I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3


 Rosaline Sharp
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1 2 The Prllel Circuit Electric Circuits: Figure 2 elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is proportionl to the vlue of the individul resistors. Higher resistnce would men less current. Lower resistnce would men more current. The sum of these splits flows from the ttery to the resistors nd then the sme totl current would gin flow ck into the negtive of the ttery. ttery Fig. 2 resistor The Kirchoff Current Lw or KCL KCL sttes tht the sum of the currents into point equls the sum of the currents out of the point. A point is defined s common tie point where wires re ctully touching or connected together. n the figure elow (Fig. 22), current flows into point nd then out. n Fig. 22, the entire current ( ) flows out into 2. We cn write KCL for Fig. 22 s follows: = 2 n Fig. 22, the current splits with some trveling down one wire nd some trveling down the other wire. We cn write KCL for Fig. 22 s follows: = 2 3 n Fig. 22c, the current comes to point from two sources, nd 2. The current trveling out from point in two wires s 3 nd 4. We cn write KCL for Fig. 22c s follows: 2 = 3 4 Fig. 22 Kirchhoff Current Lw  KCL You cn summrize KCL with the following eqution: currents into point = currents out of point KCL is crucil lw in circuit nlysis. t is simple in concept nd prolems re esy. KCL is intuitive in tht it is esy to see electrons flowing into point. Where do they go? Do they ccumulte t
2 point or do they flow out immeditely. Oviously the electrons must continue to flow out of the point s they enter. They do not ccumulte t point! An exmple of KCL: Find the vlue of the unknown in Fig. 23 elow. Applying KCL t point, write: n Fig. 23, pply KCL t point to write: 7 = 3 Therefore, = 4A 4 = 9 > = 5A n the eqution, we notice tht negtive sign occurs. The current tht flows in this rnch is opposite the ssumed direction. The circuit is redrwn in Fig. 23c showing the ctul direction of the current. The direction of current will e determined y writing the KCL eqution nd noticing the sign of ny unknown current. A minus sign shows tht current is ctully flowing in the opposite direction. KCL utomticlly determines current direction. This ws shown in the exmple in Fig. 23 nd. Fig. 23 Exmples of KCL (Kirchhoff s Current Lw) 7A 3A 4A 9A 4A 9A 5A Rememer tht KCL is true ecuse electrons do not dispper or ccumulte t point ut rther continue to trvel through the circuit. The Kirchhoff oltge Lw of KL With KCL, ll unknowns were currents. With the Kirchhoff oltge Low, unknowns re voltges. The lw sttes tht the sum of voltges round closed pth equls zero. This lw sys to strt t ny point in circuit, wlk in clockwise direction round the circuit nd write voltges of ech voltge source or lod in n eqution with the sum equl zero. A numer of exmples will give us etter understnding of this lw. Another question is why is this lw so importnt? t is importnt in tht it leds to numer of equtions with unknown voltges in ny circuit. t is lso useful to pply the voltge divider formul formul we will discuss in few pges. 2
3 A first simple exmple of KL: Fig A single source (ttery) nd single lod (resistor): Strt wlking t point ove in Fig The first voltge encountered is 2 ( resistor). Since the sign is, write () in the eqution. At 2, write: 2 Keep wlking clockwise. At write since the ttery ws entered from the negtive side: : 2 End the wlk ck t the point. There write: From this, one gets: 2 = 0 2 = Climing ldder shows n equivlent nlogy. When we clim up numer of steps, to get ck to the origin (or ground), we must clim down the equl numer of steps. Fig. 25 Going Up Ldder then Down The source t the left is the mn going up the ldder numer of steps. The resistnce or lod shows the sme numer of steps down the ldder ck to zero volts or ground. The numer of steps up must equl the numer ck down. 3
4 One source, two lods: Fig. 26 shows nother exmple of KL. Strting t point nd wlking clockwise gives the following three voltges: from which we find: = 0 or = 2 3 From these equtions, it cn e shown tht the source voltge equls equls the sum of the lod voltges 2 nd 3. Fig Fig. 27 Going Up Ldder then Down The sme energy used to go up the ldder now is divided etween the two efforts going down the ldder. Two sources, two lods Fig. 26 shows nother exmple of KL. Strting t point nd wlking clockwise gives the following four voltges: = 0 or 2 = 3 4 4
5 From these equtions, one cn see tht the source voltges nd 2 equl the sum of the lod voltges 3 nd 4. The eqution for the Kirchhoff oltge Lw (KL) sttes: oltges round closed loop = 0 How to use KL nd KCL will e the suject of prolems throughout this chpter nd the next. Find the unknown vlue of in Fig. 28: Strting t nd moving round the loop clockwise gives the following: 3 4 = 0 Solving for the unknown, = 7 Fig. 28 d c f e 2 6 The exmple gives ech voltge except one giving method for finding n unknown voltge. This method (KL) cn e used for ny loop. The next exmple shows the sme principle in use: Find the unknown vlue of in Fig. 28: Agin, using KL: giving: 3 2 = 0 = 9 5
6 Another exmple: Find the unknown vlues of nd 2 in Fig. 28c: t is noticed tht ech loop with resistors hve n unknown voltge. The first voltge to find is nd cn e found y choosing loop c which gives: Solving for we find: 3 2 = 0 = 9 The second voltge to find is 2. To find it, use pth defcd or: Solving for 2 we find: = 0 2 = 7 Rememer to e flexile with which loop to choose. We could hve found nd then use tht rnch to find 2 insted of the originl pth chosen (defcd). Either pth works. Anlysis of Series Circuits (One Pth) Rememer with series circuit tht the current is the sme t ny point in the circuit. Use this fct when solving prolems tht re series ones. This cn e summrized in generl s: = 2 This is n ppliction of KCL (current in equls current out of ny point). Also n output of the series circuit is tht the voltge rtio equls the resistnce rtio. Fig. 29, is series circuit with the R nd R 2. The equtions of voltges re: A rtio of voltges cn e written: or = R 2 = R2 = R 2 R 2 6
7 2 = R R 2 Fig. 29 c R R 2 R R 2 2 k k 50 m (d) Fig. 29c dds vlues for R, R 2 nd to llow us to write: 8 = or 2 = 4 Fig. 29d dds vlues for R, R 2 nd 2 to llow us to write: 0.00 = 0,000 = or = 0.2 = 200 m Equivlent resistnce The equivlent resistnce of two or more resistors in series circuit equls the sum of the resistnces in the circuit. Fig. 20 shows two resistors in series etween nd. The equivlent resistnce is shown in Fig. 20 nd equls: R = = 6000 Ω The equivlent resistnce is usully referred s R eq or R equivlent. R eq hs the sme resistnce s the two resistors together in series. oltge cross the two resistors equl the voltge cross the equivlent R eq. 2 = 0 = 2 7
8 Also nd = R 2 = R2 The current of the equivlent R eq is equl to the series circuit. Fig k 2k 2 6k R R 2 2 R (d) For Fig. 20d, the eqution for is: = R for the series circuit: Req = R R2 The current in the equivlent circuit of Fig. 20d equls the current in Fig. 20c. This is n exmple of n equivlent resistnce giving the sme results s the two seprte resistnces of Fig. 20c. An eqution for R eq for series circuits with more thn two resistors is cple of eing developed. This eqution is: Req = R R2 R3 Rn This is mny times referred to s Rtotl or R T. The circuit elow in Fig. 2 hs vlues of resistors: R = 20 Ω, R 2 = 50 Ω, nd R 3 = 0 Ω. The equivlent resistnce is 80 Ω. For voltge source of 8, the current cn e clculted s: = R = 8 = 0. A = 00 ma 80 8
9 R R 2 equls Fig. 2 R 3 Find the current in Fig. 32: The first step is to find the equivlent resistnce R T: RT = R R2 R3 = 2 3 = 6 Ω After finding R T, use the following eqution to find : = R T = 2 6 = 2 A The resultnt current is 2 A. This is lso the current in the originl circuit of Fig We cn use 2 A to dd the voltge through ech resistor in Fig. 22 nd dd the three voltges to find the sum or source voltge. = R = 2 x 2 = 4 2 = R2 = 3 x 2 = 6 3 = R3 = x 2 = 2 The voltges sum to 2 (the vlue of the source). This gives proof tht the current is indeed 2 A. 2 R R 2 R equls 2 Fig. 22 R T
10 An exmple of finding R T from Series Resistors: Find the totl or equivlent resistnce R T in Fig. 23 nd. Just dd the two resistnces to comine into single resistor. 5 k 3 k ==> 8 k 2 k 4 k ==> 6 k Fig. 23 (d) Another exmple of finding R T from Series Resistors, this time with multiple pths: Wht if the series resistnces re inside lrger circuit? These resistnces re to e comined first s shown in Fig. 23(e) nd (f). f the new circuit is prllel, then the prllel rules will pply to tht portion of the circuit, rules tht re to come. Fig. 23 cont 2 5 k 3 k 2 k 4 k ==> 2 8 k 6 k (e) (f) Another exmple from the Holidys: Using the older style incndescent Christms tree uls in series, form series circuit. Assume 2 lights t 00 Ω ech nd find R T y ddition. A voltge of 20 is cross the entire string. Find the current through the string. Current equls: RT = 2 x 00 =.2 kω = R = 20 = 0. A = 00 ma 200 0
11 Anlysis of Prllel Circuits (Mny Pths) For series circuits, current is the sme nd voltge splits. For prllel circuits, voltge is constnt nd current splits. oltge is equl cross prllel circuits per KL, which sttes tht the voltge round ny closed loop must sum to zero. Therefore, ny closed loop etween prllel rnches must hve the sme voltge. This cn e shown in Fig. 24 elow s: = = 2 or  = 0 nd = Around the right loop of Fig. 24 gives similr results: 2  = 0 or = 2 nd = = 2 Fig R 2 R 2 R 2 R 2 n R n Currents in Prllel Circuits t follows from the fct tht the voltge cross ech resistor is equl tht we cn write s n unknown nd find the currents in ech rnch of the prllel circuit of Fig. 34 s: = R 2 = R 2.. n = R n
12 By pplying KCL (Kirchhoff Current Lw), we cn see the sum of the currents from ech rnch is T or (Totl): T = 2 n One cn see from the following prllel equtions tht current is the inverse of resistnce. f resistnce is lrger, the current is smller. f we hve doule the resistnce, current is hlf, etc. n generl: = R 2 = R 2 We cn show the rtio of / 2 s: 2 = /R /R 2 or 2 = R 2 R We use this formul to find currents in prllel circuit in Fig. 25 elow. f we know the current through R, we cn find 2. n Fig. 25, through R equls ma: 2 = R 2 R = or 2 = A = 2 ma 00 ma ma k k 50 Fig
13 Fig. 25 is similr exmple ut with unknown. The sme eqution gives : 0. = 50 0,000 = 200 or = 0.5 ma The rtio of current is :200 sed on the rtio of resistnce. Equivlent Resistnce in Prllel Circuits For prllelcircuit nlysis, there is choice etween formuls to use. First is the reciprocl rule, the more generl rule. The development of this rule follows: Note = 2 = R nd 2 = R 2 By sustitution: = R R 2 nd: = ( R R 2 ) For the simplified circuit, = R eq The equivlent resistnce for prllel pth noted s R eq is: R eq = R R 2 (reciprocl rule) Algeric comintion of eqution s right side: 3
14 R eq = R R 2 R R 2 or R eq = R R 2 R R 2 This is the product over sum rule nd pplies to two resistors, unlike the reciprocl rule (which pplies to ny numer of resistors). Exmples Using Only Two Resistors: Fig. 26 hs two 5kΩ resistnces rrnged in prllel. We clculte R eq using the product/sum rule: Fig. 26 R eq = 5000x = 25,000,000 = 2,500 Ω 0,000 5 k 5 k ==> 2.5 k t cn e noted tht if the two resistors in prllel re equl, the comintion of the two in prllel drops the resistnce y hlf. The two resistors in Fig. 26 re not equl. Agin, we use the product/sum rule to clculte Req: Fig. 26 cont R eq = 3000x = 8,000,000 = 2,000 Ω 9,000 3 k 6 k ==> 2 k 4
15 Agin with Fig. 26c: R eq = 00x = 30, = 75 Ω Fig. 26 cont ==> 75 The reciprocl or product over sum rule my e used for two resistors ut we usully use the product over sum rule since it is usully viewed s esier. Severl Prllel Resistnces (More thn Two) When severl resistors (more thn two) re in prllel, two methods re ville to find the equivlent resistnce. One my solve the resistnces two t time. The second method will solve for ll resistnces in single eqution. This method is the reciprocl rule: R eq = R R 2 R n The product over sum rule my e used, even with lrge numer of resistors in prllel if we comine two resistors t time. The following exmple shows how this method is used with three resistors in prllel: We first comine the 3kΩ nd 6kΩ resistors into n equivlent 2kΩ resistor. We redrw the circuit in 27 nd find the equivlent resistnce R T in 27c. The vlue is kω. See Fig. 27c. Fig k 3k 6k ==> 2k 2k ==> k 5
16 The onestep pproch uses the reciprocl rule to find R eq in Fig The reciprocl eqution is: Sustituting into this eqution, we find R eq: R eq = R R 2 R n R eq = 2KΩ 3KΩ = kω 6KΩ Either method gives the sme nswer. Sometimes the reciprocl rule is quicker nd to e used. Other times, the product over sum rule is etter. You will need to choose on cseycse sis. Finding the Current in Prllel Circuits: Clculte the current in Fig. 28: Fig k 2 6k ==> 2 2k 2 4 ma 3k 2 ma 6k n Fig. 28 find nd 2. Next, use KCL to dd the two currents: = = 4 ma nd 2 = = 2 ma The vlue of T in Fig. 28 is the sme in Fig. 28 nd c. t is T = 4 ma 2 ma = 6 ma. Checking the rtio of currents s inverse of the resistnce: From Fig. 28: 2 = R 2 R 6
17 = 4 ma 2 2 ma = 2 = R 2 = 6kΩ R 3kΩ Agin, from the Holidys, nother Christms Tree Light Prolem: This time 20 uls re connected in prllel. Ech light hs resistnce of 0 kω. Find the entire string s equivlent resistnce? Apply 00 cross the string of lights nd find T (totl current). Next, find the current in ech ul. Find R eq using: R eq = 0,000 0,000 0,000 = 0,000 (totl of 20) 20 x = = 500Ω Totl current = = R = 00 = 200 ma 500 The current in ech ul is clculted s 200 ma/20 = 0 ma. Looking t Fig. 29, find R eq: R eq 0k 30k 50k Fig. 29 Usully, the onestep reciprocl rule is est. Use it to find R eq: R eq = 0kΩ 30kΩ = 6,520 Ω = 6.52 kω 50kΩ To check the vlidity of the nswer, the nswer cnnot e greter thn the smllest resistnce (0 kω). t cn lso not e less thn /3 the smllest resistnce (3.3 kω). The nswer definitely fits etween the limit of this mx nd min limit nd is vlid nswer. 7
18 Comintion Circuits Let the Rel Fun Begin Often, circuits my e comined into series/prllel comintions. These re referred to s comintion circuits. By stute ppliction of the series nd prllel rules, n equivlent resistnce my e found nd vrious currents nd voltges found throughout the circuit. The following Figure 203 will demonstrte finding R eq from seriesprllel comintion circuit: 5k 5k Fig R eq 0k 30k == > R eq 7.5k As cn e seen, the 0 kω nd 30 kω resistors re in prllel nd cn e comined using the product over sum rule. From the productoversum rule, we get 7.5 kω for the resistnce of this portion of the circuit. We oserve lso tht the 5 kω is in series with the 7.5 kω resistnce. To find R eq, we dd series resistors giving totl R eq of 2.5 kω. Another exmple: n Fig. 22 elow, there is seriesprllel comintion tht is reduced in nd. First, oserve the 6 kω nd 4 kω comined into 20 kω resistor still in prllel with the 60 kω resistor. The productoversum rule or reciprocl rule my e used to find the finl R eq of 5 kω Fig. 22 R eq 6 k 4 k 60 k ==> R eq 20 k 60 k ==> R eq 5 k Some generl rules for reducing these seriesprllel comintions re:. Look t the circuit to identifying series nd prllel comintions. 2. Replce series nd prllel comintions until one equivlent resistnce remins. 8
19 More exmples: From Fig. 222, find the equivlent resistnce of the circuit etween the terminls: Fig k 6 k 5 k R eq 60 k ==> R eq 20 k 60 k ==> 0 k 30 k 4 k 7.5 k R eq 2.5 k 20 k 60 k ==> R eq 6.82 k (d) First, oserve tht the 0 kω nd 30 kω resistors re in prllel nd the 6 kω nd 4 kω resistors re in series. Simplify oth leving results shown in. We re now presented gin with series comintion of 5 kω nd 7.5 kω resistors in series which we comine to 2.5 kω. Now we re presented with choice. Either comine two resistors in prllel nd then repet the process or use the reciprocl rule once. lwys try to do one opertion insted of two so choose the reciprocl rule which gives 6.82 kω for R eq. Rememer: Series Comintion Rule: Reciprocl Prllel Rule: Req = R R2 R3 R eq = R R 2 R n Product over Sum Prllel Rule (two resistors only): R eq = R R 2 R R 2 9
20 Prolems: 2. Find the numer of equipotentil common tiepoints in Fig. 223: Fig Clculte current nd current direction in Fig nd : Fig A 3A 2 A 7A 2.3 For Fig nd, find the unknown currents: Fig ma 0 ma ma ma
21 2.4 For Fig d, find the unknown voltges: Fig (d) 2.5 For Fig d find the unknown voltges nd currents: Fig k 3k 2 k 2k 3k k 6k 2k 9 6 2k 3k k (d) 2
22 2.6 For Fig c, find R eq t the terminls: Fig Sw A Sw B k 00 k 50k 00 2k k 50 3k 450k For Fig d find ll unknown currents (): Fig ma 4 ma 3 0 k 30 k 2k 3 k 6 k k 30 k (d) 22
23 2.8 For Fig d, find ll equivlent resistnces (Req): 6 k 2 k 3 k 9 k k k k 2 k 3 k 4 k Fig (d) (e) 2.9 For Fig d, find ll equivlent resistnces (Req): Fig k k 3 k 50 k 50 k 4 k k 2 k 4 k 0 k 2.5 k 30 k 4 k 2 k (d) 2.0 Attch 0 lod to the terminls nd find the voltge nd current cross ech resistor in the ove circuits of Fig
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