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University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 10. Resistors II 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. Follow this and additional works at: http://digitalcommons.uri.edu/elementary_physics_2 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 17. http://digitalcommons.uri.edu/elementary_physics_2/17 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.

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 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]

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]

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]

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]

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]

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]

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 1 + 1 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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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 = 159 71 pf, Q 1 = 55 71 pc, Q 2 = 104 71 pc, Q 3 = 9 71 pc, Q 4 = 64 71 pc, Q 5 = 95 71 pc. 18/9/2015 [tsl511 24/29]

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 3 + 1 R 6 «1 =, R eq = R 2 + R 36 = 4Ω. 18/9/2015 [tsl337 25/29]

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 3 + 1 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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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 1 2 18/9/2015 [tsl364 29/29]

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 1 2 18/9/2015 [tsl364 29/29]

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 1 2 18/9/2015 [tsl364 29/29]

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 1 2 18/9/2015 [tsl364 29/29]