CHAPTER 12 DIRECT CURRENT CIUITS
DIRECT CURRENT CIUITS 257 12.1 RESISTORS IN SERIES AND IN PARALLEL When wo resisors are conneced ogeher as shown in Figure 12.1 we said ha hey are conneced in series. As i is clear from he figure, any charge ha flows hrough R 1 mus equal he charge ha flows hrough R 2, ha is I eq = I 1 = I 2 a R 1 R 2 Figure 12.1 Series connecion of wo resisors. The curren in each resisor is he same. ε b Since he poenial difference beween a and b equals he sum of he poenial drop across each resisor we have V eq = V 1 + V 2 ( R ) V eq = IR + 12.1 1 + IR2 = I 1 R2 where V eq is he poenial drop across he equivalen resisor. Therefore we conclude ha R eq = R 1 + R 2 12.2 For more han wo resisors conneced in series we have R eq R + R + +L 12.3 = 1 2 R3
DIRECT CURRENT CIUITS 258 Now consider he wo resisors conneced as shown in Figure 12.2. The poenial drops across R 1 and R 2 are equal and mus equal o he poenial drop across any equivalen resisor conneced beween a and b, ha is V eq = V 1 = V 2 12.4 I 1 a I I 2 R 1 R 2 Figure 12.2 Parallel connecion of wo resisors. The poenial difference across each resisor is he same. ε b If I 1 and I 2 are he currens passing hrough R 1 and R 2, respecively, hen he ne curren of he circui is I = I 1 + I 2 12.5 Using Ohm s law and Equaion 12.4, we ge 1 1 1 R = + eq R R 12.6 1 2 In general if more ha wo resisors are conneced in parallel, hen we have 1 1 1 1 = + + +L 12.7 R eq R R R 1 2 3 Household circuis are always conneced in parallel so ha if one device is swiched off he ohers remains on. The operaing volage of all he devices is he same in such a connecion.
DIRECT CURRENT CIUITS 259 Example 12.1 Three resisors are conneced as shown in Figure 12.3. (a) Wha is he equivalen resisance of he circui? (b) If he emf of he baery is 56 V, find he curren hrough each resisor Soluion (a) The circui can be reduced, sep by sep, o a single equivalen resisance as shown in Figure 12.3. The 8.0-Ω and he 2.0-Ω are conneced in parallel, and so hey can be replaced by an equivalen resisor of 1.6 Ω, using Equaion 12.6. This resisor is conneced in series wih he 4.0-Ω resisor. The equivalen resisance of he circui is hen R eq = 4.0 + 1.6 = 5. 6Ω (b) Since he 4-Ω and he 1.6-Ω are conneced in series, hey have he same curren I 1, which mus equal o he curren of he baery. Using Ohm s law we ge I ε 56 5.6 1 = = = R eq 10.A Now he poenial difference across he 1.6-Ω is
DIRECT CURRENT CIUITS 260 8.0 Ω 4.0 Ω 4.0 Ω 1.6 Ω 5.6 Ω 2.0 Ω Figure 12.3 Example 12.1 V bc = I1 R = 10.(1.6) = 16 V This poenial difference is he same across he 8.0-Ω and he 2.0- Ω resisors due he parallel connecion beween hem. So, we can find he curren I 2 passing hrough he 8.0-Ω resisor as I 16 = V = 8.0 8.0 2 = 2.0A and he curren I 3 hrough he 2.0-Ω resisor as 16 I 3 = V = = 8.0 A 2.0 2.0
DIRECT CURRENT CIUITS 261 Example 12.2 Wha is he curren passing hrough he 6.0-Ω resisor in he circui shown in Figure 12.4? 6.0Ω 3.0 Ω 6.0 Ω 6.0 Ω 34 V 5.0 Ω 2.0 Ω 34 V 5.0 Ω 5.0 Ω 34 V 2.5 Ω Figure 12.4 Example 12.2 Soluion I is clear ha he curren I passing hrough he 6.0-Ω resisor mus pass hrough he baery. Therefore, we can wrie I ε =. R eq Now o find R eq we have o reduce he circui sep by sep as we did in he previous example. To do so we have o noe ha he 2.0- Ω and he 3.0-Ω resisors are conneced in series. Their equivalen resisor is conneced in parallel wih he 5.0-Ω resisor. The equivalen resisor of he hree resisors is conneced in series wih he 6.0-Ω resisor. The equivalen resisance of he circui R eq is hen R = 8. 5Ω. So we have eq I = 34 = 8.5 4.0A
DIRECT CURRENT CIUITS 262 6.2 KIHHOFF S RULES The series-parallel combinaions of resisors are no capable of handling some complex circuis. Gusav Rober Kirchhoff (1824-1887) developed wo rules for analyzing such circuis: I- The sum of he currens enering any juncion mus equal he sum of he currens leaving ha juncion. (A juncion is any poin in a circui where a curren can spli) II- The algebraic sum of he poenial differences across all he elemens around any loop mus be zero. The firs rule is an applicaion of he conservaion of charge principle, while he second rule is an applicaion of he conservaion of energy principle. To apply he second rule we should know he following wo remarks: 1- The change in poenial hrough any resisor is negaive for a move in he direcion of he curren and posiive for a move opposie o he direcion of he curren. This is because he curren hrough a resisor moves from he end of higher poenial o ha of lower poenial. 2- The change in poenial hrough an ideal baery is posiive for a move from he negaive o he posiive erminal of he baery and negaive for a move in he opposie direcion. Sraegy for solving problems using Kirchhoff s rules:
DIRECT CURRENT CIUITS 263 1- Draw a circui diagram and label all quaniies, known and unknown. 2- Assign a direcion for he curren in each par of he circui. Do no boher if your guess of curren direcion is incorrec; he resul will have a negaive value. 3- Apply he firs Kirchhoff s rule o any juncion in he circui. In general his rule is used one ime fewer han he number of juncions in he circui. 4- Choose any closed loop in he nework, and designae a direcion (clockwise or counerclockwise) o raverse he loop. 5- Saring from one poin in he loop, go around he loop in he designaed direcion. Sum he poenial differences across all he elemens of he chosen loop o zero. In doing so you should noe he wo remarks discussed above, ha is, he poenial difference across an emf is +ε if i is raversed from he negaive o he posiive erminal and -ε if raversed in he opposie direcion. The poenial difference across any resisor is -IR if his resisor is raversed in he direcion of he assumed curren and +IR if raversed in he opposie direcion. 6- Choose anoher loop and repea he fifh sep o ge a differen equaion relaing he unknown quaniies. Coninue unil you have as many equaions as unknowns. 7- Solve hese equaions simulaneously for he unknowns.
DIRECT CURRENT CIUITS 264 Example 12.3 In he circui shown in Figure 12.5, find he curren in each resisor. ε 1 = 16 V R 1 = 2.0 Ω a R 2 = 6.0 Ω I 1 I 2 I 3 R 3 = 3.0 Ω ε 2 = 12 V Soluion Le I 1, I 2, and I 3 Figure 12.5 Example 12.3 be he currens hrough he resisors R 1, R 2, and R 3, respecively. The direcions of he currens are assigned arbirary as shown in Figure 12.5. As i clear from he circui here are wo juncions, a and b, and wo loops, he lef loop and he righ loop. Applying Kirchhoff s firs rule o he juncion a we ge I = + (1) 1 I2 I3 Noe ha he same equaion will be go if rule is applied o juncion b. Now, we apply Kirchhoff s second rule o he lef loop and raverse he loop in he clockwise direcion, obaining ε 1 I1R1 I3R3 = 0 16 2.0I 1 3.0I3 = 0 (2) We need hree equaions o solve for he hree unknowns. To obain he hird equaion we raverse he righ loop in he counerclockwise direcion. Kirchhoff s second rule hen gives ε 2 + I2R2 I3R3 = 0 b
DIRECT CURRENT CIUITS 265 12 + 6.0I 2 3.0I3 = 0 (3) Subsiuing for I 1 from Equaion (1) ino Equaion (2) we ge 16 2.0I 2 5.0I3 = 0 (4) Solving Equaions (3) and (4) for I 2 and I 3 we ge I 2 = 0.5A, and I 3 = 3.0A Subsiuing hese wo values ino Equaion (1) we ge I 3 = 3.5A Example 12.4 Three R ideal baeries and hree 1 = 4.0 Ω resisors are conneced as shown in Figure 12.6. ε 1 = 14 V a) Find he curren in each resisor. b) Wha is he curren hrough he middle baery? c) Calculae he poenial difference Vb Va. a ε 2 = 8.0 V R 3 = 2.0 Ω b R 2 = 1.0 Ω Figure 12.6 Example 12.4. ε 3 = 10. V Soluion Le us label he currens in each branch of he circui as I 1, I 2, and I 3 wih he assumed direcions shown in Figure 12.6. Noe ha he curren I 1 ha pass hrough R 1 mus pass hrough R 3. a) If we apply Kirchhoff s second rule o he lef loop raversing i clockwise we ge
DIRECT CURRENT CIUITS 266 ε 1 I1R1 ε 2 I1R3 = 14 4.0I 8.0 2.0I1 0 1 = 0 From which we find I 1 =1.0 A Now, we choose he righ loop and raverse i in he clockwise direcion. Kirchhoff s second rule hen gives ε 2 I 2 R3 ε 3 = 0 8.0 I 2 10. = 0 or I 2 = 2.0 A b) The curren I 2 ha pass hough he middle baery can be obained by applying Kirchhoff s firs rule o he juncion a. This gives or I + 1 = I2 I3 I 2 = I1 I3 = 1.0 2.0 = 1.0A The minus sign indicaes ha I 2 should poin up c) Saring a poin a, we follow a pah oward poin b, adding poenial differences across all he elemens we encouner. If we follow he pah hrough he middle baery we obain
DIRECT CURRENT CIUITS 267 V b V a = 8.0 2.0I1 = 8.0 2.0 = 10.V The minus sign here means ha V a > V b. Try o follow anoher pahs from a o b o verify ha hey also give he same resul. 12.3 The CIUITS So far we have discussed circuis wih ime-independen currens. Now we deal wih circuis ha conain capaciors. In such circuis he curren depends on ime. Charging Process Figure 12.7 shows a capacior, iniially uncharged, conneced in series wih a resisor. If he swich S is hrown a poin 1 a = 0, he capacior will begin o charge, creaing a curren in he circui. Le I be he curren in he circui a some insan during he charging process, and q be he charge on he capacior a he same insan. circui, we obain Applying Kirchhoff s second rule o he q ε IR = 0 12.8 C The poenial difference across he capacior q C is negaive because here is a drop in poenial as we move from he posiive ε 2 1 S Figure 12.7 When he swich S is a poin 1 he capacior is charged by he baery. When S is a poin 2 he capacior discharge hrough R. R C
DIRECT CURRENT CIUITS 268 plae of he capacior o he negaive plae. Subsiuing for I wih I = dq d, in Equaion 12.8, and rearranging we obain dq εc q = d 12.9 Noing ha he charge on he capacior is iniially zero, i.e., q = 0 a = 0, we can inegrae boh sides of Equaion 12.9 as q 0 ε dq C q = εc q ln εc 0 d = q = ε C 1 e 12.10 where e is he base of he naural logarihm. To find he curren as a funcion of ime, we differeniae Equaion 12.10 wih respec o ime o ge I = ε e 12.11 R The quaniy is called he ime consan, τ, which defined as he ime required for he curren o decrease o 1 e of is iniial
DIRECT CURRENT CIUITS 269 value. Equaions 12.10 and 12.11, which are ploed in Figure 12.8, ell he following: 1- A = 0, he charge q is zero, as required, and he iniial curren I o is I o ε = 12.12 R ha is, he capacior acs as if i were a wire wih negligible resisance (shor circui). 2- As (afer a long ime), he charge has is maximum equilibrium value, Q m Q m = εc 12.13 and he curren is zero, ha is he capacior acs as i were an open swich (open circui).
DIRECT CURRENT CIUITS 270 q I Q m I o (a) (b) Figure 12.8 (a) The charge versus ime in a charging process for circui. (b) The curren versus ime in a charging process for he same circui. Discharging Process Suppose ha he capacior in Figure 6.17 is now fully charged such ha is poenial difference is equal o he emf ε. If he swich is hrown o poin 2 a a new ime = 0, he capacior begin o discharge hrough he resisor. Le I be he curren in he circui a some insan during his process, and q be he charge on he capacior a he same insan. Applying Kirchhoff s rule o he loop, we ge q C IR = 0 Subsiuing for I wih rearrange we obain 12.14 I = dq d (explain he negaive sign), and
DIRECT CURRENT CIUITS 271 dq 1 = d 12.15 q Using he iniial condiion, q = Qm a = 0 we can inegrae he las equaion o obain q Qmax dq q 1 = 0 d ln q Qm 1 = q = Q m e 12.16 The curren is he rae of decrease of he charge on he capacior, ha is I dq d = I e o = 12.17 where Q I o = 12.18
DIRECT CURRENT CIUITS 272 Example 12.5 A 2.0-kΩ resisor S 2.0 kω and a 8.0-µF capacior are conneced, in series, wih a 24-V baery as shown in Figure 12.9. The 12 V 8.0 µf capacior is iniially uncharged, and he swich S is closed a =0. Figure 12.9 Example 12.5. a) Find he ime consan of he circui, and he maximum charge on he capacior. b) Wha is he ime required for he curren o drop o half is iniial value? c) Afer being closed for a long ime, he swich is now opened a =0, wha is he ime required for he charge o decrease o onefifh is maximum value. Soluion a) The ime consan is τ = = 3 6 ( 2.0 10 )( 8.0 10 ) = 16ms The maximum charge is, from Equaion 12.13, Q m 6 ( 12)( 8.0 10 ) 96µ F = ε C = = b) From Equaions 6.11 and 6.12 we have I = I o e
DIRECT CURRENT CIUITS 273 To find he ime required for he curren o drop o half is value, we subsiue I = 1 I ino his equaion: 2 o 1 2 I o = I o e Taking he logarihm of boh sides, we have or 1 ln 2 = = ln 2 1 = 11.1ms c) In he discharging process, he charge varies wih ime according o q = Q m e Subsiuing for we ge q 1 5 = Q, and aking he logarihm of boh sides m or 1 ln 5 = = ln 5 1 = 26 ms
DIRECT CURRENT CIUITS 274 Example 12.6 In he R 1 = 6.0 Ω circui shown in Figure I 1 I 3 12.10, he capacior is ε 1 = 32 V 12 µf iniially empy and he S swich S is closed a =0. a) Find he curren in each branch of he circui a =0. b) Calculae he maximum charge on he capacior. Figure 12.10 Example 12.6. I 2 R 2 = 4.0Ω Soluion a) A =0, he capacior is reaed as if i were a wire wih negligible resisance. Therefore, we have I 2 = 0 and 32 I 1 = I3 = = 6.0 5.3 A b) The maximum charge is aained afer a long ime ( ). A his ime he capacior is reaed as if i were an open swich. So, 32 I 3 = 0 and I 1 = I2 = = 3.2A. (6.0 + 4.0) Applying Kirchhoff s second rule o righ loop we find ha he poenial difference V across he capacior is V = I R ( 3.2)( 4.0) 12 V 2 2 = = Now he maximum charge is ( 6 12 10 )( 12.8) = 1.5 10 4 C Q = CV = µ
DIRECT CURRENT CIUITS 275 6.4 ELECTRICAL INSTRUMENTS The Ammeer I is a device used o measure he curren. The curren o be measured mus pass hrough he ammeer, ha is he ammeer should be conneced in series wih he curren i is o measure. The resisance of an ammeer mus be small compared o oher resisances in he circui so as no o aler he curren being measured. The Volmeer I is a device used o measure he poenial difference. To measure a poenial difference beween wo poins he erminals of he volmeer have o be conneced, in parallel, beween hese wo poins. The resisance of a volmeer mus be large enough no o aler he volage being measured.