 # Fig 28-CO, p.858. Resistive medium Chapter 28 Direct Current Circuits 28.1 Electromotive “Force” (emf)

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Fig 28-CO, p.858

Resistive medium Chapter 28 Direct Current Circuits 28.1 Electromotive “Force” (emf)

VRVR P28.1 (p.800) VrVr

CT1: If R 1 = 2R 2, which resistor dissipates more power? A. R 1 B. R 2 C. Both the same.

CT2: If R 1 = 2R 2, which resistor dissipates more power? A. R 1 B. R 2 C. Both the same.

Chapter 28 Direct Current Circuits 28.2 Resistors in Series and Parallel

Resistors in Series A, L 1 A, L 2 R 1 =  L 1 /A R1R1 R 2 =  L 2 /A R2R2 R eq A, L 1 + L 2 R eq =  (L 1 + L 2 ) /A = R 1 + R 2 For resistors in series, the same current flows through each resistor from conservation of charge. For many resistors in series the equivalent resistance is the sum of all the individual resistances R eq = R 1 + R 2 + R 3 + …

Series Resistors 1. Current same through each resistor (conservation of charge) 2. Add

Resistors in Parallel A 1, L A 2, L 1/R 1 = A 1 / L R1R1 R2R2 R eq A 1 + A 2, L 1/R eq = (A 1 + A 2 )/ L = 1/R 1 + 1/R 2 For resistors in parallel, the voltage across each resistor is the same from conservation of energy. For many resistors in parallel the reciprocal of the equivalent resistance is the sum of all the reciprocals of the individual resistances 1/R 2 = A 2 / L 1/R eq = 1/R 1 + 1/R 2 + 1/R 3 + …

Parallel Resistors 1. Voltages same across each resistor (conservation of energy) 2. Add reciprocals

P28-15 (p.801)

Q28.11 p. 779 CT3-8 (i,ii,iii,iv,v,vi) A.increase B.decrease C.no change D.drop to zero

Circuit Elements Source of emf  = V “Perfect” conductor R = 0 Capacitor C = Q / V Resistor R = V / I Switch Open R =  Closed R = 0 + -

Junction Rule: The sum of the currents entering any junction equals the sum of the currents leaving that junction. This is due to conservation of charge. I 1 = I 2 + I 3 Loop Rule: The sum of the potential differences across each element around any closed loop is zero. This is due to the conservation of energy. 28.3 Kirchhoff’s Rules I1I1 I2I2 I3I3 R  I  - IR = 0

Kirchhoff’s Rules  Analogy: River flowing down hill through a gravel bed. R I Loop Notes 1. Go through a battery from - to +: +  2. Go through a battery from + to -: -  3. Go through a resistor with current: -IR 4. Go through a resistor against current: +IR

P28.40 (p. 804) CT9: Which circuit requires Kirchoff’s Rules to solve? A. circuit A B. circuit b C. circuit c D. circuit d

P28.40 (p. 804) CT10: Which circuit delivers the least power to the 10  resistor? A. circuit A B. circuit b C. circuit c D. circuit d

Chapter 28 Direct Current Circuits P28.17 (p.802)

P28-55 p. 805 Each bulb is 60 W at 120V Calculate P total and  V across each bulb.

Chapter 28 Direct Current Circuits 28.4 RC Circuits

Q28.14 RC Circuits p. 800 Assuming the capacitor is initially uncharged, that the time constant is several seconds, and that the bulb lights when connected directly to the battery terminals, state what happens when the switch is closed. Include in your answer a discussion of why the three assumptions are necessary.

Chapter 28 Direct Current Circuits 28.4 RC Circuits P28.27 p. 802

CT9: A fully charged capacitor stores energy U 0. How much energy remains when the charge has decreased to half its original value? A. U 0 B. U 0 /2 C. U 0 /4 D. U 0 /8

Chapter 28 Direct Current Circuits 28.5 Electrical Meters Ideal Voltmeter R =  Ideal Ammeter R = 0

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