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

Save this PDF as:
Size: px
Start display at page:

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

## Transcription

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. 2-2), current flows into point nd then out. n Fig. 2-2, the entire current ( ) flows out into 2. We cn write KCL for Fig. 2-2 s follows: = 2 n Fig. 2-2, the current splits with some trveling down one wire nd some trveling down the other wire. We cn write KCL for Fig. 2-2 s follows: = 2 3 n Fig. 2-2c, 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. 2-2c s follows: 2 = 3 4 Fig. 2-2 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. 2-3 elow. Applying KCL t point, write: n Fig. 2-3, 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. 2-3c 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. 2-3 nd. Fig. 2-3 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. 2-5 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. 2-6 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. 2-7 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. 2-6 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. 2-8: Strting t nd moving round the loop clockwise gives the following: 3 4 = 0 Solving for the unknown, = 7 Fig. 2-8 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. 2-8: Agin, using KL: giving: 3 2 = 0 = 9 5

6 Another exmple: Find the unknown vlues of nd 2 in Fig. 2-8c: 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 d-e-f-c--d 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 (d-e-f-c--d). 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. 2-9, 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. 2-9 c R R 2 R R 2 2 k k 50 m (d) Fig. 2-9c dds vlues for R, R 2 nd to llow us to write: 8 = or 2 = 4 Fig. 2-9d 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. 2-0 shows two resistors in series etween nd. The equivlent resistnce is shown in Fig. 2-0 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. 2-0d, the eqution for is: = R for the series circuit: Req = R R2 The current in the equivlent circuit of Fig. 2-0d equls the current in Fig. 2-0c. This is n exmple of n equivlent resistnce giving the sme results s the two seprte resistnces of Fig. 2-0c. 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 R-totl 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. 3-2: 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. 2-2 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. 2-2 R T

10 An exmple of finding R T from Series Resistors: Find the totl or equivlent resistnce R T in Fig. 2-3 nd. Just dd the two resistnces to comine into single resistor. 5 k 3 k ==> 8 k 2 k 4 k ==> 6 k Fig. 2-3 (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. 2-3(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. 2-3 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. 2-4 elow s: = = 2 or - = 0 nd = Around the right loop of Fig. 2-4 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. 3-4 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. 2-5 elow. f we know the current through R, we cn find 2. n Fig. 2-5, through R equls ma: 2 = R 2 R = or 2 = A = 2 ma 00 ma ma k k 50 Fig

13 Fig. 2-5 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 prllel-circuit 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. 2-6 hs two 5-kΩ resistnces rrnged in prllel. We clculte R eq using the product/sum rule: Fig. 2-6 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. 2-6 re not equl. Agin, we use the product/sum rule to clculte Req: Fig. 2-6 cont R eq = 3000x = 8,000,000 = 2,000 Ω 9,000 3 k 6 k ==> 2 k 4

15 Agin with Fig. 2-6c: R eq = 00x = 30, = 75 Ω Fig. 2-6 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 3-kΩ nd 6-kΩ resistors into n equivlent 2-kΩ resistor. We redrw the circuit in 2-7 nd find the equivlent resistnce R T in 2-7c. The vlue is -kω. See Fig. 2-7c. Fig k 3k 6k ==> 2k 2k ==> k 5

16 The one-step 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 cse-y-cse sis. Finding the Current in Prllel Circuits: Clculte the current in Fig. 2-8: Fig k 2 6k ==> 2 2k 2 4 ma 3k 2 ma 6k n Fig. 2-8 find nd 2. Next, use KCL to dd the two currents: = = 4 ma nd 2 = = 2 ma The vlue of T in Fig. 2-8 is the sme in Fig. 2-8 nd c. t is T = 4 ma 2 ma = 6 ma. Checking the rtio of currents s inverse of the resistnce: From Fig. 2-8: 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. 2-9, find R eq: R eq 0k 30k 50k Fig. 2-9 Usully, the one-step 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 2-03 will demonstrte finding R eq from series-prllel 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 product-over-sum 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. 2-2 elow, there is series-prllel 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 productover-sum rule or reciprocl rule my e used to find the finl R eq of 5 kω Fig. 2-2 R eq 6 k 4 k 60 k ==> R eq 20 k 60 k ==> R eq 5 k Some generl rules for reducing these series-prllel 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. 2-22, 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 tie-points in Fig. 2-23: 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

### Designing Information Devices and Systems I Discussion 8B

Lst Updted: 2018-10-17 19:40 1 EECS 16A Fll 2018 Designing Informtion Devices nd Systems I Discussion 8B 1. Why Bother With Thévenin Anywy? () Find Thévenin eqiuvlent for the circuit shown elow. 2kΩ 5V

### DIRECT CURRENT CIRCUITS

DRECT CURRENT CUTS ELECTRC POWER Consider the circuit shown in the Figure where bttery is connected to resistor R. A positive chrge dq will gin potentil energy s it moves from point to point b through

### Designing Information Devices and Systems I Spring 2018 Homework 8

EECS 16A Designing Informtion Devices nd Systems I Spring 2018 Homework 8 This homework is due Mrch 19, 2018, t 23:59. Self-grdes re due Mrch 22, 2018, t 23:59. Sumission Formt Your homework sumission

### 332:221 Principles of Electrical Engineering I Fall Hourly Exam 2 November 6, 2006

2:221 Principles of Electricl Engineering I Fll 2006 Nme of the student nd ID numer: Hourly Exm 2 Novemer 6, 2006 This is closed-ook closed-notes exm. Do ll your work on these sheets. If more spce is required,

### Chapter 4: Techniques of Circuit Analysis. Chapter 4: Techniques of Circuit Analysis

Chpter 4: Techniques of Circuit Anlysis Terminology Node-Voltge Method Introduction Dependent Sources Specil Cses Mesh-Current Method Introduction Dependent Sources Specil Cses Comprison of Methods Source

### Designing Information Devices and Systems I Spring 2018 Homework 7

EECS 16A Designing Informtion Devices nd Systems I Spring 2018 omework 7 This homework is due Mrch 12, 2018, t 23:59. Self-grdes re due Mrch 15, 2018, t 23:59. Sumission Formt Your homework sumission should

### 2.4 Linear Inequalities and Interval Notation

.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or

### Methods of Analysis and Selected Topics (dc)

Methods of Anlysis nd Selected Topics (dc) 8 Ojectives Become fmilir with the terminl chrcteristics of current source nd how to solve for the voltges nd currents of network using current sources nd/or

### Exam 2 Solutions ECE 221 Electric Circuits

Nme: PSU Student ID Numer: Exm 2 Solutions ECE 221 Electric Circuits Novemer 12, 2008 Dr. Jmes McNmes Keep your exm flt during the entire exm If you hve to leve the exm temporrily, close the exm nd leve

### Lecture 7 notes Nodal Analysis

Lecture 7 notes Nodl Anlysis Generl Network Anlysis In mny cses you hve multiple unknowns in circuit, sy the voltges cross multiple resistors. Network nlysis is systemtic wy to generte multiple equtions

### Physics 202, Lecture 10. Basic Circuit Components

Physics 202, Lecture 10 Tody s Topics DC Circuits (Chpter 26) Circuit components Kirchhoff s Rules RC Circuits Bsic Circuit Components Component del ttery, emf Resistor Relistic Bttery (del) wire Cpcitor

### Review of Gaussian Quadrature method

Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge

### Resistive Network Analysis

C H A P T E R 3 Resistive Network Anlysis his chpter will illustrte the fundmentl techniques for the nlysis of resistive circuits. The methods introduced re sed on the circuit lws presented in Chpter 2:

### Potential Changes Around a Circuit. You must be able to calculate potential changes around a closed loop.

Tody s gend: Potentil Chnges Around Circuit. You must e le to clculte potentil chnges round closed loop. Electromotive force (EMF), Terminl Voltge, nd Internl Resistnce. You must e le to incorporte ll

### Chapter E - Problems

Chpter E - Prolems Blinn College - Physics 2426 - Terry Honn Prolem E.1 A wire with dimeter d feeds current to cpcitor. The chrge on the cpcitor vries with time s QHtL = Q 0 sin w t. Wht re the current

### Hints for Exercise 1 on: Current and Resistance

Hints for Exercise 1 on: Current nd Resistnce Review the concepts of: electric current, conventionl current flow direction, current density, crrier drift velocity, crrier numer density, Ohm s lw, electric

### Mathematics Number: Logarithms

plce of mind F A C U L T Y O F E D U C A T I O N Deprtment of Curriculum nd Pedgogy Mthemtics Numer: Logrithms Science nd Mthemtics Eduction Reserch Group Supported y UBC Teching nd Lerning Enhncement

### How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=

### Fundamentals of Electrical Circuits - Chapter 3

Fundmentls of Electricl Circuits Chpter 3 1S. For the circuits shown elow, ) identify the resistors connected in prllel ) Simplify the circuit y replcing prllel connect resistors with equivlent resistor.

p-adic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Set-up 3 4 p-greedy Algorithm 5 5 p-egyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction

### Reading from Young & Freedman: For this topic, read the introduction to chapter 24 and sections 24.1 to 24.5.

PHY1 Electricity Topic 5 (Lectures 7 & 8) pcitors nd Dielectrics In this topic, we will cover: 1) pcitors nd pcitnce ) omintions of pcitors Series nd Prllel 3) The energy stored in cpcitor 4) Dielectrics

### Designing Information Devices and Systems I Fall 2016 Babak Ayazifar, Vladimir Stojanovic Homework 6. This homework is due October 11, 2016, at Noon.

EECS 16A Designing Informtion Devices nd Systems I Fll 2016 Bk Ayzifr, Vldimir Stojnovic Homework 6 This homework is due Octoer 11, 2016, t Noon. 1. Homework process nd study group Who else did you work

### Bridging the gap: GCSE AS Level

Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions

### Methods of Analysis and Selected Topics (dc)

8 N A Methods of Anlysis nd Selected Topics (dc) 8. INTRODUCTION The circuits descried in the previous chpters hd only one source or two or more sources in series or prllel present. The step-y-step procedure

### Designing finite automata II

Designing finite utomt II Prolem: Design DFA A such tht L(A) consists of ll strings of nd which re of length 3n, for n = 0, 1, 2, (1) Determine wht to rememer out the input string Assign stte to ech of

### Introduction to Algebra - Part 2

Alger Module A Introduction to Alger - Prt Copright This puliction The Northern Alert Institute of Technolog 00. All Rights Reserved. LAST REVISED Oct., 008 Introduction to Alger - Prt Sttement of Prerequisite

### Physics 1402: Lecture 7 Today s Agenda

1 Physics 1402: Lecture 7 Tody s gend nnouncements: Lectures posted on: www.phys.uconn.edu/~rcote/ HW ssignments, solutions etc. Homework #2: On Msterphysics tody: due Fridy Go to msteringphysics.com Ls:

### ELE B7 Power System Engineering. Unbalanced Fault Analysis

Power System Engineering Unblnced Fult Anlysis Anlysis of Unblnced Systems Except for the blnced three-phse fult, fults result in n unblnced system. The most common types of fults re single lineground

### 1B40 Practical Skills

B40 Prcticl Skills Comining uncertinties from severl quntities error propgtion We usully encounter situtions where the result of n experiment is given in terms of two (or more) quntities. We then need

Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite

### 1 From NFA to regular expression

Note 1: How to convert DFA/NFA to regulr expression Version: 1.0 S/EE 374, Fll 2017 Septemer 11, 2017 In this note, we show tht ny DFA cn e converted into regulr expression. Our construction would work

### September 13 Homework Solutions

College of Engineering nd Computer Science Mechnicl Engineering Deprtment Mechnicl Engineering 5A Seminr in Engineering Anlysis Fll Ticket: 5966 Instructor: Lrry Cretto Septemer Homework Solutions. Are

### List all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1.

Mth Anlysis CP WS 4.X- Section 4.-4.4 Review Complete ech question without the use of grphing clcultor.. Compre the mening of the words: roots, zeros nd fctors.. Determine whether - is root of 0. Show

### Lecture 2e Orthogonal Complement (pages )

Lecture 2e Orthogonl Complement (pges -) We hve now seen tht n orthonorml sis is nice wy to descrie suspce, ut knowing tht we wnt n orthonorml sis doesn t mke one fll into our lp. In theory, the process

### Lecture 2: January 27

CS 684: Algorithmic Gme Theory Spring 217 Lecturer: Év Trdos Lecture 2: Jnury 27 Scrie: Alert Julius Liu 2.1 Logistics Scrie notes must e sumitted within 24 hours of the corresponding lecture for full

### 5: The Definite Integral

5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce

### Parse trees, ambiguity, and Chomsky normal form

Prse trees, miguity, nd Chomsky norml form In this lecture we will discuss few importnt notions connected with contextfree grmmrs, including prse trees, miguity, nd specil form for context-free grmmrs

### Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions

QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.

### Torsion in Groups of Integral Triangles

Advnces in Pure Mthemtics, 01,, 116-10 http://dxdoiorg/1046/pm011015 Pulished Online Jnury 01 (http://wwwscirporg/journl/pm) Torsion in Groups of Integrl Tringles Will Murry Deprtment of Mthemtics nd Sttistics,

### Discrete Mathematics and Probability Theory Summer 2014 James Cook Note 17

CS 70 Discrete Mthemtics nd Proility Theory Summer 2014 Jmes Cook Note 17 I.I.D. Rndom Vriles Estimting the is of coin Question: We wnt to estimte the proportion p of Democrts in the US popultion, y tking

### Homework Solution - Set 5 Due: Friday 10/03/08

CE 96 Introduction to the Theory of Computtion ll 2008 Homework olution - et 5 Due: ridy 10/0/08 1. Textook, Pge 86, Exercise 1.21. () 1 2 Add new strt stte nd finl stte. Mke originl finl stte non-finl.

### Section 4: Integration ECO4112F 2011

Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic

### Lecture 08: Feb. 08, 2019

4CS4-6:Theory of Computtion(Closure on Reg. Lngs., regex to NDFA, DFA to regex) Prof. K.R. Chowdhry Lecture 08: Fe. 08, 2019 : Professor of CS Disclimer: These notes hve not een sujected to the usul scrutiny

### Regular expressions, Finite Automata, transition graphs are all the same!!

CSI 3104 /Winter 2011: Introduction to Forml Lnguges Chpter 7: Kleene s Theorem Chpter 7: Kleene s Theorem Regulr expressions, Finite Automt, trnsition grphs re ll the sme!! Dr. Neji Zgui CSI3104-W11 1

### Chapter 1: Logarithmic functions and indices

Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4

### Discrete Mathematics and Probability Theory Spring 2013 Anant Sahai Lecture 17

EECS 70 Discrete Mthemtics nd Proility Theory Spring 2013 Annt Shi Lecture 17 I.I.D. Rndom Vriles Estimting the is of coin Question: We wnt to estimte the proportion p of Democrts in the US popultion,

### Section 6: Area, Volume, and Average Value

Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find

### Things to Memorize: A Partial List. January 27, 2017

Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors - Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved

### Designing Information Devices and Systems I Anant Sahai, Ali Niknejad. This homework is due October 19, 2015, at Noon.

EECS 16A Designing Informtion Devices nd Systems I Fll 2015 Annt Shi, Ali Niknejd Homework 7 This homework is due Octoer 19, 2015, t Noon. 1. Circuits with cpcitors nd resistors () Find the voltges cross

### CS 373, Spring Solutions to Mock midterm 1 (Based on first midterm in CS 273, Fall 2008.)

CS 373, Spring 29. Solutions to Mock midterm (sed on first midterm in CS 273, Fll 28.) Prolem : Short nswer (8 points) The nswers to these prolems should e short nd not complicted. () If n NF M ccepts

### Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz

### Vectors , (0,0). 5. A vector is commonly denoted by putting an arrow above its symbol, as in the picture above. Here are some 3-dimensional vectors:

Vectors 1-23-2018 I ll look t vectors from n lgeric point of view nd geometric point of view. Algericlly, vector is n ordered list of (usully) rel numers. Here re some 2-dimensionl vectors: (2, 3), ( )

### 5. (±±) Λ = fw j w is string of even lengthg [ 00 = f11,00g 7. (11 [ 00)± Λ = fw j w egins with either 11 or 00g 8. (0 [ ffl)1 Λ = 01 Λ [ 1 Λ 9.

Regulr Expressions, Pumping Lemm, Right Liner Grmmrs Ling 106 Mrch 25, 2002 1 Regulr Expressions A regulr expression descries or genertes lnguge: it is kind of shorthnd for listing the memers of lnguge.

### Lecture Solution of a System of Linear Equation

ChE Lecture Notes, Dept. of Chemicl Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updted /) Lecture 8- - Solution of System of Liner Eqution 8. Why is it importnt to e le to solve system of liner

### 10. AREAS BETWEEN CURVES

. AREAS BETWEEN CURVES.. Ares etween curves So res ove the x-xis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in

### Fast Boolean Algebra

Fst Boolen Alger ELEC 267 notes with the overurden removed A fst wy to lern enough to get the prel done honorly Printed; 3//5 Slide Modified; Jnury 3, 25 John Knight Digitl Circuits p. Fst Boolen Alger

### 1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of the

### Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

### Jim Lambers MAT 169 Fall Semester Lecture 4 Notes

Jim Lmbers MAT 169 Fll Semester 2009-10 Lecture 4 Notes These notes correspond to Section 8.2 in the text. Series Wht is Series? An infinte series, usully referred to simply s series, is n sum of ll of

### Minimal DFA. minimal DFA for L starting from any other

Miniml DFA Among the mny DFAs ccepting the sme regulr lnguge L, there is exctly one (up to renming of sttes) which hs the smllest possile numer of sttes. Moreover, it is possile to otin tht miniml DFA

### The Regulated and Riemann Integrals

Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

### The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the

### Intermediate Math Circles Wednesday, November 14, 2018 Finite Automata II. Nickolas Rollick a b b. a b 4

Intermedite Mth Circles Wednesdy, Novemer 14, 2018 Finite Automt II Nickols Rollick nrollick@uwterloo.c Regulr Lnguges Lst time, we were introduced to the ide of DFA (deterministic finite utomton), one

### Lecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.

Lecture 3 3 Solving liner equtions In this lecture we will discuss lgorithms for solving systems of liner equtions Multiplictive identity Let us restrict ourselves to considering squre mtrices since one

### set is not closed under matrix [ multiplication, ] and does not form a group.

Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed

### POLYPHASE CIRCUITS. Introduction:

POLYPHASE CIRCUITS Introduction: Three-phse systems re commonly used in genertion, trnsmission nd distribution of electric power. Power in three-phse system is constnt rther thn pulsting nd three-phse

### Chapters Five Notes SN AA U1C5

Chpters Five Notes SN AA U1C5 Nme Period Section 5-: Fctoring Qudrtic Epressions When you took lger, you lerned tht the first thing involved in fctoring is to mke sure to fctor out ny numers or vriles

### 196 Circuit Analysis with PSpice: A Simplified Approach

196 Circuit Anlysis with PSpice: A Simplified Approch i, A v L t, min i SRC 5 μf v C FIGURE P7.3 () the energy stored in the inductor, nd (c) the instntneous power input to the inductor. (Dul of Prolem

### CS12N: The Coming Revolution in Computer Architecture Laboratory 2 Preparation

CS2N: The Coming Revolution in Computer Architecture Lortory 2 Preprtion Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes

### ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil

### 2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).

AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following

### Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

### Scientific notation is a way of expressing really big numbers or really small numbers.

Scientific Nottion (Stndrd form) Scientific nottion is wy of expressing relly big numbers or relly smll numbers. It is most often used in scientific clcultions where the nlysis must be very precise. Scientific

### 1 Nondeterministic Finite Automata

1 Nondeterministic Finite Automt Suppose in life, whenever you hd choice, you could try oth possiilities nd live your life. At the end, you would go ck nd choose the one tht worked out the est. Then you

### Math 8 Winter 2015 Applications of Integration

Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

### Special Relativity solved examples using an Electrical Analog Circuit

1-1-15 Specil Reltivity solved exmples using n Electricl Anlog Circuit Mourici Shchter mourici@gmil.com mourici@wll.co.il ISRAE, HOON 54-54855 Introduction In this pper, I develop simple nlog electricl

### Bases for Vector Spaces

Bses for Vector Spces 2-26-25 A set is independent if, roughly speking, there is no redundncy in the set: You cn t uild ny vector in the set s liner comintion of the others A set spns if you cn uild everything

### Review of Calculus, cont d

Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

### Homework Assignment 6 Solution Set

Homework Assignment 6 Solution Set PHYCS 440 Mrch, 004 Prolem (Griffiths 4.6 One wy to find the energy is to find the E nd D fields everywhere nd then integrte the energy density for those fields. We know

### MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

### f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

### CSE396 Prelim I Answer Key Spring 2017

Nme nd St.ID#: CSE96 Prelim I Answer Key Spring 2017 (1) (24 pts.) Define A to e the lnguge of strings x {, } such tht x either egins with or ends with, ut not oth. Design DFA M such tht L(M) = A. A node-rc

### Chapter 0. What is the Lebesgue integral about?

Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

### A, Electromagnetic Fields Final Exam December 14, 2001 Solution

304-351, Electrognetic Fiels Finl Ex Deceer 14, 2001 Solution 1. e9.8. In chpter9.proles.extr.two loops, e of thin wire crry equl n opposite currents s shown in the figure elow. The rius of ech loop is

### Network Theorems 9.1 INTRODUCTION 9.2 SUPERPOSITION THEOREM

9 Th Network Theorems 9.1 NTRODUCTON This chpter will introduce the importnt fundmentl theorems of network nlysis. ncluded re the superposition, Thévenin s, Norton s, mximum power trnsfer, sustitution,

### Section 3.1: Exponent Properties

Section.1: Exponent Properties Ojective: Simplify expressions using the properties of exponents. Prolems with exponents cn often e simplied using few sic exponent properties. Exponents represent repeted

### and that at t = 0 the object is at position 5. Find the position of the object at t = 2.

7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we

### Resistors. Consider a uniform cylinder of material with mediocre to poor to pathetic conductivity ( )

10/25/2005 Resistors.doc 1/7 Resistors Consider uniform cylinder of mteril with mediocre to poor to r. pthetic conductivity ( ) ˆ This cylinder is centered on the -xis, nd hs length. The surfce re of the

### State space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies

Stte spce systems nlysis (continued) Stbility A. Definitions A system is sid to be Asymptoticlly Stble (AS) when it stisfies ut () = 0, t > 0 lim xt () 0. t A system is AS if nd only if the impulse response

### Interpreting Integrals and the Fundamental Theorem

Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of

### Network Analysis and Synthesis. Chapter 5 Two port networks

Network Anlsis nd Snthesis hpter 5 Two port networks . ntroduction A one port network is completel specified when the voltge current reltionship t the terminls of the port is given. A generl two port on

### Section 1.3 Triangles

Se 1.3 Tringles 21 Setion 1.3 Tringles LELING TRINGLE The line segments tht form tringle re lled the sides of the tringle. Eh pir of sides forms n ngle, lled n interior ngle, nd eh tringle hs three interior

### Chapter 1: Fundamentals

Chpter 1: Fundmentls 1.1 Rel Numbers Types of Rel Numbers: Nturl Numbers: {1, 2, 3,...}; These re the counting numbers. Integers: {... 3, 2, 1, 0, 1, 2, 3,...}; These re ll the nturl numbers, their negtives,

### 9.4. The Vector Product. Introduction. Prerequisites. Learning Outcomes

The Vector Product 9.4 Introduction In this section we descrie how to find the vector product of two vectors. Like the sclr product its definition my seem strnge when first met ut the definition is chosen

### Chapter 6 Techniques of Integration

MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln

### I 3 2 = I I 4 = 2A

ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo 2.13 We re ske to use KCL to fin urrents 1 4. The key point in pplying KCL in this prolem is to strt with noe where only one of the urrents

### APPENDIX. Precalculus Review D.1. Real Numbers and the Real Number Line

APPENDIX D Preclculus Review APPENDIX D.1 Rel Numers n the Rel Numer Line Rel Numers n the Rel Numer Line Orer n Inequlities Asolute Vlue n Distnce Rel Numers n the Rel Numer Line Rel numers cn e represente