This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License.
|
|
|
- Erika Andrews
- 6 years ago
- Views:
Transcription
1 University of Rhode Island PHY 204: Elementary Physics II Physics Course Materials Resistors II Gerhard Müller University of Rhode Island, Creative Commons License This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License. Follow this and additional works at: Abstract Part ten of course materials for Elementary Physics II (PHY 204), taught by Gerhard Müller at the University of Rhode Island. Documents will be updated periodically as more entries become presentable. Some of the slides contain figures from the textbook, Paul A. Tipler and Gene Mosca. Physics for Scientists and Engineers, 5 th /6 th editions. The copyright to these figures is owned by W.H. Freeman. We acknowledge permission from W.H. Freeman to use them on this course web page. The textbook figures are not to be used or copied for any purpose outside this class without direct permission from W.H. Freeman. Recommended Citation Müller, Gerhard, "10. Resistors II" (2015). PHY 204: Elementary Physics II. Paper This Course Material is brought to you for free and open access by the Physics Course Materials at DigitalCommons@URI. It has been accepted for inclusion in PHY 204: Elementary Physics II by an authorized administrator of DigitalCommons@URI. For more information, please contact digitalcommons@etal.uri.edu.
2 M. C. Escher: Waterfall 18/9/2015 [tsl425 1/29]
3 Direct Current Circuit Consider a wire with resistance R = ρl/a connected to a battery. Resistor rule: In the direction of I across a resistor with resistance R, the electric potential drops: V = IR. EMF rule: From the ( ) terminal to the (+) terminal in an ideal source of emf, the potential rises: V = E. Loop rule: The algebraic sum of the changes in potential encountered in a complete traversal of any loop in a circuit must be zero: P V i = 0. physical system circuit diagram electric potential a b + emf I a ε b V a IR ε I R V b a b a 18/9/2015 [tsl143 2/29]
4 Battery with Internal Resistance Real batteries have an internal resistance r. The terminal voltage V ba V a V b is smaller than the emf E written on the label if a current flows through the battery. Usage of the battery increases its internal resistance. Current from loop rule: E Ir IR = 0 I = E R + r Current from terminal voltage: V ba = E Ir = IR I = V ba R physical system circuit diagram electric potential a b + emf a I r ε b V a IR ε Ir I R V b a b a. 18/9/2015 [tsl144 3/29]
5 Symbols Used in Cicuit Diagrams R resistor A ammeter (connect in series) C capacitor V voltmeter (connect in parallel) L inductor diode ε emf source E C transistor B 18/9/2015 [tsl158 4/29]
6 Resistor Circuit (4) Consider the resistor circuit shown. (a) Find the direction of the current I (cw/ccw). (b) Find the magnitude of the current I. (c) Find the voltage V ab = V b V a. (d) Find the voltage V cd = V d V c. d c 12V b a 18/9/2015 [tsl151 5/29]
7 Resistor Circuit (5) Consider the resistor circuit shown. (a) Find the direction (cw/ccw) of the current I in the loop. (b) Find the magnitude of the current I in the loop. (c) Find the potential difference V ab = V b V a. (d) Find the voltage V cd = V d V c. d a 12V 3Ω 7Ω 4Ω 4V b c 18/9/2015 [tsl152 6/29]
8 Resistor Circuit (6) Consider the resistor circuit shown. (a) Guess the current direction (cw/ccw). (b) Use the loop rule to determine the current. (c) Find V ab V b V a along two different paths. b 18V 6V a 12V 24V 18/9/2015 [tsl153 7/29]
9 Resistors Connected in Parallel Find the equivalent resistance of two resistors connected in parallel. Current through resistors: I 1 + I 2 = I Voltage across resistors: V 1 = V 2 = V Equivalent resistance: 1 R = 1 R R 2 1 R I V = I 1 V 1 + I 2 V 2 V 0 R 1 V 1 = V x I 1 V0 V 0 + V I 2 R 2 V 0 V = V 2 x 18/9/2015 [tsl145 8/29]
10 Resistors Connected in Series Find the equivalent resistance of two resistors connected in series. Current through resistors: I 1 = I 2 = I Voltage across resistors: V 1 + V 2 = V Equivalent resistance: R V I = V 1 I 1 + V 2 I 2 R = R 1 + R 2 V 2 V 0 V 1 x V 0 R 1 R 2 V + V 0 I I 18/9/2015 [tsl146 9/29]
11 Resistor Circuit (1) Consider the two resistor circuits shown. (a) Find the resistance R 1. (b) Find the emf E 1. (c) Find the resistance R 2. (d) Find the emf E 2. 2A 1A R 1 1A 2A 6Ω R 2 ε 1 ε 2 18/9/2015 [tsl148 10/29]
12 Resistor Circuit (2) Consider the two resistor circuits shown. (a) Find the resistance R 1. (b) Find the current I 1. (c) Find the resistance R 2. (d) Find the current I 2. R1 I 2 3Ω 3A I 1 6Ω R 2 3A 12V 12V 18/9/2015 [tsl149 11/29]
13 Resistor Circuit (3) Consider the rsistor and capacitor circuits shown. (a) Find the equivalent resistance R eq. (b) Find the power P 2, P 3, P 4 dissipated in each resistor. (c) Find the equivalent capacitance C eq. (d) Find the energy U 2, U 3, U 4 stored in each capacitor. 4Ω 2nF 4nF 3Ω 3nF 12V 12V 18/9/2015 [tsl150 12/29]
14 Power in Resistor Circuit Battery in use b Terminal voltage: V ab = E Ir = IR Power output of battery: P = V ab I = EI I 2 r Power generated in battery: EI Power dissipated in battery: I 2 r ε r R Power dissipated in resistor: P = I 2 R a Battery being charged: Terminal voltage: V ab = E + Ir Power supplied by charging device: P = V ab I Power input into battery: P = EI + I 2 r Power stored in battery: EI Power dissipated in battery: I 2 r b a ε r charging device 18/9/2015 [tsl154 13/29]
15 Resistor Circuit (7) Consider two 24V batteries with internal resistances (a) r = 4Ω, (b) r =. Which setting of the switch (L/R) produces the larger power dissipation in the resistor on the side? L R L R 4Ω 4Ω 24V 4Ω 24V (a) (b) 18/9/2015 [tsl155 14/29]
16 Impedance Matching A battery providing an emf E with internal resistance r is connected to an external resistor of resistance R as shown. For what value of R does the battery deliver the maximum power to the external resistor? Electric current: E Ir IR = 0 I = E R + r Power delivered to external resistor: P = I 2 R = Condition for maximum power: dp dr = 0 r ε R = r E2 R (R + r) 2 I R 18/9/2015 [tsl156 15/29]
17 Resistor Circuit (8) Consider the circuit of resistors shown. Find the equivalent resistance R eq. Find the currents I 1,..., I 5 through each resistor and the voltages V 1,..., V 5 across each resistor. Find the total power P dissipated in the circuit. R = 1 6Ω R = 1 2 R = 4Ω 3 R = 3Ω R = 4 5 5Ω ε = 12V 18/9/2015 [tsl157 16/29]
18 Kirchhoff s Rules Loop Rule When any closed-circuit loop is traversed, the algebraic sum of the changes in electric potential must be zero. Junction Rule At any junction in a circuit, the sum of the incoming currents must equal the sum of the outgoing currents. Strategy Use the junction rule to name all independent currents. Use the loop rule to determine the independent currents. 18/9/2015 [tsl159 17/29]
19 Applying the Junction Rule In the circuit of steady currents, use the junction rule to find the unknown currents I 1,..., I 6. 1A Ι 2 3A Ι 1 Ι 3 Ι 4 Ι 5 4A 9A Ι 6 + ε 18/9/2015 [tsl160 18/29]
20 Applying Kirchhoff s Rules Consider the circuit shown below. Junction a: I 1, I 2 (in); Junction b: I 1 + I 2 (in); I 1 + I 2 (out) I 1, I 2 (out) Two independent currents require the use of two loops. Loop A (ccw): 6V ()I 1 2V ()I 1 = 0 Loop B (ccw): (3Ω)I 2 + 1V + ()I 2 6V = 0 Solution: I 1 = 1A, I 2 = 1A I1 I 2 b I + I 1 2 2V A 6V B 1V a I + I 1 2 I 1 I 2 3Ω 18/9/2015 [tsl161 19/29]
21 Resistor Circuit (11) Consider the electric circuit shown. Identify all independent currents via junction rule. Determine the independent currents via loop rule. Find the Potential difference V ab = V b V a. a 4V 4V 2V b 18/9/2015 [tsl164 20/29]
22 Resistor Circuit (9) Use Kirchhoff s rules to find (a) the current I, (b) the resistance R, (c) the emf E, (d) the voltage V ab V b V a. b I 18V R 1A ε 6A a 18/9/2015 [tsl162 21/29]
23 Resistor Circuit (10) Consider the electric circuit shown. (a) Find the current through the 12V battery. (b) Find the current through the resistor. (c) Find the total power dissipated. (d) Find the voltage V cd V d V c. (e) Find the voltage V ab V b V a. c 4V a 12V d 3Ω b 18/9/2015 [tsl163 22/29]
24 Resistor Circuit (12) Consider the electric circuit shown. Find the equivalent resistance R eq of the circuit. Find the total power P dissipated in the circuit. 3Ω 13V 18/9/2015 [tsl165 23/29]
25 More Complex Capacitor Circuit No two capacitors are in parallel or in series. Solution requires different strategy: zero charge on each conductor (here color coded), zero voltage around any closed loop. Specifications: C 1,..., Q 5, V. Five equations for unknowns Q 1,..., Q 5 : Q 1 + Q 2 Q 4 Q 5 = 0 Q 3 + Q 4 Q 1 = 0 Q 5 C 5 + Q 3 C 3 Q 4 C 4 = 0 Q 2 C 2 Q 1 C 1 Q 3 C 3 = 0 V Q 1 C 1 Q 4 C 4 = 0 Equivalent capacitance: C eq = Q 1 + Q 2 V + + C 1 + C 4 C 2 C 3 V + + C 5 (a) C m = 1pF, m = 1,..., 5 and V = 1V: C eq = 1pF, Q 3 = 0, Q 1 = Q 2 = Q 4 = Q 5 = 1 2 pc. (b) C m = m pf, m = 1,..., 5 and V = 1V: C eq = pf, Q 1 = pc, Q 2 = pc, Q 3 = 9 71 pc, Q 4 = pc, Q 5 = pc. 18/9/2015 [tsl511 24/29]
26 Intermediate Exam II: Problem #2 (Spring 05) Consider the electrical circuit shown. (a) Find the equivalent resistance R eq. (b) Find the current I 3 through resistor R 3. R = 2 12V R = 3 3Ω R = 6 6Ω 18/9/2015 [tsl337 25/29]
27 Intermediate Exam II: Problem #2 (Spring 05) Consider the electrical circuit shown. (a) Find the equivalent resistance R eq. (b) Find the current I 3 through resistor R 3. R = 2 12V R = 3 3Ω R = 6 6Ω Solution: (a) R 36 = 1 R R 6 «1 =, R eq = R 2 + R 36 = 4Ω. 18/9/2015 [tsl337 25/29]
28 Intermediate Exam II: Problem #2 (Spring 05) Consider the electrical circuit shown. (a) Find the equivalent resistance R eq. (b) Find the current I 3 through resistor R 3. R = 2 12V R = 3 3Ω R = 6 6Ω Solution: (a) R 36 = 1 R R 6 (b) I 2 = I 36 = 12V R eq = 3A «1 =, R eq = R 2 + R 36 = 4Ω. V 3 = V 36 = I 36 R 36 = 6V I 3 = V 3 R 3 = 2A. 18/9/2015 [tsl337 25/29]
29 Intermediate Exam II: Problem #2 (Spring 06) Consider the two-loop circuit shown. (a) Find the current I 1. (b) Find the current I 2. 2V 10V I 1 3Ω I 2 18/9/2015 [tsl352 26/29]
30 Intermediate Exam II: Problem #2 (Spring 06) Consider the two-loop circuit shown. (a) Find the current I 1. (b) Find the current I 2. 2V 10V I 1 Solution: (a) ()(I 1 ) + 10V ()(I 1 ) 2V = 0 I 1 = 8V 4Ω = 2A. 3Ω I 2 18/9/2015 [tsl352 26/29]
31 Intermediate Exam II: Problem #2 (Spring 06) Consider the two-loop circuit shown. (a) Find the current I 1. (b) Find the current I 2. 2V 10V I 1 Solution: 3Ω I 2 (a) ()(I 1 ) + 10V ()(I 1 ) 2V = 0 (b) ()(I 2 ) + 10V ()(I 2 ) (3Ω)(I 2 ) = 0 I 1 = 8V 4Ω = 2A. I 2 = 10V 7Ω = 1.43A. 18/9/2015 [tsl352 26/29]
32 Unit Exam II: Problem #2 (Spring 07) Consider the electric circuit shown. (a) Find the current I when the switch S is open. (b) Find the power P 3 dissipated in resisistor R 3 when the switch is open. (c) Find the current I when the switch S is closed. (d) Find the power P 3 dissipated in resisistor R 3 when the switch is closed. R 1 = 4Ω R 2 = 4Ω I 24V S R 3 = 4Ω 18/9/2015 [tsl363 27/29]
33 Unit Exam II: Problem #2 (Spring 07) Consider the electric circuit shown. (a) Find the current I when the switch S is open. (b) Find the power P 3 dissipated in resisistor R 3 when the switch is open. (c) Find the current I when the switch S is closed. (d) Find the power P 3 dissipated in resisistor R 3 when the switch is closed. Solution: (a) I = 24V 8Ω = 3A. I R 1 = 4Ω R 2 = 4Ω 24V S R 3 = 4Ω 18/9/2015 [tsl363 27/29]
34 Unit Exam II: Problem #2 (Spring 07) Consider the electric circuit shown. (a) Find the current I when the switch S is open. (b) Find the power P 3 dissipated in resisistor R 3 when the switch is open. (c) Find the current I when the switch S is closed. (d) Find the power P 3 dissipated in resisistor R 3 when the switch is closed. Solution: (a) I = 24V 8Ω = 3A. (b) P 3 = (3A) 2 (4Ω) = 36W. I R 1 = 4Ω R 2 = 4Ω 24V S R 3 = 4Ω 18/9/2015 [tsl363 27/29]
35 Unit Exam II: Problem #2 (Spring 07) Consider the electric circuit shown. (a) Find the current I when the switch S is open. (b) Find the power P 3 dissipated in resisistor R 3 when the switch is open. (c) Find the current I when the switch S is closed. (d) Find the power P 3 dissipated in resisistor R 3 when the switch is closed. Solution: (a) I = 24V 8Ω = 3A. (b) P 3 = (3A) 2 (4Ω) = 36W. (c) I = 24V 6Ω = 4A. I R 1 = 4Ω R 2 = 4Ω 24V S R 3 = 4Ω 18/9/2015 [tsl363 27/29]
36 Unit Exam II: Problem #2 (Spring 07) Consider the electric circuit shown. (a) Find the current I when the switch S is open. (b) Find the power P 3 dissipated in resisistor R 3 when the switch is open. (c) Find the current I when the switch S is closed. (d) Find the power P 3 dissipated in resisistor R 3 when the switch is closed. Solution: (a) I = 24V 8Ω = 3A. (b) P 3 = (3A) 2 (4Ω) = 36W. (c) I = 24V 6Ω = 4A. (d) P 3 = (2A) 2 (4Ω) = 16W. I R 1 = 4Ω R 2 = 4Ω 24V S R 3 = 4Ω 18/9/2015 [tsl363 27/29]
37 Unit Exam II: Problem #2 (Spring 09) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq. (b) Find the power P supplied by the battery. (c) Find the current I 4 through the 4Ω-resistor. (d) Find the voltage V 2 across the -resistor. 3Ω 4Ω 3Ω 24V 18/9/2015 [tsl393 28/29]
38 Unit Exam II: Problem #2 (Spring 09) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq. (b) Find the power P supplied by the battery. (c) Find the current I 4 through the 4Ω-resistor. (d) Find the voltage V 2 across the -resistor. Solution: (a) R eq = 8Ω. 3Ω 4Ω 3Ω 24V 18/9/2015 [tsl393 28/29]
39 Unit Exam II: Problem #2 (Spring 09) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq. (b) Find the power P supplied by the battery. (c) Find the current I 4 through the 4Ω-resistor. (d) Find the voltage V 2 across the -resistor. Solution: (a) R eq = 8Ω. 3Ω 4Ω 3Ω (b) P = (24V)2 8Ω = 72W. 24V 18/9/2015 [tsl393 28/29]
40 Unit Exam II: Problem #2 (Spring 09) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq. (b) Find the power P supplied by the battery. (c) Find the current I 4 through the 4Ω-resistor. (d) Find the voltage V 2 across the -resistor. Solution: (a) R eq = 8Ω. 3Ω 4Ω 3Ω (b) P = (24V)2 8Ω = 72W. (c) I 4 = V 8Ω = 1.5A. 24V 18/9/2015 [tsl393 28/29]
41 Unit Exam II: Problem #2 (Spring 09) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq. (b) Find the power P supplied by the battery. (c) Find the current I 4 through the 4Ω-resistor. (d) Find the voltage V 2 across the -resistor. Solution: (a) R eq = 8Ω. 3Ω 4Ω 3Ω (b) P = (24V)2 8Ω = 72W. (c) I 4 = 1 24V 2 8Ω = 1.5A. (d) V 2 = (1.5A)() = 3V. 24V 18/9/2015 [tsl393 28/29]
42 Unit Exam II: Problem #3 (Spring 07) Consider the two-loop circuit shown. (a) Find the current I 1. (b) Find the current I 2. (c) Find the potential difference V a V b. 10V 6V a 7Ω 8V 9Ω b I I /9/2015 [tsl364 29/29]
43 Unit Exam II: Problem #3 (Spring 07) Consider the two-loop circuit shown. (a) Find the current I 1. (b) Find the current I 2. (c) Find the potential difference V a V b. 10V 6V a Solution: 7Ω 8V 9Ω (a) I 1 = 8V + 10V 7Ω = 2.57A. b I I /9/2015 [tsl364 29/29]
44 Unit Exam II: Problem #3 (Spring 07) Consider the two-loop circuit shown. (a) Find the current I 1. (b) Find the current I 2. (c) Find the potential difference V a V b. 10V 6V a Solution: 7Ω 8V 9Ω (a) I 1 = (b) I 2 = 8V + 10V = 2.57A. 7Ω 8V 6V = 0.22A. 9Ω b I I /9/2015 [tsl364 29/29]
45 Unit Exam II: Problem #3 (Spring 07) Consider the two-loop circuit shown. (a) Find the current I 1. (b) Find the current I 2. (c) Find the potential difference V a V b. 10V 6V a Solution: 7Ω 8V 9Ω 8V + 10V (a) I 1 = = 2.57A. 7Ω 8V 6V (b) I 2 = = 0.22A. 9Ω (c) V a V b = 8V 6V = 2V. b I I /9/2015 [tsl364 29/29]
M. C. Escher: Waterfall. 18/9/2015 [tsl425 1/29]
![M. C. Escher: Waterfall. 18/9/2015 [tsl425 1/29]](/thumbs/76/73138710.jpg "M. C. Escher: Waterfall. 18/9/2015 [tsl425 1/29]")
M. C. Escher: Waterfall 18/9/2015 [tsl425 1/29] Direct Current Circuit Consider a wire with resistance R = ρl/a connected to a battery. Resistor rule: In the direction of I across a resistor with resistance
This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License.
University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 11. RC Circuits Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License
This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License.
University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 18. RL Circuits Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License
This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License.
University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 07. Capacitors I Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License
This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License.
University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 08. Capacitors II Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons
This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License.
University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 16. Faraday's Law Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons
This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License.
University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 09. Resistors I Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License
20. Alternating Currents
University of hode sland DigitalCommons@U PHY 204: Elementary Physics Physics Course Materials 2015 20. lternating Currents Gerhard Müller University of hode sland, gmuller@uri.edu Creative Commons License
05. Electric Potential I
University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 05. Electric Potential I Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons
This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License.
University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 12. Magnetic Field I Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons
14. Magnetic Field III
University of Rhode sland DigitalCommons@UR PHY 204: Elementary Physics Physics Course Materials 2015 14. Magnetic Field Gerhard Müller University of Rhode sland, gmuller@uri.edu Creative Commons License
02. Electric Field II
Universit of Rhode Island DigitalCommons@URI PHY 24: Elementar Phsics II Phsics Course Materials 215 2. Electric Field II Gerhard Müller Universit of Rhode Island, gmuller@uri.edu Creative Commons License
13. Magnetic Field II
University of Rhode sland DigitalCommons@UR PHY 204: Elementary Physics Physics Course Materials 2015 13. Magnetic Field Gerhard Müller University of Rhode sland, gmuller@uri.edu Creative Commons License
Direct Current Circuits. February 18, 2014 Physics for Scientists & Engineers 2, Chapter 26 1
Direct Current Circuits February 18, 2014 Physics for Scientists & Engineers 2, Chapter 26 1 Kirchhoff s Junction Rule! The sum of the currents entering a junction must equal the sum of the currents leaving
Chapter 28. Direct Current Circuits
Chapter 28 Direct Current Circuits Circuit Analysis Simple electric circuits may contain batteries, resistors, and capacitors in various combinations. For some circuits, analysis may consist of combining
01. Introduction and Electric Field I
University of Rhode Island DigitalCommons@URI PHY 204: lementary Physics II Physics Course Materials 2015 01. Introduction and lectric Field I Gerhard Müller University of Rhode Island, gmuller@uri.edu
Chapter 2. Engr228 Circuit Analysis. Dr Curtis Nelson
Chapter 2 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 2 Objectives Understand symbols and behavior of the following circuit elements: Independent voltage and current sources; Dependent voltage and
Lecture #3. Review: Power
Lecture #3 OUTLINE Power calculations Circuit elements Voltage and current sources Electrical resistance (Ohm s law) Kirchhoff s laws Reading Chapter 2 Lecture 3, Slide 1 Review: Power If an element is
19. LC and RLC Oscillators
University of Rhode Island Digitaloons@URI PHY 204: Eleentary Physics II Physics ourse Materials 2015 19. L and RL Oscillators Gerhard Müller University of Rhode Island, guller@uri.edu reative oons License
AP Physics C. Electric Circuits III.C
AP Physics C Electric Circuits III.C III.C.1 Current, Resistance and Power The direction of conventional current Suppose the cross-sectional area of the conductor changes. If a conductor has no current,
03. Electric Field III and Electric Flux
Universit of Rhode Island DigitalCommons@URI PHY 204: lementar Phsics II Phsics Course Materials 2015 03. lectric Field III and lectric Flu Gerhard Müller Universit of Rhode Island, gmuller@uri.edu Creative
Chapter 6 DIRECT CURRENT CIRCUITS. Recommended Problems: 6,9,11,13,14,15,16,19,20,21,24,25,26,28,29,30,31,33,37,68,71.
Chapter 6 DRECT CURRENT CRCUTS Recommended Problems: 6,9,,3,4,5,6,9,0,,4,5,6,8,9,30,3,33,37,68,7. RESSTORS N SERES AND N PARALLEL - N SERES When two resistors are connected together as shown we said that
Version 001 CIRCUITS holland (1290) 1
Version CIRCUITS holland (9) This print-out should have questions Multiple-choice questions may continue on the next column or page find all choices before answering AP M 99 MC points The power dissipated
I(t) R L. RL Circuit: Fundamentals. a b. Specifications: E (emf) R (resistance) L (inductance) Switch S: a: current buildup. b: current shutdown
RL Circuit: Fundamentals pecifications: E (emf) R (resistance) L (inductance) witch : a: current buildup a b I(t) R L b: current shutdown Time-dependent quantities: I(t): instantaneous current through
Electricity & Magnetism
Electricity & Magnetism D.C. Circuits Marline Kurishingal Note : This chapter includes only D.C. In AS syllabus A.C is not included. Recap... Electrical Circuit Symbols : Draw and interpret circuit diagrams
Chapter 27. Circuits
Chapter 27 Circuits 1 1. Pumping Chagres We need to establish a potential difference between the ends of a device to make charge carriers follow through the device. To generate a steady flow of charges,
Chapter 21 Electric Current and Direct- Current Circuits
Chapter 21 Electric Current and Direct- Current Circuits Units of Chapter 21 Electric Current Resistance and Ohm s Law Energy and Power in Electric Circuits Resistors in Series and Parallel Kirchhoff s
Chapter 26 Direct-Current and Circuits. - Resistors in Series and Parallel - Kirchhoff s Rules - Electric Measuring Instruments - R-C Circuits
Chapter 26 Direct-Current and Circuits - esistors in Series and Parallel - Kirchhoff s ules - Electric Measuring Instruments - -C Circuits . esistors in Series and Parallel esistors in Series: V ax I V
physics 4/7/2016 Chapter 31 Lecture Chapter 31 Fundamentals of Circuits Chapter 31 Preview a strategic approach THIRD EDITION
Chapter 31 Lecture physics FOR SCIENTISTS AND ENGINEERS a strategic approach THIRD EDITION randall d. knight Chapter 31 Fundamentals of Circuits Chapter Goal: To understand the fundamental physical principles
Capacitance. A different kind of capacitor: Work must be done to charge a capacitor. Capacitors in circuits. Capacitor connected to a battery
Capacitance The ratio C = Q/V is a conductor s self capacitance Units of capacitance: Coulomb/Volt = Farad A capacitor is made of two conductors with equal but opposite charge Capacitance depends on shape
Circuits Practice Websheet 18.1
Circuits Practice Websheet 18.1 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. How much power is being dissipated by one of the 10-Ω resistors? a. 24
Physics 115. General Physics II. Session 24 Circuits Series and parallel R Meters Kirchoff s Rules
Physics 115 General Physics II Session 24 Circuits Series and parallel R Meters Kirchoff s Rules R. J. Wilkes Email: phy115a@u.washington.edu Home page: http://courses.washington.edu/phy115a/ 5/15/14 Phys
The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d:
PHYS 102 Exams Exam 2 PRINT (A) The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d: It is connected to a battery with constant emf V.
Physics 7B-1 (A/B) Professor Cebra. Winter 2010 Lecture 2. Simple Circuits. Slide 1 of 20
Physics 7B-1 (A/B) Professor Cebra Winter 2010 Lecture 2 Simple Circuits Slide 1 of 20 Conservation of Energy Density In the First lecture, we started with energy conservation. We divided by volume (making
IMPORTANT Read these directions carefully:
Physics 208: Electricity and Magnetism Common Exam 2, October 17 th 2016 Print your name neatly: First name: Last name: Sign your name: Please fill in your Student ID number (UIN): _ - - Your classroom
Chapter 28. Direct Current Circuits
Chapter 28 Direct Current Circuits Electromotive Force An electromotive force device, or emf device, is a source of constant potential. The emf describes the work done per unit charge and has units of
ANNOUNCEMENT ANNOUNCEMENT
ANNOUNCEMENT Exam : Tuesday September 25, 208, 8 PM - 0 PM Location: Elliott Hall of Music (see seating chart) Covers all readings, lectures, homework from Chapters 2 through 23 Multiple choice (5-8 questions)
Chapter 26 & 27. Electric Current and Direct- Current Circuits
Chapter 26 & 27 Electric Current and Direct- Current Circuits Electric Current and Direct- Current Circuits Current and Motion of Charges Resistance and Ohm s Law Energy in Electric Circuits Combination
Physics 1214 Chapter 19: Current, Resistance, and Direct-Current Circuits
Physics 1214 Chapter 19: Current, Resistance, and Direct-Current Circuits 1 Current current: (also called electric current) is an motion of charge from one region of a conductor to another. Current When
PH 222-2C Fall Circuits. Lectures Chapter 27 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition)
PH 222-2C Fall 2012 Circuits Lectures 11-12 Chapter 27 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition) 1 Chapter 27 Circuits In this chapter we will cover the following topics: -Electromotive
Physics for Scientists & Engineers 2
Review The resistance R of a device is given by Physics for Scientists & Engineers 2 Spring Semester 2005 Lecture 8 R =! L A ρ is resistivity of the material from which the device is constructed L is the
PHYSICS 171. Experiment 3. Kirchhoff's Laws. Three resistors (Nominally: 1 Kilohm, 2 Kilohm, 3 Kilohm).
PHYSICS 171 Experiment 3 Kirchhoff's Laws Equipment: Supplies: Digital Multimeter, Power Supply (0-20 V.). Three resistors (Nominally: 1 Kilohm, 2 Kilohm, 3 Kilohm). A. Kirchhoff's Loop Law Suppose that
Circuits. David J. Starling Penn State Hazleton PHYS 212
Invention is the most important product of man s creative brain. The ultimate purpose is the complete mastery of mind over the material world, the harnessing of human nature to human needs. - Nikola Tesla
PHYS 2135 Exam II March 20, 2018
Exam Total /200 PHYS 2135 Exam II March 20, 2018 Name: Recitation Section: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. For questions 6-9, solutions must
Lecture Outline Chapter 21. Physics, 4 th Edition James S. Walker. Copyright 2010 Pearson Education, Inc.
Lecture Outline Chapter 21 Physics, 4 th Edition James S. Walker Chapter 21 Electric Current and Direct- Current Circuits Units of Chapter 21 Electric Current Resistance and Ohm s Law Energy and Power
Direct Current Circuits
Name: Date: PC1143 Physics III Direct Current Circuits 5 Laboratory Worksheet Part A: Single-Loop Circuits R 1 = I 0 = V 1 = R 2 = I 1 = V 2 = R 3 = I 2 = V 3 = R 12 = I 3 = V 12 = R 23 = V 23 = R 123
Problem Solving 8: Circuits
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics OBJECTIVES Problem Solving 8: Circuits 1. To gain intuition for the behavior of DC circuits with both resistors and capacitors or inductors.
Chapter 28 Solutions
Chapter 8 Solutions 8.1 (a) P ( V) R becomes 0.0 W (11.6 V) R so R 6.73 Ω (b) V IR so 11.6 V I (6.73 Ω) and I 1.7 A ε IR + Ir so 15.0 V 11.6 V + (1.7 A)r r 1.97 Ω Figure for Goal Solution Goal Solution
Chapter 26 Direct-Current Circuits
Chapter 26 Direct-Current Circuits 1 Resistors in Series and Parallel In this chapter we introduce the reduction of resistor networks into an equivalent resistor R eq. We also develop a method for analyzing
2005 AP PHYSICS C: ELECTRICITY AND MAGNETISM FREE-RESPONSE QUESTIONS
2005 AP PHYSICS C: ELECTRICITY AND MAGNETISM In the circuit shown above, resistors 1 and 2 of resistance R 1 and R 2, respectively, and an inductor of inductance L are connected to a battery of emf e and
Class 8. Resistivity and Resistance Circuits. Physics 106. Winter Press CTRL-L to view as a slide show. Class 8. Physics 106.
and Circuits and Winter 2018 Press CTRL-L to view as a slide show. Last time we learned about Capacitance Problems Parallel-Plate Capacitors Capacitors in Circuits Current Ohm s Law and Today we will learn
Chapter 7 Direct-Current Circuits
Chapter 7 Direct-Current Circuits 7. Introduction... 7. Electromotive Force... 7.3 Resistors in Series and in Parallel... 4 7.4 Kirchhoff s Circuit Rules... 6 7.5 Voltage-Current Measurements... 8 7.6
SIMPLE D.C. CIRCUITS AND MEASUREMENTS Background
SIMPLE D.C. CICUITS AND MEASUEMENTSBackground This unit will discuss simple D.C. (direct current current in only one direction) circuits: The elements in them, the simple arrangements of these elements,
Chapter 21 Electric Current and Direct- Current Circuits
Chapter 21 Electric Current and Direct- Current Circuits 1 Overview of Chapter 21 Electric Current and Resistance Energy and Power in Electric Circuits Resistors in Series and Parallel Kirchhoff s Rules
Electric Current. Note: Current has polarity. EECS 42, Spring 2005 Week 2a 1
Electric Current Definition: rate of positive charge flow Symbol: i Units: Coulombs per second Amperes (A) i = dq/dt where q = charge (in Coulombs), t = time (in seconds) Note: Current has polarity. EECS
EXPERIMENT 12 OHM S LAW
EXPERIMENT 12 OHM S LAW INTRODUCTION: We will study electricity as a flow of electric charge, sometimes making analogies to the flow of water through a pipe. In order for electric charge to flow a complete
Chapter 25 Current Resistance, and Electromotive Force
Chapter 25 Current Resistance, and Electromotive Force 1 Current In previous chapters we investigated the properties of charges at rest. In this chapter we want to investigate the properties of charges
MasteringPhysics: Assignment Print View. Problem 30.50
Page 1 of 15 Assignment Display Mode: View Printable Answers phy260s08 homework 13 Due at 11:00pm on Wednesday, May 14, 2008 View Grading Details Problem 3050 Description: A 15-cm-long nichrome wire is
Lecture 39. PHYC 161 Fall 2016
Lecture 39 PHYC 161 Fall 016 Announcements DO THE ONLINE COURSE EVALUATIONS - response so far is < 8 % Magnetic field energy A resistor is a device in which energy is irrecoverably dissipated. By contrast,
Electrical Circuits. Winchester College Physics. makptb. c D. Common Time man. 3rd year Revision Test
Name... Set... Don.... manner~ man makptb Winchester College Physics 3rd year Revision Test Electrical Circuits Common Time 2011 Mark multiple choice answers with a cross (X) using the box below. I A B
Parallel Resistors (32.6)
Parallel Resistors (32.6) Resistors connected at both ends are called parallel resistors Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 1 / 1 Parallel Resistors (32.6)
Parallel Resistors (32.6)
Parallel Resistors (32.6) Resistors connected at both ends are called parallel resistors The important thing to note is that: the two left ends of the resistors are at the same potential. Also, the two
Physics 2020 Exam 2 Constants and Formulae
Physics 2020 Exam 2 Constants and Formulae Useful Constants k e = 8.99 10 9 N m 2 /C 2 c = 3.00 10 8 m/s ɛ = 8.85 10 12 C 2 /(N m 2 ) µ = 4π 10 7 T m/a e = 1.602 10 19 C h = 6.626 10 34 J s m p = 1.67
PHYS 1444 Section 003 Lecture #12
PHYS 1444 Section 003 Lecture #12 Monday, Oct. 10, 2005 EMF and Terminal Voltage Resisters in series and parallel Kirchhoff s Rules EMFs in series and parallel RC Circuits Today s homework is homework
4 Electric circuits. Serial and parallel resistors V 3 V 2 V Serial connection of resistors:
4 lectric circuits PHY67 Spring 006 Serial and parallel resistors Serial connection of resistors: As the current I through each of serially connected resistors is the same, one can use Ohm s law and write...
Physics 42 Exam 2 PRACTICE Name: Lab
Physics 42 Exam 2 PRACTICE Name: Lab 1 2 3 4 Conceptual Multiple Choice (2 points each) Circle the best answer. 1.Rank in order, from brightest to dimmest, the identical bulbs A to D. A. C = D > B > A
Physics 2135 Exam 2 October 20, 2015
Exam Total / 200 Physics 2135 Exam 2 October 20, 2015 Printed Name: Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. 1. A straight wire segment
Chapter 18 Electric Currents
Chapter 18 Electric Currents 1 The Electric Battery Volta discovered that electricity could be created if dissimilar metals were connected by a conductive solution called an electrolyte. This is a simple
Power lines. Why do birds sitting on a high-voltage power line survive?
Power lines At large distances, the resistance of power lines becomes significant. To transmit maximum power, is it better to transmit high V, low I or high I, low V? (a) high V, low I (b) low V, high
Circuits. PHY2054: Chapter 18 1
Circuits PHY2054: Chapter 18 1 What You Already Know Microscopic nature of current Drift speed and current Ohm s law Resistivity Calculating resistance from resistivity Power in electric circuits PHY2054:
Electromotive Force. The electromotive force (emf), ε, of a battery is the maximum possible voltage that the battery can provide between its terminals
Direct Current When the current in a circuit has a constant magnitude and direction, the current is called direct current Because the potential difference between the terminals of a battery is constant,
Physics 6B Summer 2007 Final
Physics 6B Summer 2007 Final Question 1 An electron passes through two rectangular regions that contain uniform magnetic fields, B 1 and B 2. The field B 1 is stronger than the field B 2. Each field fills
Chapter 28. Direct Current Circuits
Chapter 28 Direct Current Circuits Circuit Analysis Simple electric circuits may contain batteries, resistors, and capacitors in various combinations. For some circuits, analysis may consist of combining
Superconductors A class of materials and compounds whose resistances fall to virtually zero below a certain temperature, T C T C is called the critical temperature The graph is the same as a normal metal
Physics 102 Lab 4: Circuit Algebra and Effective Resistance Dr. Timothy C. Black Spring, 2005
Physics 02 Lab 4: Circuit Algebra and Effective Resistance Dr. Timothy C. Black Spring, 2005 Theoretical Discussion The Junction Rule: Since charge is conserved, charge is neither created or destroyed
R R V I R. Conventional Current. Ohms Law V = IR
DC Circuits opics EMF and erminal oltage esistors in Series and in Parallel Kirchhoff s ules EMFs in Series and in Parallel Capacitors in Series and in Parallel Ammeters and oltmeters Conventional Current
AP Physics C - E & M
AP Physics C - E & M Current and Circuits 2017-07-12 www.njctl.org Electric Current Resistance and Resistivity Electromotive Force (EMF) Energy and Power Resistors in Series and in Parallel Kirchoff's
Outline. Week 5: Circuits. Course Notes: 3.5. Goals: Use linear algebra to determine voltage drops and branch currents.
Outline Week 5: Circuits Course Notes: 3.5 Goals: Use linear algebra to determine voltage drops and branch currents. Components in Resistor Networks voltage source current source resistor Components in
Louisiana State University Physics 2102, Exam 2, March 5th, 2009.
PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 2, March 5th, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),
Physics 102 Spring 2007: Final Exam Multiple-Choice Questions
Last Name: First Name: Physics 102 Spring 2007: Final Exam Multiple-Choice Questions 1. The circuit on the left in the figure below contains a battery of potential V and a variable resistor R V. The circuit
Chapter 3: Electric Current and Direct-Current Circuit
Chapter 3: Electric Current and Direct-Current Circuit n this chapter, we are going to discuss both the microscopic aspect and macroscopic aspect of electric current. Direct-current is current that flows
Physics 212 Midterm 2 Form A
1. A wire contains a steady current of 2 A. The charge that passes a cross section in 2 s is: A. 3.2 10-19 C B. 6.4 10-19 C C. 1 C D. 2 C E. 4 C 2. In a Physics 212 lab, Jane measures the current versus
Slide 1 / 26. Inductance by Bryan Pflueger
Slide 1 / 26 Inductance 2011 by Bryan Pflueger Slide 2 / 26 Mutual Inductance If two coils of wire are placed near each other and have a current passing through them, they will each induce an emf on one
2/25/2014. Circuits. Properties of a Current. Conservation of Current. Definition of a Current A. I A > I B > I C B. I B > I A C. I C D. I A E.
Circuits Topics: Current Conservation of current Batteries Resistance and resistivity Simple circuits 0.1 Electromotive Force and Current Conventional current is the hypothetical flow of positive charges
Voltage Dividers, Nodal, and Mesh Analysis
Engr228 Lab #2 Voltage Dividers, Nodal, and Mesh Analysis Name Partner(s) Grade /10 Introduction This lab exercise is designed to further your understanding of the use of the lab equipment and to verify
Physics 2401 Summer 2, 2008 Exam II
Physics 2401 Summer 2, 2008 Exam II e = 1.60x10-19 C, m(electron) = 9.11x10-31 kg, ε 0 = 8.845x10-12 C 2 /Nm 2, k e = 9.0x10 9 Nm 2 /C 2, m(proton) = 1.67x10-27 kg. n = nano = 10-9, µ = micro = 10-6, m
ES250: Electrical Science. HW1: Electric Circuit Variables, Elements and Kirchhoff s Laws
ES250: Electrical Science HW1: Electric Circuit Variables, Elements and Kirchhoff s Laws Introduction Engineers use electric circuits to solve problems that are important to modern society, such as: 1.
ConcepTest PowerPoints
ConcepTest PowerPoints Chapter 19 Physics: Principles with Applications, 6 th edition Giancoli 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for
DC Circuits. Electromotive Force Resistor Circuits. Kirchoff s Rules. RC Circuits. Connections in parallel and series. Complex circuits made easy
DC Circuits Electromotive Force esistor Circuits Connections in parallel and series Kirchoff s ules Complex circuits made easy C Circuits Charging and discharging Electromotive Force (EMF) EMF, E, is the
[1] (b) Fig. 1.1 shows a circuit consisting of a resistor and a capacitor of capacitance 4.5 μf. Fig. 1.1
![[1] (b) Fig. 1.1 shows a circuit consisting of a resistor and a capacitor of capacitance 4.5 μf. Fig. 1.1](/thumbs/81/83276652.jpg "[1] (b) Fig. 1.1 shows a circuit consisting of a resistor and a capacitor of capacitance 4.5 μf. Fig. 1.1")
1 (a) Define capacitance..... [1] (b) Fig. 1.1 shows a circuit consisting of a resistor and a capacitor of capacitance 4.5 μf. S 1 S 2 6.3 V 4.5 μf Fig. 1.1 Switch S 1 is closed and switch S 2 is left
PEP 2017 Assignment 12
of the filament?.16.. Aductile metal wire has resistance. What will be the resistance of this wire in terms of if it is stretched to three times its original length, assuming that the density and resistivity
SPS Presents: A Cosmic Lunch!
SPS Presents: A Cosmic Lunch! Who: Dr. Brown will be speaking about Evolution of the Elements: from Periodic table to Standard Model and Beyond! When: October 7 th at am Where: CP 79 (by the front office)
Chapter 3: Electric Current And Direct-Current Circuits
Chapter 3: Electric Current And Direct-Current Circuits 3.1 Electric Conduction 3.1.1 Describe the microscopic model of current Mechanism of Electric Conduction in Metals Before applying electric field
Physics 102 Spring 2006: Final Exam Multiple-Choice Questions
Last Name: First Name: Physics 102 Spring 2006: Final Exam Multiple-Choice Questions For questions 1 and 2, refer to the graph below, depicting the potential on the x-axis as a function of x V x 60 40
Application of Physics II for. Final Exam
Application of Physics II for Final Exam Question 1 Four resistors are connected as shown in Figure. (A)Find the equivalent resistance between points a and c. (B)What is the current in each resistor if
A positive value is obtained, so the current is counterclockwise around the circuit.
Chapter 7. (a) Let i be the current in the circuit and take it to be positive if it is to the left in. We use Kirchhoff s loop rule: ε i i ε 0. We solve for i: i ε ε + 6. 0 050.. 4.0Ω+ 80. Ω positive value
PHY102 Electricity Course Summary
TOPIC 1 ELECTOSTTICS PHY1 Electricity Course Summary Coulomb s Law The magnitude of the force between two point charges is directly proportional to the product of the charges and inversely proportional
cancel each other out. Thus, we only need to consider magnetic field produced by wire carrying current 2.
PC1143 2011/2012 Exam Solutions Question 1 a) Assumption: shells are conductors. Notes: the system given is a capacitor. Make use of spherical symmetry. Energy density, =. in this case means electric field
Chapter 26 Direct-Current Circuits
Chapter 26 Direct-Current Circuits 1 Resistors in Series and Parallel In this chapter we introduce the reduction of resistor networks into an equivalent resistor R eq. We also develop a method for analyzing
PHYS 272 (Spring 2018): Introductory Physics: Fields Problem-solving sessions
Figure 1: Problem 1 Figure 2: Problem 2 PHYS 272 (Spring 2018): Introductory Physics: Fields Problem-solving sessions (1). A thin rod of length l carries a total charge Q distributed uniformly along its