Current, Resistance, and Electromotive Force

Σχετικά έγγραφα
[1] P Q. Fig. 3.1

Capacitors - Capacitance, Charge and Potential Difference

the total number of electrons passing through the lamp.

Potential Dividers. 46 minutes. 46 marks. Page 1 of 11

Electric Potential )الشرح واألفكار الرئيسية(

Equations of Motion. Dynamics. Week 3 )الشرح واألفكار الرئيسية( نوتات األمثلة والتمارين اإلضافية

Exercises in Electromagnetic Field

Engineering Economy. Week 12

ST5224: Advanced Statistical Theory II

Strain gauge and rosettes

2 Composition. Invertible Mappings

Surface Mount Multilayer Chip Capacitors for Commodity Solutions

EE101: Resonance in RLC circuits

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

Section 8.3 Trigonometric Equations

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

Calculating the propagation delay of coaxial cable

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

Homework 8 Model Solution Section

Areas and Lengths in Polar Coordinates

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

EE512: Error Control Coding

Mean bond enthalpy Standard enthalpy of formation Bond N H N N N N H O O O

PhysicsAndMathsTutor.com 1

Finite Field Problems: Solutions

Areas and Lengths in Polar Coordinates

Matrices and Determinants

Μηχανική Μάθηση Hypothesis Testing

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

Mock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =

(1) Describe the process by which mercury atoms become excited in a fluorescent tube (3)

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

Unshielded Power Inductor / PI Series

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

Technical Information T-9100 SI. Suva. refrigerants. Thermodynamic Properties of. Suva Refrigerant [R-410A (50/50)]

Εμπορική αλληλογραφία Παραγγελία

± 20% ± 5% ± 10% RENCO ELECTRONICS, INC.

Thi=Τ1. Thο=Τ2. Tci=Τ3. Tco=Τ4. Thm=Τ5. Tcm=Τ6

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

6.4 Superposition of Linear Plane Progressive Waves

UDZ Swirl diffuser. Product facts. Quick-selection. Swirl diffuser UDZ. Product code example:

DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C

No Item Code Description Series Reference (1) Meritek Series CRA Thick Film Chip Resistor AEC-Q200 Qualified Type

Srednicki Chapter 55

( )( ) ( ) ( )( ) ( )( ) β = Chapter 5 Exercise Problems EX α So 49 β 199 EX EX EX5.4 EX5.5. (a)

Figure 1 T / K Explain, in terms of molecules, why the first part of the graph in Figure 1 is a line that slopes up from the origin.

PHYA4/2. (JUN14PHYA4201) WMP/Jun14/PHYA4/2/E4. General Certificate of Education Advanced Level Examination June 2014

Written Examination. Antennas and Propagation (AA ) April 26, 2017.

The Simply Typed Lambda Calculus

DuPont Suva 95 Refrigerant

( ) Sine wave travelling to the right side

High Frequency Chip Inductor / CF TYPE

Study on Re-adhesion control by monitoring excessive angular momentum in electric railway traction

Lifting Entry (continued)

DuPont Suva 95 Refrigerant

Instruction Execution Times

6.003: Signals and Systems. Modulation

C.S. 430 Assignment 6, Sample Solutions

Data sheet Thick Film Chip Resistor 5% - RS Series 0201/0402/0603/0805/1206

1 String with massive end-points

Inverse trigonometric functions & General Solution of Trigonometric Equations

HOMEWORK#1. t E(x) = 1 λ = (b) Find the median lifetime of a randomly selected light bulb. Answer:

Volume of a Cuboid. Volume = length x breadth x height. V = l x b x h. The formula for the volume of a cuboid is

Prepolarized Microphones-Free Field

DuPont Suva. DuPont. Thermodynamic Properties of. Refrigerant (R-410A) Technical Information. refrigerants T-410A ENG

Homework 3 Solutions

Example Sheet 3 Solutions

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

1) Formulation of the Problem as a Linear Programming Model

RSDW08 & RDDW08 series

Terminal Contact UL Insulation Designation (provided with) style form system approval Flux tight

Numerical Analysis FMN011

65W PWM Output LED Driver. IDLV-65 series. File Name:IDLV-65-SPEC

Bulletin 1489 UL489 Circuit Breakers

ECE 308 SIGNALS AND SYSTEMS FALL 2017 Answers to selected problems on prior years examinations

TMA4115 Matematikk 3

Reminders: linear functions

Thermistor (NTC /PTC)

Second Order RLC Filters

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

wave energy Superposition of linear plane progressive waves Marine Hydrodynamics Lecture Oblique Plane Waves:

Technical Report. General Design Data of a Three Phase Induction Machine 90kW Squirrel Cage Rotor

MATSEC Intermediate Past Papers Index L. Bonello, A. Vella

AREAS AND LENGTHS IN POLAR COORDINATES. 25. Find the area inside the larger loop and outside the smaller loop

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

CMOS Technology for Computer Architects

Power Inductor LVS Series

Assalamu `alaikum wr. wb.

Chilisin Electronics Singapore Pte Ltd

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 24/3/2007

Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in

Answers to practice exercises

SCOPE OF ACCREDITATION TO ISO 17025:2005

Parametrized Surfaces

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

Metal thin film chip resistor networks

2. Chemical Thermodynamics and Energetics - I

Στο εστιατόριο «ToDokimasesPrinToBgaleisStonKosmo?» έξω από τους δακτυλίους του Κρόνου, οι παραγγελίες γίνονται ηλεκτρονικά.

Aluminum Electrolytic Capacitors

Transcript:

Week 5 Current, Resistance, and Electromotive Force )الشرح واألفكار الرئيسية( النوتات األساسية تتكون النوت من عشرة أسابيع. يحتوي كل نوت على شرح وحلول ألمثلة وتمارين من هوموركات وامتحانات سابقة. نوتات األمثلة والتمارين اإلضافية يتوفر على الموقع نوتات األسابيع لألمثلة والتمارين اإلضافية من واقع هوموركات وامتحانات سابقة. حقوق النوت لتعميم الفائدة فإني أتيح حقوق النسخ وإعادة اإلصدار وشروحات الفيديو لكل البشر سواء لبعض األجزاء أو الكل. واتساب )00965-94444260( يتم تنقحة النوت وتحديثها وإضافة الجديد إليها بشكل دوري. نتشوق النتقادك الهادف. الماضي ال يساوي المستقبل. لطلب نسخة ملونة مجانا 94444206 )1(

Current, Resistance & Electromotive Force مصطلحات وقوانين الشابتر: Charge: q (C) Current: I (A) I = q t Voltage: V (V) (also called potential) Current Density: J (A/m 2 ) V = I R n: Free electrons per cubic meter (electrons/m 3 ) (شحنة االلكترون الواحد) q: 1.6 10 19 C ρ: Material Resistivity (Ω. m) J = I/A J = n q v d v d : Dift speed (m/s) R: Resistance (Ω) ρ = A R L (Ω. m) L: Resistance length (m) R = R ο [1 + α (T T ο )] E: Electric Field (N/C or V/m) E = J ρ (V/m) ε: Electromotive force (V) ε = V + I r (internal resistance) أغلب المشكالت هي مجرد فرص لكنها ترتدي الزي الرسمي. )2(

I = Q t = 420 (80) (60) = 0.0875 C/s Or A J = n q v d 0.0875 π (0.0013) 2 = (5.8 1028 ) (1.6 10 19 ) v d v d = 1.78 10 6 m/s إذا كنت تجهل كل من عدوك ونفسك فتأكد أنك في خطر. )3(

Q = I t = (5.00) (1.00) = 5.00 C The number of electrons is Q e = 3.12 1019 c) J = = v d = = I (π 4) D 2 5.00 (π 4) (2.05 10 3 ) 2 = 1.51 106 A m 2 J n q (1.51 10 6 ) (8.5 10 28 ) (1.60 10 19 ) = 1.11 10 4 m s = 0.111 mm s d) If (I) is the same, (J = I /A) would decrease and (v d ) would decrease. The number of electrons passing through the light bulb in إذا أعطيت الفقير سمكة سددت جوعه ليوم أما إذا change. 1.00) (s would not علمته الصيد تكون قد سددت جوعه طوال العمر. )4(

I A = n q v d v d = 4.85 π (1.025 10 3 ) 2 (8.5 10 28 ) (1.60 10 19 ) = 108 10 6 m/s L = v d t t = 6,574 s = 110 min I A = n q v d v d = 26.75 10 6 m/s L = v d t t = 26,542 s = 442 min c) v d α 1 d 2 إذا أقبل الشتاء لم يعد الربيع بعيدا. )5(

I = 55 0.65 t 2 t Q = I dt 0 8.0 s = (55 0.65 t 2 ) dt 0 = [55 t 0.217 t 3 ] 0 8.0 = 329 C I = Q t = 329 8.0 = 41.1 C s = 41.1 A إذا كنت ترغب باستكشاف بحار جديدة يجب عليك أوال أن تتحلى بالشجاعة الالزمة لمغادرة الشاطئ. )6(

R = V I = 15.0 18.5 = 0.811 Ω ρ = A R L = π (0.25 10 2 ) 2 (0.811) 1.5 = 1.06 10 5 Ω. m R = R ο [1 + α (T T ο )] 15 (17.2) = (0.811) [1 + α (92 20 )] α = 1.05 10 3 (C ο ) 1 = 0.00105 (C ο ) 1 إن التفكير أصعب األعمال وهذا هو السبب في أن قليلين من يختارونه كعمل. )7(

c) E = ρ J = ρ 20 [1 + (T T 0 )] ( I π r 2) = (5.25 10 8 ) [1 + (0.0045) (120 20 12.5 )] ( π (0.0005) 2) = 1.21 V m R = ρ L A = ρ L π r 2 = (7.61 10 8 ) (0.150) π (0.0005) 2 V = E L Or = (1.21) (0.150) = 0.182 V V = I R = (12.5) (0.0145) = 0.181 V = 0.0145 Ω )8( إن المشكلة التي ال نعرفها ال يمكن تصحيحها كما المرض الذي ال يمكن تشخيصه ال يمكن عالجه.

R = ρ L A A = ρ L R Volume = A L = ρ L2 R = (1.72 10 8 ) (3.50) 2 0.125 = 1.686 10 6 m 3 m = (density) V = (8.9 10 6 ) (1.686 10 6 ) = 15 g إذا رأيت نيوب الليث بارزة فال تظن أن الليث يبتسم. )9(

E = J ρ = I A ρ I = E A ρ = (0.49) (π (0.42 10 3 ) 2 ) (2.44 10 8 ) = 11.1 A V = E L = (0.49) (6.4) = 3.136 V c) V = I R R = V I = 3.136 إذا طلب الصديق شيئا فال = 0.282 Ω 11.1 مجال لقول أنظرني غدا. )11(

= 0.0004 (C 0 ) 1 (Table 25.2) Let T 0 = 0 C and T = 11.5 C R = R 0 [1 + (T T 0 )] R 0 = = R 1 + (T T 0 ) 100.0 1 + (0.0004) (11.5 0 ) = 99.54 Ω = 0.0005 (C 0 ) 1 (Table 25.2) Let T 0 = 0 C and T = 25.8 C R = R 0 [1 + (T T 0 )] = (0.0160) [1 + ( 0.0005) (25.8 0 )] = 0.0158 Ω Nichrome, like most metallic conductors, has a positive (α) and its resistance increases with temperature. For carbon, (α) is negative and its resistance decreases as (T) increases. إذا فكرت أكثر فغالبا ستختصر مسافات كبيرة. )11(

E = 24.0 V I = 4.00 A V ab = E I r r = E V ab I (24.0) (21.2) = 4.00 = 0.700 Ω V ab I R = 0 R = V ab I = 21.2 4.00 = 5.30 Ω The voltage drop across the internal resistance of the battery causes the terminal voltage of the battery to be less than its (emf). The total resistance in the circuit is (R + r = 6.00 Ω) I = 24.0 6.00 = 4.00 A إذا فكرررت فرري أهررداف صررغيرة فتوقررع إنجررازات ضئيلة فكر في أهداف كبيرة وستفوز بنجاح هائل. )12(

E = J ρ = I A ρ = 125 π (0.05) 2 (1.72 10 8 ) = 2.74 10 4 V/m V = E L = (2.74 10 4 ) (100,000) = 27.4 V E = P t = V I t = (27.4) (125) (60 60) = 12.3 10 6 J = 12.3 MJ أجه ل الناس م ن ترك يقين ما عنده لظن ما عند الناس. )13(

Open switch: V ab = ε = 3.08 V Closed switch: 3.08 = V + I r = 2.97 + 1.65 r r = 0.067 Ω V = I R R = V I = 1.8 Ω )14( أخطر األضرار التي يمكن أن تصيب اإلنسان هو ظنه السيئ بنفسه.

There is a single current path so the current is the same at all points in the circuit. Assume the current is counterclockwise and apply Kirchhoff s loop rule. 16.0 I (1.6 + 5.0 + 1.4 + 9.0) 8.0 = 0 I = (16.0 8.0) 17.0 = 0.471 A Our calculated (I) is positive so (I) is counterclockwise, as we assumed. V b + 16.0 I (1.6) = V a V ab = 16.0 (0.471) (1.6) = 15.2 V Note: If we traveled around the circuit in the direction opposite to the current, the final answers would be the same. )15( ما تخاف منه سيقع إذا داومت على تفكيرك فيه.

P = V2 R 100 = (120)2 R R = 144 Ω c) P = V2 R 60 = (120)2 R R = 240 Ω I 100 W = P V = 100 120 = 0.833 A I 60 W = 60 أعط اهتمام أكثر لما تقوم به = 0.5 A 120 وسيخرج على نحو أفضل. )16(

Europe: P = V2 R 100 = (220)2 R R = 484 Ω United States: P = V2 R = (120)2 484 = 29.8 W I = P V = 29.8 إذا كنت ذا رأي فكن = 0.248 A 120 فيه مقداما فإن فساد الرأي أن تترددا. )17(

By definition p = )18( P, use P = V I, V = E L and I = J A L A (E) is related to (V) and (J) is related to (I), so use (P = V I) p = V I L A V L = E and I A = J So p = E J (J) is related to (I) and (ρ) is related to (R), so use (p = I 2 R) p = I2 R L A I = J A and R = ρ L A So p = J2 A 2 ρ L L A 2 = ρ J 2 c) (E) is related to (V) and (ρ) is related to (R), so use (p = V 2 R) V2 p = R L A V = E L and R = ρ L A So p = E2 L 2 L A ( A ρ L ) = E2 ρ For a given material (ρ constant), p is proportional to J 2 or to E 2 أغلب من يعملون أفضل ال يعملون ساعات أطول.

P = V I = (12) (60) = 720 W U = P t = (720) (1 60 60 ) = (720) (3,600 ) = 2.6 10 6 J For gasoline the heat of combustion is L c = 46 10 6 J kg ρ = m V m = (2.6 106 ) (46 10 6 ) = 0.0565 kg V = m ρ = 0.0565 900 = (6.3 10 5 ) ( 1,000 L ) = 0.063 L 1 c) Energy = (power) (time) U = P t t = U P = (2.6 106 ) 450 = 5,800 s = 97 min = 1.6 h The battery discharges at a rate of (720 W for 60 A) and is charged at a rate of (450 W), so it takes longer to charge than to discharge. إذا لم تتمتع بما لديك فلن تتمتع بما عند غيرك. )19(

V = I R 3.5 Ω 12 V I = V R P useful = I 2 R = 12 25 = 0.48 A = (0.48) 2 (25) = 5.76 W 25 Ω P loss = I 2 r = (0.48) 2 (3.5) = 0.8064 W loss% = 0.8064 (5.76 + 0.8064) = 12.3% )21( بداخل كل صعوبة تواجهها تكمن فرصة أو فائدة متساوية أو أكبر.

E I r I R = 0 I = E r + R = 12.0 1.0 + 5.0 = 2.00 A P = E I = (12.0) (2.00) = 24.0 W P = I 2 r = (2.00) 2 (1.0) = 4.0 W c) P = I 2 R = (2.00) 2 (5.0) = 20.0 W Note: The rate of production of electrical energy in the circuit is (24.0 W). The total rate of consumption of electrical energy in the circuit is (4.00 W + 20.0 W = 24.0 W). So both are equal. إذا لم تقفز فلن تعرف أبدا معنى الطيران. )21(

P = V2 R R = V2 P = (120)2 540 = 26.7 Ω P = V I I = P V = 540 120 = 4.5 A c) Assuming that (R = 26.7 Ω) d) P new = V2 R = (110)2 26.7 = 453 W P = V2 R P α 1 R (larger) بداية النجاح أن ينبع من إمكاناتك أنت دون التطلع إلمكانات وممتلكات اآلخرين. )22(

ρ = A R L = ((π/4) (0.0025)2 ) (0.104) 14.0 = 36.5 10 9 Ω. m V = E L = I R c) I = E L R = (1.28) (14) 0.104 = 172.3 A J = n q v d J = E ρ = 1.28 (3.65 10 8 ) = 3.51 107 A m 2 v d = = J n q (3.51 10 7 ) (8.5 10 28 ) (1.602 10 19 ) = 2.58 mm/s تذكر أن الشمس حين تغرب تأخذ جزء منك كل يوم إلى غير رجعة. )23(

I 1 = I 2 = 2.5 ma I = 2.5 ma Copper c) d) E = J ρ = ρ I A E 1 = (2.5 10 3 ) (π 4) (1.6 10 3 ) 2 (1.72 10 8 ) = 2.14 10 5 V/m (2.5 10 3 ) E 2 = (π 4) (0.8 10 3 ) 2 (1.72 10 8 ) = 8.55 10 5 V/m V = E L = E 1 (1.2) + E 2 (1.8) أحب دائما أن يقول لي الناس انك ال تستطيع أن تفعل ذلك = 0.18 mv ألنهم كلما يقولون لي ذلك أعمله بجدارة. )24( L1 = 1.2 m L 2 = 1.8 m V =?

I = E R + r 96.0 = (28.0 + 1.2) = 3.288 A P = I 2 R = (3.288) 2 (28.0) = 302.7 W The total heat needed is: Cup: Q = m c T = (0.130) (910) (34.5 21.2 ) = 1,573.39 J Water: Q = m c T = (0.200) (4,190) (34.5 21.2 ) = 11,145.4 J Total: Q = 12,718.79 J t = Q P = 12,718.79 302.7 = 42.02 s إذا كان هم الربان المحافظة على سالمه السفينة فسفينته لن تبرح الميناء. )25(

The battery with the (1.5 A) current is sketched in figure shown: V ab = 8.4 V V ab = E I r E (1.5) r = 8.4 The battery with the (3.5 A) current is sketched in figure shown: V ab = 9.4 V V ab = E + I r E + (3.5) r = 9.4 Solving, E = 8.4 + (1.5) r 8.4 + (1.5) r + (3.5) r = 9.4 r = 1.0 5.0 = 0.20 Ω E = 8.4 + (1.5) (0.20) = 8.7 V إذا ما أتيت األمر من غير بابه ضللت وإن تقصد إلى الباب تهتد. )26(

R = ρ L A = (5.0) (1.6) π (0.050) 2 = 1.0 103 Ω = 1 kω V = I R = (100 10 3 ) (1.0 10 3 ) = 100 V c) P = I 2 R = (100 10 3 ) 2 (1.0 10 3 ) = 10 W أساسيات السعادة في هذه الحياة ثالثة: شيء نفعله وشيء نحبه وشيء نأمل في حدوثه. )27(

E 1 E 2 I (r 1 + r 2 + R) = 0 I = (E 1 E 2 ) (r 1 + r 2 + R) = (12.0 8.0) (1.0 + 1.0 + 8.0) = 0.4 A P = I 2 R + I 2 r 1 + I 2 r 2 = I 2 (R + r 1 + r 2 ) = (0.4) 2 (8.0 + 1.0 + 1.0) = 1.6 W إذا ما لم يكن لك حسن فهم أسأت إجابة وأسأت سمعا. )28(

Continued Solution: c) p 12.0 V = E 1 I = (12.0) (0.4) = 4.8 W d) p 8.0 V = E 2 I = (8.0) (0.4) = 3.2 W e) Total rate of production of electrical energy = 4.8 W. Total rate of consumption of electrical energy = 1.6 + 3.2 = 4.8 W, which equals the rate of production, as it must. إما أن تتعب لتحصل على ما تحب أو ستجبر نفسك على حب ما تحصل عليه. )29(

No current flows through the capacitor when it is fully charged. With the capacitor fully charged. I = E R 1 + R 2 V C = Q C = (36.0 10 6 ) (9.00 10 6 ) = 4.00 V V R1 = V C = 4.00 V I = V R 1 R 1 = 4.00 6.00 = 0.667 A V R2 = I R 2 = (0.667) (4.00) = 2.668 V E = V R1 + V R2 = 4.00 + 2.668 = 6.67 V إن المستحيل كلمة غير فرنسية. )31(