1.vire is...

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Unformatted text preview: 26.1. IDENIIFY: The newly—formed 1.vire is a combination of series and parallel resistors. SET UP: Each of the three linear segments has resistance REE . The circle is two R36 resistors in parallel. EEECITE: The resistance of the circle is R-"ll since it consists of two R-"Ifi resistors in parallel. The equivalent resistance is rvvo RE resistors in series with an Rs'IS resistor. givhig flqfi = iii-'3 - R6 + 3512 = 31-1-4 BALE-111:: The equivalent resistance of the original svire has been reduced because the circle's resistance is less than it was as a linear 1tvire. 26.4. IDES‘IJIY: For resistors in parallel the voltages are the same and equal to the voltage across the equivalent resistance. . 1 1 1 SETLP: I"=lFt.—=—+—. R... 31 R; _2 1 1 '3-1 EJctctTr: o .= +— =12.3§2. {JRT' [3311 20:14 {h}1=l—= ‘4?" =lsja. Rm 124:: _t’_24ov__,- _ _it-'_24D“..-'_ 1 “fl "'j‘l‘Js‘i'E‘fi'l'A' ETALL'ATE: More current flows through the resistor that has the smaller R. 26.11. IDENTIFY: Replace the series combinations of resistors bv their equivalents. In the resulting parallel network the battery voltage is the voltage across each resistor. SET UP: The circuit is sketched in Figure 25.12:]. 2' = 4M] ‘v' EKECITE: R, and H2 in series have an equivalent “I m” 51 R1 = 3'0“ n resistance of R.1 = R, +RE =4.00 r: R, and R4 in series have an equivalent resistance of R“ =3. —i =110 E2 R; = Tilflll R4 = 5.001! Figure 26.12:] The circuit is equivalent to the circuit sketched in Figure 26.12b. '5 ' ‘3'“ 1"" R1: and H54 in parallel are equivalent to Rm givean i=i+i= Rd _H“ REL: RI: Ran R123“ s = Sis. l1 . = [4.00 flab: 1.»..0 fl] =3-m n a =I1ntt 4.00§2+11.Ufl _'I-| Tig'nl'e 16.11b The voltage across each branch ofthe parallel combination is 5'. so 5' — TEE” = 0. _ _ 43.0 V R... 4.00 [1 r. =12oa The cun'ent is 11.0 A through the 1.0-0 ['1 and 3.00 D. resistors, and it is 4.0 A through the 100' E1 and 5.00 G resistors. ETALL'ATE: The current through die battery is JT = I: - I"; = 11.0 :‘L + 4.0 A = 10.0 FL, and this is equal to £3th — 43.0 $53.00 [2 =16.0 A. 26.1.4. 26.2 0. 26.39. 26.42. IDENIIFY: Use Kn'chhoff’s RuJes to find the currents. SETUP: Sinee the 1.0V battery has the largervoltage, assume Il is tothelefttlrroughthe IDVbattery, I2 is to the right through the 5 V battery, and Iais to lheright tln'ough Ihe 10$] resistor. Go around each loop in the counterclotkudsedireclion. Enema: Upperloup:10.0v-{2.00:z+3.00o]r,—[1.00o+4.00o}1,—5.00v:0.Injsgives 5.0 V—[iflfl [1:11, —|[5.00 {1}}: =0 , and :aI.+11=1.0I0d . lower loop: 5.001? +[1.00 (2+ 4.1:!) (2)1} —I[10.0 (2)13 = 0 . This gives 5.00v+{5.00 o}I,—{10.0 (2)13 = 0 , and 11—21,=—1.00A Along 1with 1. =11 + 1,, we can solve for the three currents and find: I, = 0.000 as, = 0.200 an, = 0.500 a. [0) 0;, = 40.200 attain o}—[0.300 a}(s.00 n] = —s.20 v. EYALUATE: Traveling from I: to a through the 4.00 H and 3.00 (1 resistors yonpasa through the resistors in the diJection of the current and the potential decreases; point it is at higherpotential thanpoint a. IDENTIFY: We need to use Kirchhoff‘s rules. SET UP: Take a loop around the outside ofthe circuit use the current at the upperjunction, and then take a loop around the right side ofthe circuit. EEECITI: The outside loop gives ?5.0 ‘3 — {12.0 fl]{1.i0 A) — {40.0 5131+; = 0 , so I.“ = 1.138 A. At a junction we have 1.50 A = I,+ 1.130 .5. , and L= 0.313 A. A loop around the right part ofthe en'euit gives 5 — {40 {130:1 .1 SE A} - (15.0 fl){0.313.-’-‘c.j. E = 52.30, with the polarity shots-n in the figure intbe problem. ETALL'ATI: The unknown battery has a smaller emf than the known one, so the cun‘eut though it goes against its polarity. IDENTIFY: The capacitor discharges exponentially through the voltmeter. Since the potential difference across the capacitor is directly proportional to the charge on the plates, the voltage across the plates decreases eaponeutialh' with the same time constant as the charge. r -I'.FI'Ii.' SET UP: The reading of the voltmeter obeys the equation F’ = Fus , where RC is the time constant. EXECITE: {0] Solving for C and evaluating the result when t= 4.00 s gives - ,_,, , 1,1, R 1110" to] (3.40x10é ma 1"“ l" '- -, 3.00 V} {b} r=RC= [3.40 x 10ls HHS-49 x104?) =139 s ETALI'ATE: Iu roost laboratory circuits, time constants are much shorter than this one. IDENTIFY: In r = RC use the equivalent capacitance of the two capacitors. SET UP: For capacitors in series, CL = 4% + Cl . For capacitors in parallel. Cul = C1 — C1 . Originally, a. I 0 r =HC = 0.070 s. . . . . . . . . . 1 1 1 2 EIEECITE: (a) The combmed capacltance of the two 1d.eut1cal capacuors JJJ. senes 1s given by C— = E + E = E , vi _, so CE.l = g . The new time constant is thus 1005.52] = ll :3 5 = L a”. 0.435 s. {b} With die two capacitors in parallel the new total capacitance is simply 2C. Thus the time constant is REEC] = 2(0.E?0s] =1.?4 s. ETALI'ATI: The time constant is proportional to C“, . For capacitors in series the capacitance is decreased and for capacitors in parallel the capacitance is increased. 26.45. IIIEFIIFY: The stored energy i'E proportional to the square of the charge on the capacitor, so it will obey an exponential equation. but not the same equation as, theficharge. __ SET UP: _The energy stored in the capacitor is E = QTEC and the charge on the plates ii Q.) a'm'. Ihe current is I = In 3—H“: Exact-m t-'= @1526 = (pg emit-2c: tag “" 1ill-‘hen the capacitor has lost 30% of its stored energy. the energy is 20% ofthe initial energy. which is, Uy'S. Lia-'3 = Ht: 9 “1 give: I = [RC-'2) ln 5 = {23.0 QJHoE ijfln_5].-'] = 92.9 p5,. At this time. the current is I = In a “‘1' = [Ea-RC) a ‘ "r': so I = {3.3 nC'].-'[[25.IC| rotate: pI'J] a “‘39P”: “15-” “3*”: fl = 13.5 a. ETALI'ATE: 1ithen the energy reduced E‘DE’o, neither the current nor the charge are reduced that percent. ...
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Assignment8 - 26.1. IDENIIFY: The newly—formed 1.vire is...

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