Chapter 3: Resistive Circuits

Transcription

1 Chapter 3: esistive Circuits KICHHOFF S LAWS Kirchhoff s laws are the goverig laws of electric circuits. All techiques are derived from Kirchhoff s laws (VD, CD, Node ad Mesh equatios) ad Ohm s law. I collectively called these Simple Circuit Techiques (SCT). This chapter is o idetifyig series ad parallel o steroids. Defiitio A ode (equipotetial) is a juctio of 2 or more elemets. Whe a circuit is limited to the absolute miimum of coectios, the oly odes would be formed at juctios. For eample, how may odes are there i each circuit below? Simple words of advice: util you become i epert circuit solver (which you will ad fully epect), I strogly recommed that you use multiple colors to idetify odes. Kirchhoff s Curret Law (KCL): the algebraic sum of currets eterig ay ode is zero. Stated aother way is the currets eterig is equal to the currets comig out of the ode. N i 0 i i i out Sig covetio: currets ito a ode (+) while currets out of a ode ( ) As a eample, apply KCL to fid currets i 3 ad i 4. Applyig KCL to each of the 3 odes: Node : i i 0 i i Node y: i i i 0 Node z: i i 2 i 3 i i i i i i 0 i i5 i i2 7 ( ) 5 3A i4 Defiitio A loop is a closed path formed withi a circuit. Loops are closed potetial paths so that the work doe aroud a closed loop by a coservative force is always zero. How may possible loops are there withi this circuit? There are 6 loops but oly 3 idepedet oes.

2 Kirchhoff s Voltage Law (KVL): the sum of voltage drops aroud ay closed loop is zero. N v 0 closed loop To determie the sig of the voltage, oe should use the assive sig covetio. However, it turs out that this sig covetio is arbitrary. I am goig to suggest that you use the followig covetio, which is differet from the book covetio. I fid that this is much easier overall. Eample 3. Determie the power absorbed by each resistor. Always look for elemets that are i series or parallel with voltage ad curret sources. Immediately we see that the 4resistor is i series with the 2A-source ad the 8-resistor is parallel with the 8V source. So we kow that i 2A i 2 4 6W v 8 8 v 8V 8W As for the 6-resistor, if I apply a KVL loop (eed to assume the sig of the 6 voltage drop): 2 2 v6 2 v v6 2V 6 24W SEIES & AALLEL EQUIVALENTS The circuits we ow ecouter are much more comple, ad util you lear to see these, at times, you will fid that they ca be overwhelmig. This chapter will require that you see series ad parallel at a higher level ad to do this, you will eed to lear how to see circuits as subcircuits or the series or parallel equivalets. Aalogy: it s like solvig algebra problems for the first time. This is oe of the most importat skills you will lear i this course. That is, beig able to look at a circuit ad fid the easiest way to reduce it to maageable parts so that it is solvable with SCT. Aother aalogy comes to mid oe does ot just eat a whole sadwich i oe bite, you break it dow by bitig smaller pieces so that you ca chew ad shallow. I comple circuits, we will isolate a portio of a circuit (subcircuit) ito a series or parallel equivalet. Ad the repeat the process to reveal its weakess to gettig the desired curret or voltage. Warig: your ability to see series ad parallel is of upmost importace ad you should sped a lot of time learig "how to see." Whe the course is most challegig i Chapter 9 (derivig 2d order equatios) you will do so usig these techiques

3 etesively. If you do ot see series ad parallel by the, you should drop the course because it will be too late to pass the secod ad fial eam. Do I have to covice you ay more tha this? Here is the mai approach to this chapter: is the primary behavior series or parallel? As soo as you idetify the primary behavior of the circuit, metally you will have a procedure that will take complicated circuits ad solve them very quickly (aalogy: solvig algebra equatios). However, after almost two decades of teachig circuits it is very clear ow that without recogizig such behaviors, you will oly look at a circuit with blided eyes ad o solutio will come to you o matter how much time you sped o it. You MUST lear to see series ad parallel to move to the et level. Series Equivalets As soo as oe idetifies the primary behavior of the circuit to be SEIES the these are the characteristics of the brach i questio:. Series esistace add: Series = Same curret through series elemets: i = i 2 = = i Brach 3. Voltages are divided amog series elemets accordig to (i) KVL: v Brach = v + + v (ii) VD: v = ( / Series ) v Brach 4. Series Voltage sources added ad all series curret sources must be the same. Elemets are i series because each adjacet elemet has oly oe commo ode. Sice charge does ot collect at ay poit i a wire carryig a steady curret, the series resistors must carry the same curret i. Usig KVL, v v v v i i i ( ) i s What the battery sees is a effective resistace of Series = Geeralizig this to resistors, Series The equivalet series subcircuit ca be draw as From the series equivalet circuit, KVL ad Ohm s law, we ca derive VD (oe of the most useful equatios for solvig for voltages): v v v v 2 S VD i v vbrach (VD) 2 eq Series same curret through series elemets VD shows how the voltages are divided up i direct proportio to their resistace: percet of voltage drop larger larger voltage drop across series resistors smaller smaller voltage drop eq Voltage Sources i Series There are N voltage sources i a series brach, applyig KVL aroud the closed loop to determie v oc, we get v v v v v 0 OC 2 3 N OC sum of sources that sum of sources that absorb rg (passive) deliver rg (active) v sum of active sum of passive v et 3.3

4 That is, series voltage sources act as a sigle equivalet voltage source. For eample, a brach of a circuit has series characteristics, which meas that we ca reduce that brach dow to a sigle resistace ad voltage source as arallel Equivalets As soo as oe idetifies the primary behavior of the circuit to be AALLEL the these are the characteristics of the braches i questio:. arallel esistace add as iverses: / arallel = / + / 2 + Trick:,,, Same voltages across parallel elemets: v = v 2 = = v Brach 3. Currets are divided accordig to (i) KCL ad (ii) CD i. KCL: i Brach = i + + i ii. CD: i = ( arallel / ) i Brach For two resistors: i = 2 /( + 2 ) i Brach ad i 2 = /( + 2 ) i Brach 4. arallel curret sources added ad all parallel voltage sources must be the same. Elemets are i parallel because each elemet has two commo odes. Sice each ode represets a equipotetial surface, all parallel elemets betwee these two odes have the same voltage. Usig KCL, i is i i2 i3 2 v 3 S vs What the battery sees is a effective resistace value of parallel p 2 3 or The equivalet series subcircuit ca be draw as S 2 3 For two parallel resistors, For purposes of checkig results, Summary: is less tha the smallest resistor value Whe we add resistors i series, we icrease Whe we add resistors i parallel, we decrease Whe dealig with sums, we are able to iterchage resistors at will to make our calculatios easier. For eample, resistors ca be reduced quickly by movig resistors aroud to make the calculatios easier: eq eq 3.4

5 From the fact that all parallel elemets have the same voltage ad Ohm s law, we ca solve for the idividual currets i ad i 2 : v v S S vs i i i ad i i S S 2 S 2 2 So the idividual currets obey the Curret Divider ule (CD): i Similar to VD, CD tells us the percetage of the total curret which will flow through ad 2 is give by /. A very hady relatio to remember is that for two resistors i parallel: 2 The currets flowig through resistor & 2 are i S 2 2 i i i i ( ) eq 2 2 S S S 2 2 % of curret flowig through i i i i ( ) eq 2 2 S S S % of curret flowig through 2 If 2, the curret i 2 i. The curret follows the path of least resistace. Eample arallel circuits ted to cause most of the problems is SCT. Here are some tricks. Determie ad the curret through each resistor. Wheever there are more tha two resistors i parallel, here is a importat short cut for dealig with these situatios. if oe resistor is half as larger as aother the Eample: So the equivalet resistace is oe-third the value of the largest resistor value. if oe resistor is oe-third as larger as aother the Eample: So the equivalet resistace is oe-fourth the value of the largest resistor value. if oe resistor is oe-fourth as larger as aother the 4 3.5

6 Eample: So the equivalet resistace is oe-fifth the value of the largest resistor value. Usig these ideas, look at how quick I ca derive the parallel resistace for the circuit. I demad that you be able to reduced parallel resistors this quickly i your head, of course, after practicig it a lot. Curret Sources i arallel There are N curret sources i parallel braches, applyig KCL at oe of the odes, we get Ieq I I I I 2 3 N sum of sources ito the ode sum of sources out of the ode That is, parallel curret sources act as a sigle equivalet curret source. Aside: voltage sources caot be i parallel ad have differet voltage values There are may ways to determie the currets of the three resistors. (i) hysically approach is to realize that we start off with two 4 resistors that equally split the curret: each gets 3 A. The left-haded 4 resistor is a 6 ad 2 resistors splittig this 3A. Sice the 2W is twice as large, it gets half the curret so I ca immediately write i 6 = 2A ad i 2 = A. (ii) Ca use CD: 2 6 i 3A 2A i ad i 3A A i KCL : i i i i 2 3 6A S Curret Sources i Series/arallel Cosider a series of N curret sources. KCL tells us that the curret i is equal to the curret out. The curret sources i ad i 2 at the ceter ode must be the same, otherwise it violates KCL. Therefore, series curret sources must have the same curret value, otherwise, the curret sources "discharge." However, curret sources add i parallel Summary ALICATIONS OF SIMLE CICUIT TECHNIQUES (SCT) We will ow start to look at circuits that have both series ad parallel behaviors. However, the questio still stads is the primary behavior series or parallel? There are a zillio ways to solve circuits ad o oe way is more correct tha the other. Noetheless, differet solvig processes take loger tha others ad this may lead to trouble o quizzes ad eams. Therefore, I will try to teach faster techiques that immediately lead to the heart of the problem. 3.6

7 Eample 3.2 Fid i ad ab for the circuit. How may odes are there? 5! Use color pes to help you idetify them i the begiig. Note that the curret i is opposite of the voltage source, so keep i mid that there will be a egative sig to the curret. There are two ways of solvig this problem; i terms of series or parallel thikig. a. edrawig the circuit immediately leads to eq 242 eq b. Solvig this circuit both ways, we arrive at the followig. Series Thikig. Usig VD, the voltage across the equivalet 4 resistor is 4 Ohm's law 6.67 v 40V 6.7 V i 0.84A across The two parallel resistors have the same voltage of 6.7V. However, each brach has two series resistors (3ad5) ad therefore, have the same curret. Usig Ohm's law to determie this curret across the 8, we get 6.67 i i i 0.84A arallel Thikig. To use CD, we eed to determie the source curret. From the equivalet resistace ad Ohm's law, we get S i v 40 5 A S ab 8 Solvig for the curret i usig CD: i A 3 A However, the currets are divided across both 8 resistors, so the curret through the 5 resistor is i 8 i A 6 A i Eample 3.3 Use Simple Circuit Methods (KVL, KCL, CD, VD, ad Ohm s law) oly to determie i ad v whe v ab = 2V. emember that odes are equipotetial surfaces ad therefore, oly potetial differeces (differece i ode voltages) produces a curret. 3.7

8 How may odes are there? 3! I typically look at a circuit ad immediately I start to fid series ad parallel equivalets to simplify the circuits as much as possible. I see two sets of parallel resistors that I ca reduce to oe. First of all, ote that the top ode has o potetial differece across resistors 6 ad 30. Therefore, the short kills-off the 6//30 resistors sice there is o curret flowig through those resistors. We say that the 6//30 resistors are shorted out ad removed from the circuit sice they are oly there for decoratios. This short makes it easier to fid the equivalet resistace ab for the circuit: 36 ( ) eq The curret i is through the 8-resistor, so usig Ohm's law we get vs 2 i A i 8 To solve this voltage v is to realize that the 36 ad the right brach are i parallel, therefore, they each have the same voltage of 2V. This immediately tells us that we ca drop the 36 resistor from the circuit. Usig series techiques to get voltage v, ote (i) 9 is /8 of 72, so 72/9 = 8 ad (ii) apply VD across resistor v = v 8 : v vs 2V V v Eample 3.4 Fid I 0 i the circuit. eq How may odes are there? 4! Which resistors are i parallel or series? The 6k ad 2k are i parallel betwee odes A ad B, so replace 6k 2k 4k The 4k ad 2k are i parallel betwee odes B ad C, so replace 4k 2k 3k However, ote that if I do combie the 4kΩ 2kΩ = 3kΩ, I immediately loose the iformatio o I

9 Oce agai, there are multiple ways to solve for the curret I 0. As soo as I combied the resistors 4kΩ 2kΩ = 3kΩ, I am forced to focus o the voltage v = v 3kΩ. I have the tedecy to use VD sice there are still o currets i the circuit (although I could calculate the curret easy eough). To use VD we eed to setup the circuit to be i series resistors. The first step the is to idetify this combiatio; combied (9kΩ + 3kΩ) 4kΩ = 3kΩ. Applyig VD to these two resistors i series gives 3k vab 2V 6V 3k 3k We ow epad the resistor ito the parallel resistors ad reapply VD: 3k 3k v3k v 3 CB vab 6V 2 V v4k v 2k 3k 9k 3k 9k Sice the 4kΩ ad2kω are parallel to each other, they have the same voltages. We apply Ohm s law ad get the curret through the 4kΩ resistor: 3 v4k 2 V 3 I ma I Eample 3.5 Fid i 0 i the circuit k 4k How may odes are there i the circuit? 3! Therefore, these may parallel elemets will go through a massive reductio to oly a few resistors. edrawig the circuit will especially be key here i solvig for the curret i 0. Nodes & 2: the four left 2-resistors ad the right most 3-resistor all share the same two odes they are all i parallel. Get their equivalet resistaces Nodes & 3: there are 3 parallel elemets: the 6mA,2, 6. Oe should ot reduce the 2 ad 6resistors because you lose the details about the 6 curret i 0. Nodes 2 & 3: there is oly oe resistor i series 3. edrawig the circuit results i 3.9

10 As you ca see this circuit is trivial, oce you've see it. So ow we have a circuit that has 3 resistors i parallel ad should use CD to solve for the curret i 0 through the 6 resistor. educig the 4.5 ad 2 resistors ito oe resistace, CD gives us i 0 : CD i0 6mA 2. A i ractical Applicatio: Not the Earth Groud from Geology (see page 65) Up to ow, we have bee drawig circuit schematics i a fashio Depedet Sources i Simple Circuits Whe oe ecouters a depedet source i a circuit, a good rule of thumb is to thik i steps util you see how to solve depedet sources:. Apply simple circuit ideas, treatig the depedet source like ay other source. 2. Apply the DEENDENT SOUCE CONDITION (DSC or Dep Coditio): i. Fid the cotrollig elemet i the circuit ad see how it is defied. The most commo way the cotrollig elemet is defied is the curret through or voltage across a resistor. However, the cotrollig variable could be defied as the curret at a ode or a voltage aroud a loop. ii. Determie a additioal equatio that defies the cotrollig variable (curret/voltage). If cotrollig variable is the voltage or curret through a resistor, use Ohm s law: v c = i c. If the cotrollig variable is defied by a curret at a ode, apply KCL or CD. If defied by a voltage aroud some loop, apply KVL or VD. 3. Warig: do ot place the cotrollig elemet ito the equivalet resistaces for the circuit. Why? All the iformatio about the cotrollig variable is lost. Eample 3.6 Fid v ad the power delivered to the 0 resistor. The depedet source is a VCVS. We are iterested i the voltage v across the 5 resistor ad the power 0. Note that the depedet source is ot defied i terms of a ope circuit or a short circuit it is defied by a resistor. Step : Simple Circuit Methods Sice this is a series circuit, reduce all resistors (ecept the cotrollig resistor), ad voltage sources ito equivalets. However, because the cotrollig voltage is across the 5W-resistor, that resistor is left utouched. The equivalets are S ad vet v 27 3v The result is Applyig KVL aroud the loop leads us to 3v 27 25i v 0 4v 25i 27 reorgaizig the equatio This equatio is usolvable because it ivolves oe equatio with two ukows. The secod step i this process is applyig the DC. 3.0

11 Step 2: Depedet Coditio (DC) Go to where the cotrollig voltage is defied ad use Ohm s law to get the secod equatio: v 5i v 5i 0 Now we have the two equatios ad two ukows: 25i 4v 27 multiply lower eq by 5 ad add 5i 5v 0 9v 27 v 3V b. Usig v to fid the curret i ad power i the circuit, we get Eample 3.7 Determie i 3Ω. v 2 ower 2 i 3 3 5A 0 i W 0 5 v 3V Sice this is parallel, I immediately reduce resistors ad sources accordigly. However, I eed to be careful ot to iclude the 2 resistor sice it is the cotrollig curret (I would lose all of the iformatio o the cotrollig curret i ). The oly parallel equivalets are the curret sources: i arallel = 3i = 3i 4. The ew equivalet circuit is Step. Treat the circuit usig SCT. That is, how do we solve for ay curret whe there is a curret source i the circuit with parallel resistors? We use CD! So usig CD to calculate i 3Ω, we get 2 2 i 3i 4 3i 4 i variables ad equatio (ot solvable) Step 2. The DC tells me that I have to look at the circuit ad relate the curret i 3Ω with i by ay meas possible. I ca relate these two usig KCL o the top ode: KCL: 3i 4 i i i 2i secod equatio that relates i3ad i Now I take these two equatios ad solve for i 3Ω : 5 i 2 3 3i 4 solve for i i 3 2A 3 i3 2i 4 Wrap up of this chapter Clearly, there are may ways to solve circuits. A key to solvig circuits is to realize that each circuit has its ow characteristics ad oe has to lear how to idetify those characteristics. If a portio of the circuit is i series, the thik alog those ideas combie series voltage sources, series resistaces, ad apply KVL & VD. If a portio of the circuit is i parallel the thik alog those ideas combie parallel curret sources, parallel resistaces, ad apply KCL & CD. If the circuit is ot simple, the use the two-step process i solvig depedet source problems. 3.