# Solucionario de Circuitos de Corriente Continua

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FISICA II CIRCUITOS DE CORRIENTE CONTINUA _______________________________________________________________________________________________________

1 1 26.1: a) Req = + = 12.3 . 32 20 240 V V = = 19.5 A. b) I = Req 12.3

1

c) I 32 =

V 240 V V 240 V = = 7.5 A; I 20 = = = 12 A. R 32 R 20

26.2:

1 R + R2 R1 R2 1 Req = + = 1 R R R R Req = R + R . 2 1 2 1 1 2 R1 R2 Req = R1 < R1 and Req = R2 < R2 . R1 + R2 R1 + R2

1

1

26.3: For resistors in series, the currents are the same and the voltages add. a) true. 2 b) false. c) P = I R. i same, R different so P different; false. d) true. e) V = IR. I same, R different; false. f) Potential drops as move through each resistor in the direction of the current; false. g) Potential drops as move through each resistor in the direction of the current, so Vb > Vc ; false. h) true.

26.4: a) False, current divides at junction a. b) True by charge conservation. 1 c) True. V1 = V2 , so I R d) False. P = IV .V1 = V2 , but I1 I 2 , so P P2 . 1 e) False. P = IV = VR . Since R2 > R1 , P2 < P1 . f) True. Potential is independent of path. g) True. Charges lose potential energy (as heat) in R1 . h) False. See answer to (g). i) False. They are at the same potential.2

___________________________________________________________________________Ms. Moiss Enrique BELTRAN LAZARO

FISICA II CIRCUITOS DE CORRIENTE CONTINUA _______________________________________________________________________________________________________

1 1 1 26.5: a) Req = 2.4 + 1.6 + 4.8 = 0.8 . b) I 2.4 = R2.4 = (28 V) (2.4 ) = 11.67 A; I1.6 = R1.6 = (28 V) (1.6 ) = 17.5 A; I 4.8 = R4.8 = (28 V) (4.8 ) = 5.83 A. c) I total = Rtotal = (28 V) (0.8 ) = 35 A. d) When in parallel, all resistors have the same potential difference over them, so here all have V = 28 V. e) P2.4 = I 2 R2.4 = (11.67 A) 2 (2.4 ) = 327 W; P1.6 = I 2 R1.6 = (17.5 A) 2 (1.6 ) =490 W; P4.8 = I 2 R4.8 = (5.83 A) 2 (4.8 ) = 163 W. f) For resistors in parallel, the most power is dissipated through the resistor with the V2 , with V = constant. least resistance since P = I 2 R = R

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26.6: a) Req = Ri = 2.4 + 1.6 + 4.8 = 8.8 .

28 V = = 3.18 A. Req 8.8 c) The current through the battery equals the current of (b), 3.18 A. d) V2.4 = IR2.4 = (3.18 A)(2.4 ) = 7.64 V; V1.6 = IR1.6 = (3.18 A)(1.6 ) = 5.09 V; V4.8 = IR4.8 = (3.18 A)(4.8 ) = 15.3 V.b) The current in each resistor is the same and is I = e) P2.4 = I 2 R2.4 = (3.18 A) 2 (2.4 ) = 24.3 W; P .6 = I 2 R1.6 = (3.18 A) 2 (1.6 ) = 116.2 W; P4.8 = I 2 R4.8 = (3.18 A) 2 (4.8 ) = 48.5 W. f) For resistors in series, the most power is dissipated by the resistor with the greatest resistance since P = I 2 R with I constant.

26.7: a) P =

V2 V = PR = (5.0 W )(15,000 ) = 274 V. R V 2 (120 V) 2 = = 1.6 W. b) P = R 9,000

26.8:

1 1 1 1 1 1 + Req = + 12.0 + 4.00 = 5.00 . 3.00 6.00

I total = Rtotal = (6.00 V) (5.00 ) = 12.0 A 4 12 I 12 = (12.0) = 3.00 A; I 4 = (12.0) = 9.00 A; 12 + 4 12 + 4 6 3 I3 = (12.0) = 8.00 A; I 6 = (12.0) = 4.00 A . 3+6 3+6

___________________________________________________________________________Ms. Moiss Enrique BELTRAN LAZARO

FISICA II CIRCUITOS DE CORRIENTE CONTINUA _______________________________________________________________________________________________________

26.9:

1 1 Req = 3.00 + 1.00 + 5.00 + 7.00 = 3.00 . I total = Rtotal = (48.0 V ) (3.00 ) = 16.0 A . 4 12 I5 = I7 = (16.0) = 4.00 A; I 1 = I 3 = (16.0) = 12.0 A . 4 + 12 4 + 12

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26.10: a) The three resistors R2 , R3 and R4 are in parallel, so:

R234

1 1 1 1 1 1 = R + R + R = 8.20 + 1.50 + 4.50 3 4 2 Req = R1 + R234 = 3.50 + 0.99 = 4.49 .

1

1

= 0.99

b) I1 =

6.0 V = = 1.34 A V1 = I1R1 = (1.34 A) (3.50 ) = 4.69 V. Req 4.49 VR234 R2 = 1.33 V = 0.162 A, 8.20

VR234 = I1 R234 = (1.34 A) (0.99 ) = 1.33 V I 2 = I3 = VR234 R3 =

VR 1.33 V 1.33 V = 0.887 A and I 4 = 234 = = 0.296 A. R4 4.50 1.50

26.11: Using the same circuit as in Problem 27.10, with all resistances the same: 1 3 1 1 Req = R1 + R234 = R1 + + + = 4.50 + 4.50 R 2 R3 R4 1 9.00 V = = 1.50 A, I 2 = I 3 = I 4 = I1 = 0.500 A. a) I1 = Req 6.00 31 1

= 6.00 .

1 P1 = 1.125 W. 9 c) If there is a break at R4 , then the equivalent resistance increases:

b) P1 = I 1 R1 = (1.50 A) 2 (4.50 ) = 10.13 W, P2 = P3 = P4 =

2

1 1 Req = R1 + R23 = R1 + + R 2 R3 And so:I1 =2

1

2 = 4.50 + 4.50

1

= 6.75 .

1 9.00 V = = 1.33 A, I 2 = I 3 = I1 = 0.667 A. Req 6.75 2

1 P1 = 1.99 W. 4 e) So R2 and R3 are brighter than before, while R1 is fainter. The amount of current flow is all that determines the power output of these bulbs since their resistances are equal.

d) P1 = I 1 R1 = (1.33 A) 2 (4.50 ) = 7.96 W, P2 = P3 =

___________________________________________________________________________Ms. Moiss Enrique BELTRAN LAZARO

26.12: From Ohms law, the voltage drop across the 6.00 resistor is V = IR = (4.00 A)(6.00 ) = 24.0 V. The voltage drop across the 8.00 resistor is the same, since these two resistors are wired in parallel. The current through the 8.00 resistor is then I = V R = 24.0 V 8.00 = 3.00 A. The current through the 25.0 resistor is the sum of these two currents: 7.00 A. The voltage drop across the 25.0 resistor is V = IR = (7.00 A)( 25.0 ) = 175 V, and total voltage drop across the top branch of the circuit is 175 + 24.0 = 199 V, which is also the voltage drop across the 20.0 resistor. The current through the 20.0 resistor is then I = V R = 199 V 20 = 9.95 A.

26.13: Current through 2.00- resistor is 6.00 A. Current through 1.00- resistor also is 6.00 A and the voltage is 6.00 V. Voltage across the 6.00- resistor is 12.0 V + 6.0 V = 18.0 V. Current through the 6.00- resistor is (18.0V) (6.00) = 3.00 A. The battery voltage is 18.0 V.26.14: a) The filaments must be connected such that the current can flow through each separately, and also through both in parallel, yielding three possible current flows. The parallel situation always has less resistance than any of the individual members, so it will give the highest power output of 180 W, while the other two must give power outputs of 60 W and 120 W. V2 (120 V) 2 V2 (120 V) 2 60 W = R1 = = 240 , and 120 W = R2 = = 120 . R1 60 W R2 120 W Check for parallel: P =V2 (120 V) 2 (120 V) 2 = 1 = = 180 W. 1 1 1 ( R1 + R2 ) 1 ( 240 + 120 ) 1 80

b) If R1 burns out, the 120 W setting stays the same, the 60 W setting does not work and the 180 W setting goes to 120 W: brightnesses of zero, medium and medium. c) If R2 burns out, the 60 W setting stays the same, the 120 W setting does not work, and the 180 W setting is now 60 W: brightnesses of low, zero and low.

26.15: a) I =

120 V = = 0.100 A. R (400 + 800 )

b) P400 = I 2 R = (0.100 A) 2 (400 ) = 4.0 W; P800 = I 2 R = (0.100 A) 2 (800 ) = 8.0 W Ptotal = 4 W + 8 W = 12 W. c) When in parallel, the equivalent resistance becomes:

1 1 120 V Req = 400 + 800 = 267 I total = R = 267 = 0.449 A. eq 400 800 I 400 = (0.449 A) = 0.150 A. (0.449 A) = 0.30 A; I 800 = 400 + 800 400 + 800 d) P400 = I 2 R = (0.30 A) 2 (400 ) = 36 W; P800 = I 2 R = (0.15 A) 2 (800 ) = 18 W Ptotal = 36 W + 18 W = 54 W. e) The 800 resistor is brighter when the resistors are in series, and the 400 is brighter when in parallel. The greatest total light output is when they are in parallel.V 2 (120 V) 2 V 2 (120 V) 2 = = 240 ; R200 W = = = 72 . P P 60 W 200 W 240 V I 60 W = I 200 W = = = 0.769 A. R (240 + 72 )

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26.16: a) R60 W =

b) P60 W = I 2 R = (0.769 A) 2 (240 ) = 142 W; P200 W = I 2 R = (0.769 A) 2 (72 ) = 42.6 W. c) The 60 W bulb burns out quickly because the power it delivers (142 W) is 2.4 times its rated value.

26.17:

30.0 V I (20.0 + 5.0 + 5.0 ) = 0; I = 1.00 A For the 20.0- resistor thermal energy is generated at the rate 2 P = I R = 20.0 W. Q = Pt and Q = mcT gives mcT (0.100 kg) (4190 J kg K ) (40.0 C) t= = = 1.01 103 s P 20.0 W

26.18: a)

P1 = I 12 R1 20 W = (2A) 2 R1 R1 = 5.00 R1 and 10 in parallel: (10 ) I 10 = (5 ) (2 A)

I 10 = 1 A So I 2 = 0.50 A. R1 and R2 are in parallel, so (0.50 A) R2 = (2 A) (5 ) R2 = 20.0 b) = V1 = (2A)(5 ) = 10.0 V c) From (a): I 2 = 0.500 A, I 10 = 1.00 A P1 = 20.0 W (given) d)2 P2 = i2 R2 = (0.50 A) 2 (20 ) = 5.00 W 2 P10 = i10 R10 = (1.0 A) 2 (10 ) = 10.0 W PResist = 20 W + 5 W + 10 W = 35.0 W PBattery = I = (3.50 A) (10.0 V) = 35.0 W

PResist = PBattery, which agrees with the conservation of energy.

26.19: a) I R = 6.00 A 4.00 A = 2.00 A. b) Using a Kirchhoff loop around the outside of the circuit:

28.0 V (6.00 A) (3.00 ) (2.00 A) R = 0 R = 5.00 . c) Using a counterclockwise loop in the bottom half of the circuit: (6.00 A) (3.00 ) (4.00 A) (6.00 ) = 0 = 42.0 V. d) If the circuit is