Parallel Circuits PDF

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r INTRODUCTION In Chapter 5, you learned about series circuits and how to apply Ohm's law and Kirchhoff's voltage law You also saw how a seriescircuit can be used as a voltage divider to obtain several specified voltages from a single source voltage. The effects of opens and shorts in series...

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r INTRODUCTION

PARALLEL CIRCUITS 6-1 6-2 6-3 6-4 6-5 6-6 6-7 6-8 6-9 6-10 6-11

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Resistors in Parallel VoltageDrop in ParallelCircuits Kirchhoff'sCurrentLaw TotalParallelResistance Ohm'sLawin ParallelCircuits CurrentSourcesin Parallel CurrentDividers Powerin ParallelCircuits Examples of ParallelCircuit Applications Troubleshooting TechnologyTheoryinto Practice Electronics Workbench(EWB)and PSpiceTutorialsavailableat http://www.prenhalLcom/floyd

In Chapter 5, you learned about series circuits and how to apply Ohm's law and Kirchhoff's voltage law You also saw how a seriescircuit can be used as a voltage divider to obtain several specified voltages from a single source voltage. The effects of opens and shorts in seriescircuits were also examined. In this chapter,you will see how Ohm's law is used in parallel circuits; and you will learn Kirchhoff's current law. Also, several applications of parallel circuits, including automotive lighting, residential wiring, and the internal wiring of analog ammeters are presented.You will learn how to determine total parallel resistanceand how to troubleshoot for open resistors. When resistors are connectedin parallel and a voltage is applied acrossthe parallel circuit, each resistor provides a separatepath for cuffent. The total resistanceof a parallel circuit is reduced as more resistors are connectedin parallel. The voltage across each of the parallel resistors is equal to the voltage applied acrossthe entire parallel circuit. In the TECH TIP, Section6-11, imagine you are employed by an electronic instruments company. Your first job is to troubleshoot any defective instrument that fails the routine line check. In this particular assignment,you must determine the problem(s) with a defective five-range milliammeter so that it can be repaired. The knowledge of parallel circuits and of basic ammetersthat you will acquire in this chapter plus your understandingof Ohm's law, current dividers, and the resistor color code will be put to good use.

I TECHnology Theory Into Practice

r CHAPTER OBIECTIVES O Identifya parallelcircuit 0 Determinethe voltage acrosseachparallel branch O Apply Kirchhoff's cur:rentlaw 0 Determinetotal parallel resistance 0 Apply Ohm's law in a parallel circuit

HIST(DR,ICAI,

NOTD

Ernst Werner von Siemens 1816-1872 Emst Wemer von Siemens was born i n l 8 l 6 i n P n r s s i aW . hile in prison for acting as a second in a duel, he began to experiment with chemistry. Theseexperimentseventually led to his invention of the first electroplatingsystem. In 1837, Siemens began making improvementsin the eariy telegraph and contributed greatlyto the development of telegraphic systems. Emst had a younger brotheq Wilhelm, who was also a noted engineer. After moving to England, Wilhelm founded the companywhich today bears the family name. Unfortunately,it is not certain for which brother the unit of conductanceis named, although it is most likely Ernst. Photo credit: AIP Emilio Segri\4sualArchives, E. Scott Barr Collection.

Q Determine the total effect of current sourcesin parallel tr Use a parallel circuit as a current divider O Determinepower in a parallel circuit tr Describesomebasic applicationsof parallel circuits tr Troubleshootoarallel circuits

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PARALLEL CIRCUITS

6-1 r RESISTORS IN PARALIEL When two or more resistors are individually connected between the same two separate points, they are in parallel with each other. A parallel circuit provi.des more thqn one path for current. After completing this section, you should be uble to r Identify a parallel circuit . Translate a physical arrangementof resistors into a schematic

Each current path is called a branch. A parallel circuit is one that has more than one branch. Two resistors connectedin parallel are shown in Figure 6-1(a). As shown in part (b), the current out of the source divides when it gets to point A. Part of it goes through R1 and part through Rr. If additional resistors are connected in parallel with the first two, more current paths are provided between point A and point B as shown in Figure 6-1(c). All points along the top shown in blue are electrically the same as point A, and all points along the bottom shown in green are electrically the same as point B.

Rr

V^

(a)

FIGURE6-1 Resistorsin parallel.

ldentifyingParallelCircuits In Figure 6-1, it is obvious that the resistorsare connectedin parallel. Often, in actual circuit diagrams,the parallel relationship is not as clear. It is important that you learn to recognize parallel circuits regardlessof how they may be drawn. A rule for identifying parallel circuits is as follows: If there is more than one current path (branch) between two separate points (nodes), then there is a parallel circuit between those two points. Figure 6-2 shows parallel resistors drawn in different ways between two separate points labeled A and B. Notice that in each case, the current "travels" two paths going from A to B, and the voltage across each branch is the same.Although these examples in Figure 6-2 show only two parallel paths, there can be any number of resistors in parallel.

RESISTORS IN PARALLELT

163

6-2 of circuits with two paths.

- B (a

EXAMPLE 6-1

Bo-

)

B

)

Supposethat there are five resistors positioned on a circuit board as shown in Figure 6-3. Show the wiring required to connect all of the resistors in parallel, keeping in mind the positions of the + and - signs. Draw a schematic and label each of the resistors with its value. FIGURE6-3

Solution Wires are connected as shown in the assembly diagram of Figure 64(a). The schematicis shown in Figure 6-4(b). Again, note that the schematicdoes not necessarily have to show the actual physical arrangementof the resistors.The schematic shows how componentsare connectedelectrically.

R5

1.0ko

(a) Assembly wiring diagram

(b) Schematic

FICURE6-4 Related Problem

How would the circuit have to be rewired if Rr is removed?

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P A R A L L ECLI R C U I T S

EXAMPLE6-2

Determine the parallel groupings in Figure 6-5 and the value of each resistor.

Pin 3 Pin 4

FIGURE6-5 Solution Resistors R1 through Ra and R11 and R12 are all in parallel. This parallel combination is connectedto pins 1 afi 4. Each resistor in this group is 56 kO. Resistors R5 through R1e are all in parallel. This combination is connected to pins 2 and 3. Each resistorin this group is 100 kf2. Related Problem allel?

sEciloN 6-1 REVIEW

How would you connect all of the resistors in Figure 6-5 in par-

1. How are the resistors connectedin a parallel circuit? 2. How do you identify a parallel circuit? 3. Complete the schematicsfor the circuits in each part of Figure 6-6 by connecting the resistorsin parallel betweenpointsA and B. 4. Now connect each group of parallel resistors in Figure 6-6 in parallel with each other. FICURE6-6

R7

R^

o--rMf

+j, ?.+^,

ao-*.lf/------o

I

**'

"--{id%

(a)

(b)

R8

ryV/------o Re

o-A1\FB (cl

VOLTACE D R O PI N P A R A L L ECLI R C U I T ST

165

r VOLTACE DROP IN PARALTEL CIRCUITS As mentioned in the previous section, each current path in a parallel circuit is called s brunch. The voltage &cross any given brunch of a parallel circuit is equal to the voltage across each of the other branches in parallel. After completing this section, you should be able to I Determine the voltage across each parallel branch . Explain why the voltage is the same across all parallel resistors

To illustrate voltage drop in a parallel circuit, Iet's examine Figure 6-7(a). Points A, B, C, and D along the left side of the parallel circuit are electrically the same point and form one node becausethe voltage is the same along this line. You can think of all of these points as being connected by a single wire to the negative terminal of the battery. The points E, E G, and H along the right side of the circuit form another node and are all at a voltage equal to that of the positive terminal of the source. Thus, voltage across each parallel resistor is the same, and each is equal to the source voltage. Figure 6-1(b) is the same circuit as in part (a), drawn in a slightly different way. Here the left side of each resistor is connectedto a single point, which is the negativebattery terminal. The right side of each resistor is connectedto a single point, which is the positive battery terminal. The resistors are still all in parallel acrossthe source. In Figure 6-8, a 12 V battery is connectedacrossthree parallel resistors.When the voltage is measuredacrossthe battery and then acrosseach of the resistors,the readings are the same.As you can see, the same voltage appearsacross each branch in a parallel circuit.

ys= 12V (a) Pictorial b-/

acrossparallel branches is the same,

(b) Schematic

FIGURE6-B The samevoltageappearsacrosseach resistor in parallel.

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P A R A L L ECLI R C U I T S

EXAMPLE 6-3

Determine the voltage acrosseach resistor in Figure 6-9.

FICURE6-9

R5

3.3ko

Solution The flve resistors are in parallel, so the voltage drop across each one is equal to the applied source voltage. There is insignificant voltage drop acrossthe fuse. Vr = Vz = V3 = V+ = Vs = Vs = 25 Y Related Prohlem

SECTION6-2 REVIEW

If Ra is removed from the circuit, what is the voltage acrossR3?

1. A 10 O and a22 C) resistor are connectedin parallel with a 5 V source.What is the voltage acrosseach of the resistors?

) A voltmeter is connectedacrossR1 in Figure 6-10. It measures118 V. If you move the meter and connectit acrossR2.how much voltagewill it indicate?What is the sourcevoltage?

3. In Figure 6-1 1, how much voltage does voltmeter 1 indicate? Voltmeter 2? 4. How are voltages acrosseach branch of a parallel circuit related?

F I G U R E6 - 1 0

F I G U R E6 - 1 1

6-3 r KIRCHHOFF,S CURRENTtAW In the last chapter, you learned Kirchhoff\ voltage law that deult with voltages in a closed series circuit. Now, you will leurn Kirchhoff\ carrent luw that deals with currents in a parallel circuit. After completing this section, you should be able to r Apply Kirchhoff's current law . State Kirchhoff's current law . Determine the total current by adding the branch currents . Determine an unknown branch current

K I R C H H O F FC , SU R R E N T LAW .

167

Kirchhoff's current law, often abbreviatedKCL, is statedas follows: The sum of the currents into a junction (total current in) is equal to the sum of the currents out of that junction (total current out). A junction is any point in a circuit where two or more componentsare connected. So, in a parallel circuit, a junction is where the parallel branches come together. For example,in the circuit of Figure 6-12,point A is one junction and point B is another. Let's start at the positive terminal of the source and follow the current. The total current ft from the sourceis into the jtnction at point A. At this point, the current splits up among the three branchesas indicated. Each of the three branch currents (11, 12,and \) is out of junction A. Kirchhoff's current law says that the total current into junction A is equal to the total current out of junction A,' that is, 1 1 = 1 1 +1 2 +1 3 Now following the currents in Figure 6-12 through the three branches,you seethat they come back together at point B. Currents 11,12,and 13 are into junction B, and ft is out of junction B. Kirchhoff's current law formula at this iunction is therefore the same as at junction A. h = Ir-l Iz -f Iz

F I G U R E6 - 1 2 Total current into junction A equalsthe sum of currentsout of junction A, The sum of currentsinto junction B equalsthe total current out of junction B.

GeneralFormulafor Kirchhoff'sCurrentLaw The previous discussionused a specific example to illustrate Kirchhoff's current law. Figure 6-13 shows a generalizedcircuit junction where a number of branchesare connected to a point in the circuit. Currents !y11, through 4Nr1,)are into the junction (n can be any

RE6-13 circuit junction illustrating Kirchhoff \

,II N ( 1 )

law, /rN(z)

1rN(:)

I rIN(a)

4N(l) + 4N(2)+ 1rN(:)+'

+4N(r)=lour(r)+lour(z)+lour(:)+

lour@)

168 I

P A R A L L ECLI R C U I T S number). Currents lour(r) through lour(*) are out of the junction (m can be any number, but not necessarilyequal to n ). By Kirchhoff's current law, the sum of the currents into a junction must equal the sum of the currents out of the junction. With reference to Figure 6-13, the general formula for Kirchhoff's current law is 1 n u,1* I w r z r * ' ' ' *

/ n q r r=t I o u r , r r* l o u r r z +t . . . + l o u r i - ,

(6-1)

If all the terms on the right side of Equation (6-1) are brought over to the left side, their signs changeto negative, and a zero is left on the right side as follows: 4N(r)*1rN(z;* "'*ItN(,)

-Iour(r) - Ioure)

Iour@)=0

Kirchhoff's current law can also be statedin this way: The algebraic sum of all the currents entering and leaving a junction is equal to zero. You can verify Kirchhoff's current law by connecting a circuit and measuring each branch current and the total current from the source, as illustrated in Figure 6-14. When the branch currents are added together, their sum will equal the total current. This rule applies for any number of branches. The following three examplesillustrate use of Kirchhoff's current law.

FICURE6-14 Illustration of verifying Kirchhoff 's currenl Iaw.

EXAMPLE 6-4

The branch currents in the circuit of Figure 6-15 are known. Determine the total current entering junction A and the total current leaving junction B.

F I C U R E6 - 1 5

I2

12mA

Solution The total current out of junction A is the sum of the two branch cunents. So the total current into A is

I^r= Ir *12 = 5 mA + 12 mA= L7 mA

K I R C H H O F FC , SU R R E N T LAW .

169

The total cuffent entering junction B is the sum of the two branch currents. So the total current out of B is h = Ir * Iz= 5 mA + 12 mA = 17 mA t n e c a l c u t a t osr e q u e n ctes

@GB@@ @o@@m@ GFIEF:] Related Problem If a third branch is addedto the circuit in Figure 6-15, and its current is 3 mA, what is the total current into junction A and out of junction B?

Determine the current 12through R2 in Figure 6-16. F I G U R E6 - 1 6

I,,

R3

Solution The total current into the junction of the three branchesis 1.. = I, + I, + Ir. From Figure 6-16, you know the total current and the branch currents through R1 and R3. Solve for 12as follows: Iz= Ir - Ir - Iz= 100 mA - 30 mA - 20 mA = 50 mA Related Prohlem Determine 1t and 12 if a fourth branch is added to the circuit in Figure 6-16 and it has 12 mA through it.

EXAMPTE 6-6

Use Kirchhoff's current law to find the current measuredby ammetersAl and A2 in Figure 6-17.

F I G U R E6 - 1 7

Solution The total cur:rentinto junction X is 5 A. Two currents are out of junction X: 1.5 A through resistorR1 and the curent throughA1. Kirchhoff's current law applied at junction X gives 5A=1.5A+1ot

170 T

P A R A L L ECLI R C U I T S Solving for 1a1yields ler=5A-1.5A=3.5A The total current into junction Y is 1a1= 3.5 A. Two cuffents are out of junction I: I A through resistor R2 and the current through A2 and R3. Kirchhoff's current law applied at junction Igives 3.5A=lA+I^2 Solving for 1a2yields I n z = 3 . 5A - I A = 2 . 5 A Related Prohlem How much current will an arnmetermeasurewhen it is placed in the circuit right below R3? Below the negative battery terminal?

sEcTroN6-3

1. State Kirchhoff's cunent law in two ways.

REVIEW

2. There is a total current of 2.5 A into the junction of three parallel branches.What is the sum o[ all three branchcurrents? 3. ln Figure6-18. 100 mA and 300 mA are into the junction.What is the amountof curent out of the junction? 4. Determine/1 in the circuit of Figure6-19. 5. Two branch cuffents enter a junction, and two branch currents leave the samejunction. One of the currentsenteringthe junction is I A, and one of the currentsleaving the junction is 3 A. The total current entering and leaving the junction is 8 A. Determine the value of the unknown current entering the junction and the value of the unknown cuffent leavingthe junction.

F I G U R E6 - 1 8

FICURE6-19

6_4 . TOTATPARALLEL RESISTANCE When resistors are connected in parallel, the total resistsnce of the circuit decreases. The total resistance of a parallel circait is always less thun the value of the smallest resiston For example, if a 10 {l resistor and a 100 {L resistor sre connected in parallel, the total resistance is less than 10 Q. After completing this section, you should be uble to I Determine total parallel resistance . Explain why resistancedecreasesas resistors are connectedin parallel . Apply the parallel-resistanceformula

TOTALPARALLEL . 171 RESISTANCE

The Number of CurrentPathsAffectsTotalResistance As you know, when resistors are connected in parallel, the current has more than one path. The number of current paths is equal to the number of parallel branches. For example,in Figure 6-20(a), there is only one cutrent path since it is a seriescircuit. There is a certain amount of current, 11,through R1.If resistor R2 is connectedin parallel with R1, as shown in Figure 6-20(b), there is an additional amount of current, 12, through R2.The total current from the source has increasedwith the addition of the parallel resistor.Assuming that the sourcevoltage is constant,an increasein the total current from the source means that the total resistancehas decreased,in accordancewith Ohm's law. Additional resistors connected in parallel will further reduce the resistance and increasethe total current.

ys

Rl

(aJ

(b)

FIGURE6*20 Additinn of resistors in parallel reducestotel resistqnceand increasestotal current.

Formulafor TotalParallelResistance The circuit in Figure 6-21 shows a general case of n resistors in parallel (n can be any number). From Kirchhoff's current law, the current equation is Ir=It*12+Ir+"'lI,

Rl

R2

.l

t,

.l

F I C U R E6 - 2 1 Circuit with n resistors in parallel.

Since v5 is the voltage across each of the parallel resistors, by ohm's Iaw, I, = VslRb 12=VslRz,and so on. By substitutioninto the cur:rentequation,

5Rr= &R1+ 4 + v r * . . . + Y ' R2

R3

Rn

The term V5 can be factored out of the right side of the equation and canceledwith V5 on the left side, leaving only the resistanceterms.

-1 : - = - : 1- * - - * 1 Rr

Rl

R2

1 R3

1 R,

(6_2)

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P A R A L L ECLI R C U I T S Recall that the reciprocal of resistance(l/R) is called conductance and is symbolized by G. The unit of conductanceis the siemens(S). Equation (6-2) can be expressedin terms of conductanceas