Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 CHPTE9 1 When the uls re onnete in series, the equivlent resistne is series = i = 4 ul = 4(140 Ω) = 560 Ω When the uls re onnete in prllel, we fin the equivlent resistne from 1/ prllel = (1/ i ) = 4/ ul = 4/(140 Ω), whih gives prllel = 35 Ω 2 () When the uls re onnete in series, the equivlent resistne is series = i = 3 3 = 3(40 Ω) 3(80 Ω) = 360 Ω () When the uls re onnete in prllel, we fin the equivlent resistne from 1/ prllel = (1/ i ) = (3/ ) (3/ ) = [3/(40 Ω)] [3/(80 Ω)], whih gives prllel = 89 Ω 3 f we use them s single resistors, we hve = 30 Ω = 50 Ω When the uls re onnete in series, the equivlent resistne is series = i = = 30 Ω 50 Ω = 80 Ω When the uls re onnete in prllel, we fin the equivlent resistne from 1/ prllel = (1/ i ) = (1/ ) (1/ ) = [1/(30 Ω)] [1/(50 Ω)], whih gives prllel = 19 Ω 4 Beuse resistne inreses when resistors re onnete in series, the mximum resistne is series = 3 = 500 Ω 900 Ω 1400 Ω = 2800 Ω = 280 kω Beuse resistne ereses when resistors re onnete in prllel, we fin the minimum resistne from 1/ prllel = (1/ ) (1/ ) (1/ 3 ) = [1/(500 Ω)] [1/(900 Ω)] [1/(1400 Ω)], whih gives prllel = 261 Ω 5 The voltge is the sme ross resistors in prllel, ut is less ross resistor in series onnetion We onnet three 10-Ω resistors in series s shown in the igrm Eh resistor hs the sme urrent n the sme voltge: i = @ = @(60 ) = 20 Thus we n get 40- output etween n Pge 19 1
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 6 When the resistors re onnete in series, s shown in, we hve = i = 3 = 3(240 Ω) = 720 Ω When the resistors re onnete in prllel, s shown in B, we hve 1/ B = (1/ i ) = 3/ = 3/(240 Ω), so B = 80 Ω n iruit C, we fin the equivlent resistne of the two resistors in prllel: 1/ = (1/ i ) = 2/ = 2/(240 Ω), so = 120 Ω C This resistne is in series with the thir resistor, so we hve C = = 120 Ω 240 Ω = 360 Ω n iruit D, we fin the equivlent resistne of the two resistors in series: = = 240 Ω 240 Ω = 480 Ω This resistne is in prllel with the thir resistor, so we hve 1/ D = (1/ ) (1/) = (1/480 Ω) (1/240 Ω), so D = 160 Ω D B 7 We n reue the iruit to single loop y suessively omining prllel n series omintions We omine n, whih re in series: 7 = = 28 kω 28 kω = 56 kω We omine 3 n 7, whih re in prllel: 1/ 8 = (1/ 3 ) (1/ 7 ) = [1/(28 kω)] [1/(56 kω)], whih gives 8 = 187 kω We omine 4 n 8, whih re in series: 9 = 4 8 = 28 kω 187 kω = 467 kω We omine 5 n 9, whih re in prllel: 1/0 = (1/ 5 ) (1/ 9 ) = [1/(28 kω)] [1/(467 kω)], whih gives 0 = 175 kω We omine 0 n 6, whih re in series: eq = 0 6 = 175 kω 28 kω = 46 kω B B 3 4 5 6 D B 8 4 C 5 6 7 B 3 4 C 5 6 9 D C 5 6 D D C 0 6 D eq D Pge 19 2
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 8 () n series the urrent must e the sme for ll uls f ll uls hve the sme resistne, they will hve the sme voltge: ul = /N = (110 )/8 = 138 () We fin the resistne of eh ul from ul = ul / = (138 )/(040 ) = 34 Ω The power issipte in eh ul is P ul = ul = (040 )(138 ) = 55 W 9 For the prllel omintion, the totl urrent from the soure is = N ul = 8(0240 ) = 192 The voltge ross the les is les = les = (192 )(15 Ω) = 29 The voltge ross eh of the uls is ul = les = 110 288 = 107 We fin the resistne of ul from ul = ul / ul = (107 )/(0240 ) = 450 Ω The power issipte in the les is les n the totl power use is, so the frtion wste is les / = les / = (29 )/(110 ) = 0026 = 26% ul les 10 n series the urrent must e the sme for ll uls f ll uls hve the sme resistne, they will hve the sme voltge: ul = /N = (110 )/8 = 138 We fin the resistne of eh ul from P ul = ul 2 / ul 70 W = (138 ) 2 / ul, whih gives ul = 27 Ω 11 Fortuntely the require resistne is less We n reue the resistne y ing prllel resistor, whih oes not require reking the iruit We fin the neessry resistne from 1/ = (1/ ) (1/ ) 1/(320 Ω) = [1/(480 Ω)] (1/ ), whih gives = 960 Ω in prllel 12 The equivlent resistne of the two resistors onnete in series is s = We fin the equivlent resistne of the two resistors onnete in prllel from 1/ p = (1/ ) (1/ ), or p = /( ) The power issipte in resistor is P = 2 /, so the rtio of the two powers is P p /P s = s / p = ( ) 2 / = 4 When we expn the squre, we get 2 2 2 = 4, or 2 2 2 = ( ) 2 = 0, whih gives = = 220 kω Pge 19 3
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 13 With the two uls onnete in prllel, there will e 110 ross eh ul, so the totl power will e 75 W 40 W = 115 W We fin the net resistne of the uls from P = 2 / 115 W = (110 ) 2 /, whih gives = 105 Ω 14 We n reue the iruit to single loop y suessively omining prllel n series omintions We omine n, whih re in series: 7 = = 220 kω 220 kω = 440 kω We omine 3 n 7, whih re in prllel: 1/ 8 = (1/ 3 ) (1/ 7 ) = [1/(220 kω)] [1/(440 kω)], whih gives 8 = 147 kω We omine 4 n 8, whih re in series: 9 = 4 8 = 220 kω 147 kω = 367 kω We omine 5 n 9, whih re in prllel: 1/0 = (1/ 5 ) (1/ 9 ) = [1/(220 kω)] [1/(367 kω)], whih gives 0 = 138 kω We omine 0 n 6, whih re in series: eq = 0 6 = 138 kω 220 kω = 358 kω The urrent in the single loop is the urrent through 6 : 6 = = / eq = (12 )/(358 kω) = 336 m For C we hve C = 0 = (336 m)(138 kω) = 463 This llows us to fin 5 n 4 5 = C / 5 = (463 )/(220 kω) = 211 m 4 = C / 9 = (463 )/(367 kω) = 126 m For B we hve B = 4 8 = (126 m)(147 kω) = 185 This llows us to fin 3, 2, n 1 3 = B / 3 = (185 )/(220 kω) = 1 = 2 = B / 7 = (185 )/(440 kω) = From ove, we hve B = 185 084 m 042 m 2 8 5 5 0 B 3 3 4 4 5 C 5 6 D 4 4 C 6 D C 6 7 5 9 3 5 eq D B 3 2 B 4 5 4 C 6 D C 6 D D Pge 19 4
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 15 () When the swith is lose the ition of to the prllel set will erese the equivlent resistne, so the urrent from the ttery will inrese This uses n inrese in the voltge ross, n orresponing erese ross 3 n 4 The voltge ross inreses from zero Thus we hve 1 n 2 inrese 3 n 4 erese () The urrent through hs inrese This urrent is now split into three, so urrents through 3 n 4 erese Thus we hve 1 = n 2 inrese 3 n 4 erese () The urrent through the ttery hs inrese, so the power output of the ttery inreses () Before the swith is lose, 2 = 0 We fin the resistne for 3 n 4 in prllel from 1/ = (1/ i ) = 2/ 3 = 2/(100 Ω), whih gives = 50 Ω For the single loop, we hve = 1 = /( ) = (450 )/(100 Ω 50 Ω) = 0300 This urrent will split evenly through 3 n 4 : 3 = 4 =! =!(0300 ) = 0150 fter the swith is lose, we fin the resistne for, 3, n 4 in prllel from 1/ B = (1/ i ) = 3/ 3 = 3/(100 Ω), whih gives B = 333 Ω For the single loop, we hve = 1 = /( B ) = (450 )/(100 Ω 333 Ω) = 0338 This urrent will split evenly through, 3, n 4 : 2 = 3 = 4 = @ = @(0338 ) = 0113 2 S 3 3 3 3 4 4 4 4 Pge 19 5
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 16 () When the swith is opene, the removl of resistor from the prllel set will inrese the equivlent resistne, so the urrent from the ttery will erese This uses erese in the voltge ross, n orresponing inrese ross The voltge ross 3 ereses to zero Thus we hve 1 n 3 erese 2 inreses () The urrent through hs erese The urrent through hs inrese The urrent through 3 hs erese to zero Thus we hve 1 = n 3 erese 2 inreses () Beuse the urrent through the ttery ereses, the r term ereses, so the terminl voltge of the ttery will inrese () When the swith is lose, we fin the 1 resistne for n 3 in prllel from 1/ = (1/ i ) = 2/ = 2/(550 Ω), whih gives = 275 Ω For the single loop, we hve = /( r) = (180 )/(550 Ω 275 Ω 050 Ω) = 206 For the terminl voltge of the ttery, we hve = r = 180 (206 )(050 Ω) = 170 When the swith is opene, for the single loop, we hve = /( r) = (180 )/(550 Ω 550 Ω 050 Ω) = 157 For the terminl voltge of the ttery, we hve = r = 180 (157 )(050 Ω) = 172 r r 2 S 3 3 3 17 We fin the resistne for n in prllel from 1/ p = (1/ ) (1/ ) = [1/(28 kω)] [1/(21 kω)], whih gives p = 12 kω Beuse the sme urrent psses through p n 3, the higher resistor will hve the higher power issiption, so the limiting resistor is 3, whih will hve power issiption of! W We fin the urrent from P 3mx = 3mx2 3 050 W = 3mx 2 (18 10 3 Ω), whih gives 3mx = 00167 The mximum voltge for the network is mx = 3mx ( p 3 ) = (00167 )(12 10 3 Ω 18 10 3 Ω) = 50 1 2 4 3 3 3 3 18 () For the urrent in the single loop, we hve = /( r) = (850 )/(810 Ω 0900 Ω) = 0104 For the terminl voltge of the ttery, we hve = r = 850 (0104 )(0900 Ω) = 841 () For the urrent in the single loop, we hve = /( r) = (850 )/(810 Ω 0900 Ω) = 00105 For the terminl voltge of the ttery, we hve = r = 850 (00105 )(0900 Ω) = 849 Pge 19 6
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 Pge 19 7
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 19 The voltge ross the ul is the terminl voltge of the four ells: = ul = 4( r) (062 )(12 Ω) = 4[20 (062 )r], whih gives r = 023 Ω 20 We fin the resistne of ul from the nominl rting: ul = nominl 2 /P nominl = (120 ) 2 /(30 W) = 48 Ω We fin the urrent through eh ul when onnete to the ttery from: ul = / ul = (118 )/(48 Ω) = 0246 Beuse the uls re in prllel, the urrent through the ttery is = 2 ul = 2(0246 ) = 0492 We fin the internl resistne from = r 118 = [120 (0492 )r], whih gives r = 04 Ω 21 f we n ignore the resistne of the mmeter, for the single loop we hve = /r 25 = (15 )/r, whih gives r = 0060 Ω 22 We fin the internl resistne from = r 88 = [120 (60 )r], whih gives r = 0053 Ω Beuse the terminl voltge is the voltge ross the strter, we hve = 88 = (60 ), whih gives = 015 Ω 23 0 0 8 6 8 4 6 5 r 6 7 6 5 r From the results of Exmple 197, we know tht the urrent through the 60-Ω resistor, 6, is 048 We fin from = 6 7 = (048 m)(27 Ω) = 130 This llows us to fin 8 8 = / 8 = (130 )/(80 kω) = 016 Pge 19 8
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 24 For the urrent in the single loop, we hve = /( r) = (90 )/(80 Ω 120 Ω 20 Ω) = 041 For the terminl voltge of the ttery, we hve = r = 90 (041 )(20 Ω) = 818 The urrent in resistor goes from high to low potentil For the voltge hnges ross the resistors, we hve = = (041 )(120 Ω) = 491 = = (041 )(80 Ω) = 327 For the sum of the voltge hnges, we hve = 818 491 327 = 0 r 25 For the loop, we strt t point : 2 r 2 1 r 1 = 0 (66 Ω) 12 (2 Ω) 18 (1 Ω) = 0, whih gives = 0625 The top ttery is ishrging, so we hve 1 = 1 r 1 = 18 (0625 )(1 Ω) = 174 The ottom ttery is hrging, so we hve 2 = 2 r 2 = 12 (0625 )(2 Ω) = 133 r 1 r 2 1 2 26 To fin the potentil ifferene etween points n, we n use ny pth The simplest one is through the top resistor From the results of Exmple 198, we know tht the urrent 1 is 087 We fin from = = 1 = ( 087 )(30 Ω) = 26 h 1 3 r 2 2 3 1 r 2 1 g f e 27 From the results of Exmple 198, we know tht the urrents re 1 = 087, 2 = 26, 3 = 17 On the iruit igrm, oth tteries re ishrging so we hve = 2 3 r 2 = 45 (17 )(1 Ω) = 43 eg = 1 2 r 1 = 80 (26 )(1 Ω) = 77 h 1 3 r 2 2 3 1 r 2 1 g f e Pge 19 9
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 28 For the onservtion of urrent t point, we hve in = out 1 = 2 3 For the two loops inite on the igrm, we hve loop 1: 1 2 1 = 0 90 2 (15 Ω) 1 (22 Ω) = 0 loop 2: 3 2 = 0 60 2 (15 Ω) = 0 When we solve these equtions, we get 1 = 068, 2 = 040, 3 = 108 Note tht 2 is opposite to the iretion shown 1 1 1 2 3 3 2 29 For the onservtion of urrent t point, we hve in = out 1 = 2 3 When we the internl resistne terms for the two loops inite on the igrm, we hve loop 1: 1 2 1 1 r 1 = 0 90 2 (15 Ω) 1 (22 Ω) 1 (12 Ω) = 0 loop 2: 3 2 3 r 3 = 0 60 2 (15 Ω) 2 (12 Ω) = 0 When we solve these equtions, we get 1 = 060, 2 = 033, 3 = 093 1 1 1 2 3 3 2 30 For the onservtion of urrent t point, we hve in = out 1 3 = 2 For the two loops inite on the igrm, we hve loop 1: 1 1 2 = 0 90 1 (15 Ω) 2 (20 Ω) = 0 loop 2: 2 2 3 3 = 0 120 2 (20 Ω) 2 (30 Ω) = 0 When we solve these equtions, we get 1 = 0156 right, 2 = 0333 left, 3 = 0177 up We hve rrie n extr eiml ple to show the greement with the juntion eqution 1 2 1 2 2 1 3 3 Pge 19 10
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 31 For the onservtion of urrent t point, we hve in = out 1 3 = 2 When we the internl resistne terms for the two loops inite on the igrm, we hve loop 1: 1 1 1 r 1 2 = 0 90 1 (10 Ω) 1 (15 Ω) 2 (20 Ω) = 0 loop 2: 2 3 r 2 2 3 3 = 0 120 3 (10 Ω) 2 (20 Ω) 2 (30 Ω) = 0 When we solve these equtions, we get 1 = 015 right, 2 = 033 left, 3 = 018 up 1 2 1 2 2 1 3 3 32 When we inlue the urrent through the ttery, we hve six unknowns For the onservtion of urrent, we hve juntion : = 1 2 juntion : 1 = 3 5 juntion : 2 5 = 4 For the three loops inite on the igrm, we hve loop 1: 1 5 5 2 = 0 1 (20 Ω) 5 (10 Ω) 2 (25 Ω) = 0 loop 2: 3 3 4 4 5 5 = 0 4 3 (2 Ω) 4 (2 Ω) 5 (10 Ω) = 0 3 loop 3: 2 4 4 = 0 60 2 (25 Ω) 4 (2 Ω) = 0 When we solve these six equtions, we get 1 = 0274, 2 = 0222, 3 = 0266, 4 = 0229, 5 = 0007, = 0496 We hve rrie n extr eiml ple to show the greement with the juntion equtions 1 3 5 1 2 5 2 4 3 33 When the 25-Ω resistor is shorte, points n eome the sme point n we lose 2 For the onservtion of urrent, we hve juntion : 5 = 1 4 juntion : 1 = 3 5 For the three loops inite on the igrm, we hve loop 1: 1 5 5 = 0 1 (20 Ω) 5 (10 Ω) = 0 loop 2: 3 3 4 4 5 5 = 0 3 (2 Ω) 4 (2 Ω) 5 (10 Ω) = 0 loop 3: 4 4 = 0 60 4 (2 Ω) = 0 When we solve these six equtions, we get 1 = 023, 3 = 069, 4 = 300, 5 = 046, = 369 Thus the urrent through the 10-Ω resistor is 046 up 3 1 5 5 2 1 3 4 4 3 Pge 19 11
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 34 For the onservtion of urrent t point, we hve 2 3 = 1 For the two loops inite on the igrm, we hve loop 1: 1 1 r 1 1 3 2 2 r 2 2 1 = 0 120 1 (10 Ω) 1 (80 Ω) 120 2 (10 Ω) 2 (10 Ω) 1 (12 Ω) = 0 loop 2: 3 3 r 3 3 5 2 2 2 r 2 3 4 = 0 60 3 (10 Ω) 3 (18 Ω) 120 2 (10 Ω) 3 (15 Ω) = 0 When we solve these equtions, we get 1 = 077, 2 = 071, 3 = 0055 For the terminl voltge of the 60- ttery, we hve fe = 3 3 r 3 = 60 (0055 )(10 Ω) = 595 3 e r 1 1 1 2 r 2 2 g 2 4 2 5 3 1 3 r 3 f 35 The upper loop eqution eomes loop 1: 1 1 r 1 1 3 2 2 r 2 2 1 = 0 120 1 (10 Ω) 1 (80 Ω) 120 2 (10 Ω) 2 (10 Ω) 1 (12 Ω) = 0 The other equtions re the sme: 2 3 = 1 loop 2: 3 3 r 3 3 5 2 2 2 r 2 3 4 = 0 60 3 (10 Ω) 3 (18 Ω) 120 2 (10 Ω) 3 (15 Ω) = 0 When we solve these equtions, we get 1 = 130, 2 = 112, 3 = 018 36 For the onservtion of urrent t point, we hve = 1 2 For the two loops inite on the igrm, we hve loop 1: 1 1 r 1 = 0 20 1 (010 Ω) (40 Ω) = 0 loop 2: 2 2 r 2 = 0 30 2 (010 Ω) (40 Ω) = 0 When we solve these equtions, we get 1 = 531, 2 = 469, = 062 For the voltge ross we hve = = (062 )(40 Ω) = 25 Note tht one ttery is hrging the other with signifint urrent 1 1 r 1 2 2 r 2 2 1 37 We fin the equivlent pitne for prllel onnetion from C prllel = C i = 6(37 µf) = 22 µf When the pitors re onnete in series, we fin the equivlent pitne from 1/C series = (1/C i ) = 6/(37 µf), whih gives C series = 062 µf 38 Fortuntely the require pitne is greter We n inrese the pitne y ing prllel pitor, whih oes not require reking the iruit We fin the neessry pitor from C = C 1 C 2 16 µf = 50 µf C 2, whih gives C 2 = 11 µf in prllel Pge 19 12
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 39 We n erese the pitne y ing series pitor We fin the neessry pitor from 1/C = (1/C 1 ) (1/C 2 ) 1/(3300 pf) = [1/(4800 pf)] (1/C 2 ), whih gives C 2 = 10,560 pf Yes, it is neessry to rek onnetion to series omponent 40 The pitne inreses with prllel onnetion, so the mximum pitne is C mx = C 1 C 2 C 3 = 2000 pf 7500 pf 00100 µf = 00020 µf 00075 µf 00100 µf = 00195 µf The pitne ereses with series onnetion, so we fin the minimum pitne from 1/C min = (1/C 1 ) (1/C 2 ) (1/C 3 ) = [1/(2000 pf)] [1/(7500 pf)] [1/(00100 µf)] = [1/(20 nf)] [1/(75 nf)] [1/(100 nf)], whih gives C min = 14 nf 41 The energy store in pitor is U =!C 2 To inrese the energy we must inrese the pitne, whih mens ing prllel pitor Beuse the potentil is onstnt, to hve three times the energy requires three times the pitne: C = 3C 1 = C 1 C 2, or C 2 = 2C 1 = 2(150 pf) = 300 pf in prllel 42 () From the iruit, we see tht C 2 n C 3 re in series n fin their equivlent pitne from 1/C 4 = (1/C 2 ) (1/C 3 ), whih gives C 4 = C 2 C 3 /(C 2 C 3 ) From the new iruit, we see tht C 1 n C 4 re in prllel, with n equivlent pitne C eq = C 1 C 4 = C 1 [C 2 C 3 /(C 2 C 3 )] = (C 1 C 2 C 1 C 3 C 2 C 3 )/(C 2 C 3 ) () Beuse is ross C 1, we hve Q 1 = C 1 = (125 µf)(450 ) = 563 µc Beuse C 2 n C 3 re in series, the hrge on eh is the hrge on their equivlent pitne: Q 2 = Q 3 = C 4 = C 2 C 3 /(C 2 C 3 ) = [(125 µf)(625 µf)/(125 µf 625 µf)](450 ) = 188 µc C 2 C 1 C 3 C 1 C 4 43 From Prolem 42 we know tht the equivlent pitne is C eq = (C 1 C 2 C 1 C 3 C 2 C 3 )/(C 2 C 3 ) = 3C 1 /2 = 3(88 µf)/2 = 132 µf Beuse this pitne is equivlent to the three, the energy store in it is the energy store in the network: U =!C eq 2 =!(132 10 6 F)(90 ) 2 = 0053 J Pge 19 13
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 44 () We fin the equivlent pitne from 1/C series = (1/C 1 ) (1/C 2 ) = [1/(040 µf)] [1/(050 µf)], whih gives C series = 0222 µf The hrge on the equivlent pitor is the hrge on eh pitor: Q 1 = Q 2 = Q series = C series = (0222 µf)(90 ) = 200 µc We fin the potentil ifferenes from Q 1 = C 1 1 200 µc = (040 µf) 1, whih gives 1 = 50 Q 2 = C 2 2 200 µc = (050 µf) 2, whih gives 2 = 40 () s we foun ove Q 1 = Q 2 = 20 µc () For the prllel network, we hve 1 = 2 = 90 We fin the two hrges from Q 1 = C 1 1 = (040 µf)(90 ) = 36 µc Q 2 = C 2 2 = (050 µf)(90 ) = 45 µc C 1 C 2 C 1 C 2 45 For the prllel network the potentil ifferene is the sme for ll pitors, n the totl hrge is the sum of the iniviul hrges We fin the hrge on eh from Q 1 = C 1 = (Å 0 1 / 1 ) Q 2 = C 2 = (Å 0 2 / 2 ) Q 3 = C 3 = (Å 0 3 / 3 ) Thus the sum of the hrges is Q = Q 1 Q 2 Q 3 = [(Å 0 1 / 1 )] [(Å 0 2 / 2 )] [(Å 0 3 / 3 )] The efinition of the equivlent pitne is C eq = Q/ = [(Å 0 1 / 1 )] [(Å 0 2 / 2 )] [(Å 0 3 / 3 )] = C 1 C 2 C 3 46 The potentil ifferene must e the sme on eh hlf of the pitor, so we n tret the system s two pitors in prllel: C = C 1 C 2 = [K 1 Å 0 (!)/] [K 2 Å 0 (!)/] = (Å 0!/)(K 1 K 2 ) =!(K 1 K 2 )(Å 0 /) =!(K 1 K 2 )C 0 K 1 K 2 47 f we think of lyer of equl n opposite hrges on the interfe etween the two ieletris, we see tht they re in series For the equivlent pitne, we hve 1/C = (1/C 1 ) (1/C 2 ) = (!/K 1 Å 0 ) (!/K 2 Å 0 ) = (/2Å 0 )[(1/K 1 ) (1/K 2 )] = (1/2C 0 )[(K 1 K 2 )/K 1 K 2 ], whih gives C = 2C 0 K 1 K 2 /(K 1 K 2 ) K 1 K 2 Pge 19 14
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 48 For series network, we hve Q 1 = Q 2 = Q series = 125 pc We fin the equivlent pitne from Q series = C series 125 pc = C series (250 ), whih gives C series = 500 pf We fin the unknown pitne from 1/C series = (1/C 1 ) (1/C 2 ) 1/(500 pf) = [1/(200 pf)] (1/C 2 ), whih gives C 2 = 513 pf 49 () We fin the equivlent pitne of the two in series from 1/C 4 = (1/C 1 ) (1/C 2 ) = [1/(70 µf)] [1/(30 µf)], whih gives C 4 = 21 µf This is in prllel with C 3, so we hve C eq = C 3 C 4 = 40 µf 21 µf = 61 µf () We fin the hrge on eh of the two in series: Q 1 = Q 2 = Q 4 = C 4 = (21 µf)(24 ) = 504 µc We fin the voltges from Q 1 = C 1 1 C 3 C 1 C 2 504 µc = (70 µf) 1, whih gives 1 = 72 Q 2 = C 2 2 C 3 C 4 504 µc = (30 µf) 2, whih gives 2 = 168 The pplie voltge is ross C 3 : 3 = 24 C 50 Beuse the two sies of the iruit re ientil, we fin the resistne from the time onstnt: τ = C 30 s = (30 µf), whih gives = 10 MΩ Pge 19 15
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 51 () We know from Exmple 197 tht the equivlent resistne of the two resistors in prllel is 27 Ω There n e no stey urrent through the pitor, so we n fin the urrent in the series resistor iruit: = /( 6 7 5 r) = (90 )/(60 Ω 27 Ω 50 Ω 050 Ω) = 0634 We use this urrent to fin the potentil ifferene ross the pitor: = ( 6 7 ) = (0634 )(60 Ω 27 Ω) = 552 The hrge on the pitor is Q = C = (75 µf)(552 ) = 41 µc () s we foun ove, the stey stte urrent through the 60-Ω n 50-Ω resistors is 063 The potentil ifferene ross the 27-Ω resistor is = 7 = (0634 )(27 Ω) = 17 We fin the urrents through the 80-Ω n 40-Ω resistors from = 8 8 17 = 8 (80 Ω), whih gives 8 = 021 = 4 4 17 = 4 (40 Ω), whih gives 4 = 042 C 8 6 8 4 6 4 5 r C 6 7 5 r 52 () We fin the pitne from τ = C 35 10 6 s = (15 10 3 Ω)C, whih gives C = 23 10 9 F = 23 nf () The voltge ross the pitne will inrese to the finl C stey stte vlue The voltge ross the resistor will strt t the ttery voltge n erese exponentilly: = e t/τ S 16 = (24 )e t/(35 µs), or t/(35 µs) = ln(24 /16 ) = 0405, whih gives t = 14 µs 53 The time onstnt of the iruit is τ = C = (67 10 3 Ω)(30 10 6 F) = 00201 s = 201 ms The pitor voltge will erese exponentilly: C = 0 e t/τ 001 0 = 0 e t/(201 ms), or t/(201 ms) = ln(100) = 461, whih gives t = 93 ms C S Pge 19 16
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 54 () n the stey stte there is no urrent through the pitors Thus the urrent through the resistors is = /( ) = (24 )/(88 Ω 44 Ω) = 182 The potentil t point is = = = (182 )(44 Ω) = 80 () We fin the equivlent pitne of the two series pitors: 1/C = (1/C 1 ) (1/C 2 ) = [1/(048 µf)] [1/(024 µf)], = 24 whih gives C = 016 µf We fin the hrge on eh of the two in series: Q 1 = Q 2 = Q = C = (016 µf)(24 ) = 384 µc The potentil t point is = 0 = = Q 2 /C 2 = (384 µc)/(024 µf) = 16 () With the swith lose, the urrent is the sme Point must hve the sme potentil s point : = = 80 () We fin the hrge on eh of the two pitors, whih re no longer in series: Q 1 = C 1 = (048 µf)(24 80 ) = 768 µc Q 2 = C 2 = (024 µf)(80 ) = 192 µc When the swith ws open, the net hrge t point ws zero, euse the hrge on the negtive plte of C 1 h the sme mgnitue s the hrge on the positive plte of C 2 With the swith lose, these hrges re not equl The net hrge t point is Q = Q 1 Q 2 = 768 µc 192 µc = 58 µc, whih flowe through the swith C 1 S C 2 55 We fin the resistne on the voltmeter from = (sensitivity)(sle) = (30,000 Ω/)(250 ) = 750 10 6 Ω = 750 MΩ 56 We fin the urrent for full-sle efletion of the mmeter from = mx / = mx /(sensitivity) mx = 1/(10,000 Ω/) = 100 10 4 = 100 µ 57 () We mke n mmeter y putting resistor in prllel with the glvnometer For full-sle efletion, we hve meter = G r = s s (50 10 6 )(30 Ω) = (30 50 10 6 ) s, whih gives s = 50 10 6 Ω in prllel () We mke voltmeter y putting resistor in series with the glvnometer For full-sle efletion, we hve meter = ( x r) = G ( x r) 1000 = (50 10 6 )( x 30 Ω), whih gives x = 20 10 6 Ω = 20 MΩ in series s G r G x s r G G Pge 19 17
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 58 () The urrent for full-sle efletion of the glvnometer is = 1/(sensitivity) = 1/(35 kω/) = 285 10 2 m = 285 µ We mke n mmeter y putting resistor in prllel with the glvnometer For full-sle efletion, we hve meter = G r = s s (285 10 6 )(200 Ω) = (20 285 10 6 ) s, whih gives s = 29 10 5 Ω in prllel () We mke voltmeter y putting resistor in series with the glvnometer For full-sle efletion, we hve meter = ( x r) = G ( x r) 100 = (285 10 6 )( x 20 Ω), whih gives x = 35 10 4 Ω = 35 kω in series s G r G x s r G G 59 We n tret the millimmeter s glvnometer oil, n fin its resistne from 1/ = (1/ s ) (1/r) = (1/020 Ω) (1/30 Ω), whih gives = 0199 Ω We mke voltmeter y putting resistor in series with the glvnometer For full-sle efletion, we hve meter = ( x ) = G ( x ) 10 = (10 10 3 )( x 0199 Ω), whih gives x = 10 10 3 Ω = 10 kω in series The sensitivity of the voltmeter is Sensitivity = (1000 Ω)/(10 ) = 100 Ω/ x s G r s G Pge 19 18
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 60 Before onneting the voltmeter, the urrent in the series iruit is 0 = /( ) = (45 )/(37 kω 28 kω) = 0692 m The voltges ross the resistors re 01 = 0 = (0692 m)(37 kω) = 256 02 = 0 = (0692 m)(28 kω) = 194 When the voltmeter is ross, we fin the equivlent resistne of the pir in prllel: 1/ = (1/ ) (1/ ) = (1/37 kω) (1/100 kω), whih gives = 270 kω The urrent in the iruit is 1 = /( ) = (45 )/(27 kω 28 kω) = 0818 m The reing on the voltmeter is 1 = 1 = (0818 m)(270 kω) = 221 = 22 When the voltmeter is ross, we fin the equivlent resistne of the pir in prllel: 1/ B = (1/ ) (1/ ) = (1/28 kω) (1/100 kω), whih gives B = 219 kω The urrent in the iruit is 2 = /( B ) = (45 )/(37 kω 22 kω) = 0764 m The reing on the voltmeter is 1 = 1 B = (0764 m)(219 kω) = 167 = 17 We fin the perent inuries introue y the meter: (256 221 )(100)/(256 ) = 14% low (194 167 )(100)/(194 ) = 14% low 0 1 2 61 We fin the voltge of the ttery from the series iruit with the mmeter in it: = ( ) = (525 10 3 )(60 Ω 700 Ω 400 Ω) = 609 Without the meter in the iruit, we hve = 0 ( ) 609 = 0 (700 Ω 400 Ω), whih gives 0 = 554 10 3 = 554 m 0 Pge 19 19
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 62 We fin the equivlent resistne of the voltmeter in prllel with one of the resistors: 1/ = (1/ ) (1/ ) = (1/90 kω) (1/15 kω), whih gives = 563 kω The urrent in the iruit, whih is re y the mmeter, is = /(r ) = (120 )/(10 Ω 050 Ω 563 kω 90 kω) = 082 m The reing on the voltmeter is = = (082 m)(563 kω) = 46 r 63 n iruit 1 the voltmeter is ple in prllel with the resistor, so we fin their equivlent resistne from 1/ eq1 = (1/) (1/ ), or eq1 = /( ) The mmeter mesures the urrent through this equivlent resistne n the voltmeter mesures the voltge ross this equivlent 1 resistne, so we hve = / = eq1 = /( ) n iruit 2 the mmeter is ple in series with the resistor, so we fin their equivlent resistne from eq2 = The mmeter mesures the urrent through this equivlent resistne n the voltmeter mesures the voltge ross this equivlent resistne, so we hve = / = eq2 = 2 () For iruit 1 we get = (200 Ω)(100 10 3 Ω)/(200 Ω 100 10 3 Ω) = 200 Ω For iruit 2 we get = (200 Ω 100 Ω) = 300 Ω Thus iruit 1 is etter () For iruit 1 we get = (100 Ω)(100 10 3 Ω)/(100 Ω 100 10 3 Ω) = 99 Ω For iruit 2 we get = (100 Ω 100 Ω) = 101 Ω Thus oth iruits give out the sme inury () For iruit 1 we get = (50 kω)(100 kω)/(50 kω 100 kω) = 33 kω For iruit 2 we get = (50 kω 100 Ω) = 50 kω Thus iruit 2 is etter Ciruit 1 is etter when the resistne is smll ompre to the voltmeter resistne Ciruit 2 is etter when the resistne is lrge ompre to the mmeter resistne e f Pge 19 20
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 64 The resistne of the voltmeter is = (sensitivity)(sle) = (1000 Ω/)(30 ) = 30 10 3 Ω = 30 kω We fin the equivlent resistne of the resistor n the voltmeter from 1/ eq = (1/) (1/ ), or eq = /( ) = (74 kω)(30 kω)/(74 kω 30 kω) = 213 kω The voltmeter mesures the voltge ross this equivlent resistne, so the urrent in the iruit is = / = (20 )/(213 kω) = 0937 m For the series iruit, we hve = ( eq ) = (0937 m)(74 kω 213 kω) = 89 65 We know from Exmple 1914 tht the voltge ross the resistor without the voltmeter onnete is 40 Thus the minimum voltmeter reing is = (097)(40 ) = 388 We fin the mximum urrent in the iruit from = / = (80 388 )/(15 kω) = 0275 m Now we n fin the minimum equivlent resistne for the voltmeter n : eq = / = (388 )/(0275 m) = 141 kω For the equivlent resistne, we hve 1/ eq = (1/ ) (1/ ) 1/(141 kω) = [1/(15 kω)] (1/ ), whih gives = 240 kω We see tht the minimum eq gives the minimum, so we hve ³ 240 kω 66 We fin the resistnes of the voltmeter sles: 100 = (sensitivity)(sle) = (20,000 Ω/)(100 ) = 20 10 6 Ω = 2000 kω 30 = (sensitivity)(sle) = (20,000 Ω/)(30 ) = 60 10 5 Ω = 600 kω The urrent in the iruit is = ( / ) ( / ) For the series iruit, we hve = When the 100-volt sle is use, we hve = [(25 )/(2000 kω)] [(25 )/(120 kω)] = 0221 m = 25 (0221 m) When the 30-volt sle is use, we hve = [(23 )/(600 kω)] [(23 )/(120 kω)] = 0230 m = 23 (0230 m) We hve two equtions for two unknowns, with the results: = 741, n Without the voltmeter in the iruit, we fin the urrent: = ( ) 741 = (120 kω 222 kω), whih gives = 0217 m Thus the voltge ross is = = (0217 m)(120 kω) = 26 = 222 kω Pge 19 21
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 67 When the voltmeter is ross, for the juntion, we hve 1 1 = 1 [(55 )/ ] [(55 )/(150 kω)] = (120 55 )/ [(55 )/ ] 0367 m = (65 )/ When the voltmeter is ross, for the juntion e, we hve 2 = 2 2 (120 40 )/ = [(40 )/ ] [(40 )/(150 kω)] [(80 )/ ] = [(40 )/ ] 0267 m We hve two equtions for two unknowns, with the results: = 94 kω, n = 68 kω 1 2 1 1 2 e 2 f 68 The voltge is the sme ross resistors in prllel, ut is less ross resistor in series onnetion We onnet two resistors in series s shown in the igrm Eh resistor hs the sme 1 urrent: oy = /( ) = (90 )/( ) f the esire voltge is ross, we hve = = (90 ) /( ) 025 = (90 ) /( ) = (90 )/[1 ( / )], whih gives / = 35 When the oy is onnete ross, we wnt very negligile urrent through the oy, so the potentil ifferene oes not hnge This requires oy = 2000 Ω f we lso o not wnt lrge urrent from the ttery, possile omintion is = 2 Ω, = 70 Ω 69 Beuse the voltge is onstnt n the power is itive, we n use two resistors in prllel For the lower rtings, we use the resistors seprtely for the highest rting, we use them in prllel The rotry swith shown llows the B ontt to suessively onnet to C n D The ontt onnets to C n D for the prllel onnetion We fin the resistnes for the three settings from P = 2 / 50 W = (120 ) 2 /, whih gives = 288 Ω 100 W = (120 ) 2 /, whih gives = 144 Ω 150 W = (120 ) 2 / 3, whih gives 3 = 96 Ω s expete, for the prllel rrngement we hve 1/ eq = (1/ ) (1/ ) 1/ eq = [1/(288 Ω)] [1/(144 Ω)], whih gives eq = 96 Ω = 3 Thus the two require resistors re 288 Ω, 144 Ω D C B Pge 19 22
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 70 The voltge rop ross the two wires is rop = = (30 )(00065 Ω/m)(2)(95 m) = 37 The pplie voltge t the pprtus is = 0 rop = 120 37 = 116 71 We fin the urrent through the ptient (n nurse) from the series iruit: = /( motor e nurse ptient ) = (220 )/(10 4 Ω 0 10 4 Ω 10 4 Ω) = 73 10 3 = 73 m 72 () When the pitors re onnete in prllel, we fin the equivlent pitne from C prllel = C 1 C 2 = 040 µf 060 µf = 100 µf The store energy is U prllel =!C prllel 2 =!(100 10 6 F)(45 ) 2 = 10 10 3 J () When the pitors re onnete in series, we fin the equivlent pitne from 1/C series = (1/C 1 ) (1/C 2 ) = [1/(040 µf)] [1/(060 µf)], whih gives C series = 024 µf The store energy is U series =!C series 2 =!(024 10 6 F)(45 ) 2 = 24 10 4 J () We fin the hrges from Q = C eq Q prllel = C prllel = (100 µf)(45 ) = 45 µc Q series = C series = (024 µf)(45 ) = 11 µc 73 The time etween firings is t = (60 s)/(72 ets) = 0833 s We fin the time for the pitor to reh 63% of mximum from = 0 (1 e t/τ ) = 063 0, whih gives e t/τ = 037, or t = τ = C 0833 s = (75 µc), whih gives = 011 MΩ 74 We fin the require urrent for the hering i from P = 2 W = (40 ), whih gives = 050 With this urrent the terminl voltge of the three merury ells woul e merury = 3( merury r merury ) = 3[135 (050 )(0030 Ω)] = 401 With this urrent the terminl voltge of the three ry ells woul e ry = 3( ry r ry ) = 3[15 (050 )(035 Ω)] = 398 Thus the merury ells woul hve higher terminl voltge 75 () We fin the urrent from 1 = / oy = (110)/(900 Ω) = 012 () Beuse the lterntive pth is in prllel, the urrent is the sme: 012 () The urrent restrition mens tht the voltge will hnge Beuse the voltge will e the sme ross oth resistnes, we hve 2 = oy oy 2 (40 ) = oy (900 ), or 2 = 225 oy For the sum of the urrents, we hve 2 oy = 235 oy = 15, whih gives oy = 64 10 2 = 64 m Pge 19 23
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 Pge 19 24
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 76 () When there is no urrent through the glvnometer, we hve BD = 0, urrent 1 through n, n urrent 3 through 3 n x Thus we hve D = B 1 = 3 3, n BC = DC 1 = 3 x When we ivie these two equtions, we get / = x / 3, or x = ( / ) 3 () The unknown resistne is x = ( / ) 3 = (972 Ω/630 Ω)(426 Ω) = 657 Ω 1 3 B 3 x G S 1 D C 77 The resistne of the pltinum wire is x = ( / ) 3 = (460 Ω/380 Ω)(348 Ω) = 421 Ω We fin the length from = ρl/ = (106 10 8 Ω m)l/¹(0460 10 3 m)2, whih gives L = 264 m 78 () When there is no urrent through the glvnometer, the urrent must pss through the long resistor, so the potentil ifferene etween n C is C = Beuse there is no urrent through the mesure emf, for the ottom loop we hve = C B When ifferent emfs re lne, the urrent is the sme, so we hve S s = s, n x = x When we ivie the two equtions, we get G s / x = s / x, or x = ( x / s ) s x () Beuse the resistne is proportionl to the length, we hve or x = ( x / s ) s = (458 m/254 m)(10182 ) = 1836 S () f we ssume tht the urrent in the slie wire is muh greter thn the glvnometer urrent, the unertinty in the voltge is Æ = ± G G = ± (30 Ω)(0015 m) = ± 045 m Beuse this n our for eh setting n there will e unertinties in mesuring the istnes, the minimum unertinty is ± 090 m () The vntge of this metho is tht there is no effet of the internl resistne, euse there is no urrent through the ell 79 () We see from the igrm tht ll positive pltes re onnete to the positive sie of the ttery, n tht ll negtive pltes re onnete to the negtive sie of the ttery, so the pitors re onnete in prllel () For prllel pitors, the totl pitne is the sum, so we hve C min = 7(Å 0 min /) = 7(885 10 12 C 2 /N m 2 )(20 10 4 m 2 )/(20 10 3 m) = 62 10 12 F = 62 pf C mx = 7(Å 0 min /) = 7(885 10 12 C 2 /N m 2 )(120 10 4 m 2 )/(20 10 3 m) = 37 10 11 F = 37 pf Thus the rnge is 62 pf ² C ² 37 pf Pge 19 25
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 Pge 19 26
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 80 The terminl voltge of ishrging ttery is = r For the two onitions, we hve 408 = (740 )r 473 = (220 )r We hve two equtions for two unknowns, with the solutions: = 501, n r = 125 Ω 81 One rrngement is to onnet N resistors in series Eh resistor will hve the sme power, so we nee N = P totl /P = 5 W/! W = 10 resistors We fin the require vlue of resistne from totl = N series 22 kω = 10 series, whih gives series = 022 kω Thus we hve 10 022-kΩ resistors in series nother rrngement is to onnet N resistors in series Eh resistor will gin hve the sme power, so we nee the sme numer of resistors: 10 We fin the require vlue of resistne from 1/ totl = (1/ i ) = N/ prllel 1/22 kω = 10/ prllel, whih gives prllel = 22 kω Thus we hve 10 22-kΩ resistors in prllel 82 f we ssume the urrent in 4 is to the right, we hve 4 = 4 4 = (350 m)(40 kω) = 140 4 5 We n now fin the urrent in 8 : 8 = / 8 = (140 )/(80 kω) = 175 m 8 From onservtion of urrent t the juntion, we hve 8 = 4 8 = 350 m 175 m = 525 m r f we go lokwise roun the outer loop, strting t, we hve 5 4 4 r = 0, or = (525 m)(50 kω) (350 m)(40 kω) 120 (525 m)(10 Ω) = 52 f we ssume the urrent in 4 is to the left, ll urrents re reverse, so we hve = 140 8 = 175 m, n = 525 m f we go ounterlokwise roun the outer loop, strting t, we hve r 4 4 5 r = 0, or = = r 4 4 5 r = (525 m)(10 Ω) 120 (350 m)(40 kω) (525 m)(50 kω) = 28 The negtive vlue mens the ttery is fing the other iretion Pge 19 27
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 83 When the les re shorte ( x = 0), there will e mximum urrent in the iruit, inluing the glvnometer We hve,mx = G,mx r = (35 10 6 )(25 Ω) = 875 10 4 We n now fin the urrent in the shunt, sh : G sh,mx =,mx / sh = (875 10 4 )/ sh From onservtion of urrent t the juntion, we hve mx = G,mx sh,mx = 35 10 6 (875 10 4 )/ sh sh f we go lokwise roun the outer loop, strting t, ser we hve,mx mx ser = 0 875 10 4 30 [35 10 6 (875 10 4 )/ sh ] ser = 0 35 10 6 (875 10 4 )/ sh = (30 875 10 4 )/ ser 35 10 6 (875 10 4 )/ sh Å (30 )/ ser When the les re ross x = 30 kω, ll urrents will e one-hlf their mximum vlues We hve = G r =!(35 10 6 )(25 Ω) = 4375 10 4 The urrent in the shunt is sh = / sh = (4375 10 4 )/ sh From onservtion of urrent t the juntion, we hve = G sh = 175 10 6 (4375 10 4 )/ sh f we go lokwise roun the outer loop, strting t, we hve ( x ser ) = 0 4375 10 4 30 [175 10 6 (4375 10 4 )/ sh ](30 10 3 Ω ser )= 0 175 10 6 (4375 10 4 )/ sh = (30 4375 10 4 )/(30 10 3 Ω ser ) 175 10 6 (4375 10 4 )/ sh Å (30 )/(30 10 3 Ω ser ) We hve two equtions for two unknowns, with the solutions: sh = 13 Ω, n ser = 30 10 3 Ω = 30 kω r G sh x Pge 19 28
Solutions to Physis: Priniples with pplitions, 5/E, Ginoli Chpter 19 84 The resistne long the potentiometer is proportionl to the length, so we fin the equivlent resistne etween points n : 1/ eq = (1/x pot ) (1/ ul ), or eq = x pot ul /(x pot ul ) We fin the urrent in the loop from = /[(1 x) pot eq ] The potentil ifferene ross the ul is = eq, so the power expene in the ul is P = 2 / ul () For x = 1 we hve eq = (1)(100 Ω)(200 Ω)/[(1)(100 Ω) 200 Ω] = 667 Ω = (120 )/[(1 1)(100 Ω) 667 Ω] = 180 = (180 )(667 Ω) = 120 P = (120 ) 2 /(200 Ω) = 720 W () For x =! we hve eq = (!)(100 Ω)(200 Ω)/[(!)(100 Ω) 200 Ω] = 400 Ω x pot ul = (120 )/[(1!)(100 Ω) 400 Ω] = 133 = (133 )(400 Ω) = 533 P = (533 ) 2 /(200 Ω) = 142 W () For x = # we hve eq = (#)(100 Ω)(200 Ω)/[(#)(100 Ω) 200 Ω] = 222 Ω = (120 )/[(1 #)(100 Ω) 667 Ω] = 123 = (123 )(222 Ω) = 274 P = (274 ) 2 /(200 Ω) = 375 W 85 () Normlly there is no DC urrent in the iruit, so the voltge of the ttery is ross the pitor When there is n interruption, the pitor voltge will erese exponentilly: C = 0 e t/τ We fin the time onstnt from the nee to mintin 70% of the voltge for 020 s: 070 0 = 0 e (020 s)/τ, or (020 s)/τ = ln(143) = 357, whih gives τ = 056 s We fin the require resistne from τ = C 056 s = (22 10 6 F), whih gives = 25 10 4 Ω = 25 kω () n norml opertion, there will e no voltge ross the resistor, so the evie shoul e onnete etween n 86 () Beuse the pitor is isonnete from the power supply, the hrge is onstnt We fin the new voltge from Q = C 1 1 = C 2 2 (10 pf)(10,000 ) = (1 pf) 2, whih gives 2 = 10 10 5 = 010 M () mjor isvntge is tht, when the store energy is use, the voltge will erese exponentilly, so it n e use for only short ursts Pge 19 29