# Physics exercises_solution: Chapter 26

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## Physics exercises_solution: Chapter 26

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 26

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## Nội dung Text: Physics exercises_solution: Chapter 26

1. 1  1 1  26.1: a) Req      12.3 .  32 20  V 240 V b) I    19.5 A. Req 12.3  V 240 V V 240 V c) I 32    7.5 A; I 20    12 A. R 32  R 20  1 1 1 1   R  R2  R1 R2 26.2: Req       1 R R   R R   Req  R  R .   1 2   1 2  1 2 R2 R1  Req  R1  R1 and Req  R2  R2 . R1  R2 R1  R2 26.3: For resistors in series, the currents are the same and the voltages add. a) true. b) false. c) P  I R. i same, R different so P different; false. d) true. e) V = IR. I 2 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  R . Since R2  R1 , P2  P1 . V2 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. 1  1 1 1  26.5: a) Req    2.4  1.6  4.8    0.8 .      b) I 2.4  ε R2.4  (28 V) (2.4 )  11.67 A; I1.6  ε R1.6  (28 V) (1.6 )  17.5 A; I 4.8  ε R4.8  (28 V) (4.8 )  5.83 A. c) I total  ε 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 least resistance since P  I 2 R  , with V  constant. R 26.6: a) Req   Ri  2.4   1.6   4.8   8.8 . ε 28 V b) The current in each resistor is the same and is I    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. 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 )  1 16.2 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. V2 26.7: a) P   V  PR  (5.0 W )(15,000 )  274 V. R V 2 (120 V) 2 b) P    1.6 W. R 9,000 
3.  1 1  1  1 1   1     26.8: Req      3.00  6.00    12.0   4.00     5.00  .       I total  ε 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 1  1 1  26.9: Req    3.00   1.00   5.00   7.00    3.00  .    I total  ε 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 26.10: a) The three resistors R2 , R3 and R4 are in parallel, so: 1 1  1 1 1   1 1 1  R234   R  R  R    8.20   1.50   4.50       0.99   2 3 4     Req  R1  R234  3.50   0.99   4.49  . ε 6.0 V b) I1    1.34 A  V1  I1R1  (1.34 A) (3.50 )  4.69 V. Req 4.49  VR234 1.33 V  VR234  I1 R234  (1.34 A) (0.99 )  1.33 V  I 2    0.162 A, R2 8.20  VR234 1.33 V VR 1.33 V I3    0.887 A and I 4  234   0.296 A. R3 1.50  R4 4.50 
4. 26.11: Using the same circuit as in Problem 27.10, with all resistances the same: 1 1  1 1 1   3  Req  R1  R234  R1    R    4.50      4.50     6.00  .  2 R3 R4    ε 9.00 V 1 a) I1    1.50 A, I 2  I 3  I 4  I1  0.500 A. Req 6.00  3 2 1 b) P1  I 1 R1  (1.50 A) 2 (4.50 )  10.13 W, P2  P3  P4  P1  1.125 W. 9 c) If there is a break at R4 , then the equivalent resistance increases: 1 1  1 1   2  Req  R1  R23  R1   R  R   4.50     4.50     6.75 .  2 3    And so: ε 9.00 V 1 I1    1.33 A, I 2  I 3  I1  0.667 A. Req 6.75  2 2 1 d) P1  I 1 R1  (1.33 A) 2 (4.50 )  7.96 W, P2  P3  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. 26.12: From Ohm’s 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.
5. 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 V2 (120 V) 2 (120 V) 2 Check for parallel: P   1   180 W. ( R1  R2 ) 1 ( 240   120  ) 1 1 1 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. ε 120 V 26.15: a) I    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 1  ε 120 V Req    400   800    267   I total  R  267   0.449 A.    eq 800 400 I 400  (0.449 A)  0.30 A; I 800  (0.449 A)  0.150 A. 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.
6. V 2 (120 V) 2 V 2 (120 V) 2 26.16: a) R60 W    240 ; R200 W    72 . P 60 W P 200 W ε 240 V  I 60 W  I 200W    0.769 A. R (240   72 ) 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 P  I 2 R  20.0 W. Q  Pt and Q  mcT gives mcT (0.100 kg) (4190 J kg  K ) (40.0 C) t   1.01  10 3 s P 20.0 W
7. 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) ε  V1  (2A)(5 )  10.0 V c) From (a): I 2  0.500 A, I 10  1.00 A d) P1  20.0 W (given) P2  i2 R2  (0.50 A) 2 (20 )  5.00 W 2 P10  i10 R10  (1.0 A) 2 (10 )  10.0 W 2 PResist  20 W  5 W  10 W  35.0 W PBattery  I ε  (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: ε  (6.00 A) (3.00 )  (4.00 A) (6.00 )  0  ε  42.0 V. d) If the circuit is broken at point x, then the current in the 28 V battery is: ε 28.0 V I   3.50 A.  R 3.00   5.00  26.20: From the given currents in the diagram, the current through the middle branch of the circuit must be 1.00 A (the difference between 2.00 A and 1.00 A). We now use Kirchoff’s Rules, passing counterclockwise around the top loop: 20.0 V  (1.00 A) 6.00 Ω  1.00 Ω   1.00 A 4.00 Ω  1.00 Ω   ε1  0  ε1  18.0 V. Now traveling around the external loop of the circuit: 20.0 V  1.00 A 6.00   1.00    2.00 A 1.00   2.00    ε2  0  ε2  7.0 V. And Vab  1.00 A 4.00   1.00    18.0 V  13.0 V, so Vba  13.0 V.
8. 26.21: a) The sum of the currents that enter the junction below the 3 -  resistor equals 3.00 A + 5.00 A = 8.00 A. b) Using the lower left loop: ε1  4.00  3.00 A   3.00  8.00 A   0  ε1  36.0 V. Using the lower right loop: ε2  6.00  5.00 A   3.00  8.00 A   0  ε2  54.0 V. c) Using the top loop: 18.0 V 54.0 V  R2.00 A   36.0 V  0  R   9.00 . 2.00 A 26.22: From the circuit in Fig. 26.42, we use Kirchhoff’s Rules to find the currents, I1 to the left through the 10 V battery, I 2 to the right through 5 V battery, and I 3 to the right through the 10  resistor: Upper loop: 10.0 V  2.00   3.00  I1  1.00   4.00  I 2  5.00 V  0  5.0 V  5.00  I1  5.00  I 2  0  I1  I 2  1.00 A. Lower loop: 5.00 V  1.00   4.00  I 2  10.0  I 3  0  5.00 V  5.00  I 2  10.0  I 3  0  I 2  2 I 3  1.00 A Along with I1  I 2  I 3 , we can solve for the three currents and find: I 1  0.800 A, I 2  0.200 A, I 3  0.600 A. b) Vab   0.200 A 4.00    0.800 A 3.00    3.20 V. 26.23: After reversing the polarity of the 10-V battery in the circuit of Fig. 26.42, the only change in the equations from Problem 26.22 is the upper loop where the 10 V battery is: Upper loop:  10.0 V  2.00   3.00  I1  1.00   4.00  I 2  5.00 V  0  15.0 V  5.00  I1  5.00  I 2  0  I1  I 2  3.00 A . Lower loop: 5.00 V  1.00   4.00  I 2  10.0  I 3  0  5.00 V  5.00  I 2  10.0  I 3  0  I 2  2 I 3  1.00 A . Along with I1  I 2  I 3 , we can solve for the three currents and find: I 1  1.60 A, I 2  1.40 A, I 3  0.200 A. b) Vab   1.40 A 4.00    1.60 A 3.00    10.4 V.
9. 26.24: After switching the 5-V battery for a 20-V battery in the circuit of Fig. 26.42, there is a change in the equations from Problem 26.22 in both the upper and lower loops: Upper loop: 10.0 V  2.00   3.00  I1  1.00   4.00  I 2  20.00 V  0  10.0 V  5.00  I1  5.00  I 2  0  I1  I 2  2.00 A . Lower loop: 20.00 V  1.00   4.00  I 2  10.0  I 3  0  20.00 V  5.00  I 2  10.0  I 3  0  I 2  2 I 3  4.00 A . Along with I1  I 2  I 3 , we can solve for the three currents and find: I 1  0.4 A, I 2  1.6 A, I 3  1.2 A. b) I 2 4    I1 3    1.6 A 4    0.4 A 3    7.6 V 26.25: The total power dissipated in the four resistors of Fig. 26.10a is given by the sum of: P2  I 2 R2  0.5 A  2    0.5 W, P3  I 2 R3  0.5 A  3    0.75 W, 2 2 P4  I 2 R4  0.5 A  4    1 W, P7  I 2 R7  0.5 A  7    1.8 W. 2 2  Ptotal  P2  P3  P4  P7  4 W. 26.26: a) If the 12-V battery is removed and then replaced with the opposite polarity, the current will flow in the clockwise direction, with magnitude;  ε 12 V  4 V I   1 A. R 16  b) Vab  R4  R7 I  ε 4  4   7   1 A   4 V  7 V.
10. 26.27: a) Since all the external resistors are equal, the current must be symmetrical through them. That is, there can be no current through the resistor R for that would imply an imbalance in currents through the other resistors. With no current going through R, the circuit is like that shown below at right. So the equivalent resistance of the circuit is 1  1 1  13 V Req   2   2    1   I total  1   13 A.    1  I each leg  I total  6.5 A, and no current passes through R. 2 b) As worked out above, Req  1  . c) Vab  0, since no current flows. d) R does not show up since no current flows through it.
11. 26.28: Given that the full-scale deflection current is 500 A and the coil resistance is 25.0  : a) For a 20-mA ammeter, the two resistances are in parallel: Vc  Vs  I c Rc  I s Rs  500  10 6 A 25.0    20  10 3 A  500  10 6 A Rs  Rs  0.641  b) For a 500-m voltmeter, the resistances are in series: V Vab  I Rc  Rs   Rs  ab  Rc I 3 500  10 V  Rs   25.0   975 . 500  10  6 A
12. 26.29: The full-scale deflection current is 0.0224 A, and we wish a full-scale reading for 20.0 A. 0.0224 A 9.36   R   20.0 A  0.0224 A 0.0250   0.499 A R  9.36   12.9 . 0.0224 A ε 90 V 26.30: a) I    0.208 A Rtotal 8.23   425    V  ε  Ir  90 V  0.208 A 8.23    88.3 . εr εRV ε r ε b) V  ε  Ir  ε       1. r  RV r  RV r / RV   1 RV V r 90 Now if V is to be off by no more than 4% it requires:   1  0.0416. RV 86.4 26.31: a) When the galvanometer reading is zero: R x ε2  IRcb and ε1  IRab  ε2  ε1 cb  ε1 . Rab l b) The value of the galvanometer’s resistance is unimportant since no current flows through it. x 0.365 m c) ε2  ε1  9.15 V   3.34 V. l 1.000 m
13. 26.32: Two voltmeters with different resistances are connected in series across a 120-V V 120 V line. So the current flowing is I    1.20  10 3 A. But the current Rtotal 100  10  3 required for full-scale deflection for each voltmeter is: 150 V 150 V I fsd (10 k )   0.0150 A and I fsd 90 k    1.67  10 3 A. 10,000  90,000  So the readings are:  1.20  10 3 A   1.20  10 3 A  V10 kΩ  150 V  0.0150 A    12 V and V90 kΩ  150 V  1.67  10 3 A   108 V.      26.33: A half-scale reading occurs with R  600 . So the current through the galvanometer is half the full-scale current.  3.60  103 A   ε  I Rtotal  1.50 V    15.0   600   Rs   Rs  218 .   2  26.34: a) When the wires are shorted, the full-scale deflection current is obtained: ε  IRtotal  1.52 V  2.50  10 3 A 65.0   R   R  543 . V 1.52 V b) If the resistance Rx  200  : I    1.88 mA. Rtotal 65.0   543   Rx ε 1.52 V 1.52 V c) I x    Rx   608 . Rtotal 65.0   543   Rx Ix 1 1.52 V So: I x  I fsd  6.25  10 4 A  Rx   608   1824 . 4 6.25  10 4 A 1 1.52 V I x  I fsd  1.25  10 3 A  Rx   608   608 . 2 1.25  10 3 A 3 1.52 V I x  I fsd  1.875  10 3 A  Rx   608   203 . 4 1.875  10 3 A V Q   Q   Q  26.35: RC        t   I V   I  Q t    
14. 26.36: An uncharged capacitor is placed into a circuit. a) At the instant the circuit is completed, there is no voltage over the capacitor, since it has no charge stored. b) All the voltage of the battery is lost over the resistor, so VR  ε  125 V. c) There is no charge on the capacitor. ε 125 V d) The current through the resistor is i    0.0167 A. Rtotal 7500  e) After a long time has passed: The voltage over the capacitor balances the emf: Vc  125 V. The voltage over the resister is zero. The capacitor’s charge is q  Cvc  (4.60  10 6 F) (125 V)  5.75  10 4 C. The current in the circuit is zero. q 6.55  108 C 26.37: a) i   10  1.12  10 4 A. RC (1.28  10 ) (4.55  10 F) 6 10 4 b) τ  RC  (1.28  10 ) (4.55  10 F)  5.82  10 s. 6 26.38: τ 4.00 s v  v0 e  τ / RC  C    8.49  10 7 F. R ln(v0 / v) (3.40  10 ) (ln (12 / 3)) 6
15. 26.39: a) The time constant RC  (0.895  106 ) (12.4  106 F)  11.1 s. So at : t  0 s : q  Cε (1  e t / RC )  0. t  5 s : q  Cε (1  e  t / RC )  (12.4  106 F) (60.0 V) (1  e  ( 5.0 s ) /(11.1 s ) )  2.70  10  4 C. t  10 s : q  Cε (1  e  t / RC )  (12.4  10 6 F) (60.0 V) (1  e  (10.0 s) /(11.1 s ) )  4.42  10  4 C. t  20 s : q  Cε (1  e  t / RC )  (12.4  106 F) (60.0 V) (1  e  ( 20.0 s ) /(11.1 s ) )  6.21  10 4 C. t  100 s : q  Cε (1  e  t / RC )  (12.4  106 F) (60.0 V) (1  e  (100 s ) /(11.1 s ) )  7.44  10 4 C. ε  t / RC b) The current at time t is given by: i  e . So at : R 60.0 V t  0 s:i  e 0 /11.1  6.70  10 5 A. 8.95  105  60.0 V t  5s:i  e 5 / 11.1  4.27  10 5 A. 8.95  10  5 60.0 V t  10 s : i  e 10 / 11.1  2.27  10 5 A. 8.95  105  60.0 V t  20 s : i  e  20 / 11.1  1.11  105 A. 8.95  10  5 60.0 V t  100 s : i  e 100 / 11.1  8.20  109 A. 8.95  105  c) Charge against time: Current against time:
16. 26.40: a) Originally,   RC  0.870 s. The combined capacitance of the two identical capacitors in series is given by 1 1 1 2 C    ; Ctot  Ctot C C C 2 The new time constant is thus R ( C )  0.870 s  0.435 s. 2 2 b) With the two capacitors in parallel the new total capacitane is simply 2 C. Thus the time constant is R(2C )  2(0.870 s)  1.74 s. 26.41: ε  VR  VC  0 ε  120 V, VR  IR  (0.900 A) (80.0 )  72V, so VC  48 V Q  CV  (4.00  10 6 F) (48V)  192 μC) 26.42: a) Q  CV  (5.90  10 6 F) (28.0 V)  1.65  10 4 C. q t b) q  Q (1  e t / RC )  e  t / RC  1   R  . Q C ln(1  q / Q)  3  103 s After t  3  10 3 s : R   463 . (5.90  10 6 F) (ln(1  110 / 165)) c) If the charge is to be 99% of final value: q  (1  e t / RC )  t   RC ln(1  q / Q) Q   (463 ) (5.90  10 6 F) ln(0.01)  0.0126 s. 26.43: a) The time constant RC  (980 ) (1.50  10 5 F)  0.0147 s. t  0.05 s : q  Cε (1  e  t / RC )  (1.50  105 F) (18.0 V) (1  e 0.010 / 0.0147 )  1.33  104 C. ε 18.0 V  0.10 / 0.0147 b) i  e t / RC  e  9.30  103 A. R 980   VR  IR  (9.30  10 3 A) (980 )  9.11 V and VC  18.0 V  9.11 V  8.89 V. c) Once the switch is thrown, VR  VC  8.89 V. d) After t  0.01 s : q  Q0 e  t / RC  (1.50  10 5 F) (8.89 V)e 0.01 / 0.0147  6.75  10 5 C.
17. P 4100 W 26.44: a) I    17.1 A. So we need at lest 14-gauge wire (good up to 18 V 240 V A). 12 gauge is ok (good up to 25 A). V2 V 2 (240 V) 2 b) P  R   14  R P 4100 W c) At 11c /kWhr  in 1 hour, cost (11c/kWhr) (1 hr )(4.1 kW)  45c.    26.45: We want to trip a 20-A circuit breaker: 1500 W P 1500 W 900 W I   With P  900 W : I   20 A. 120 V 120 V 120 V 120 V 26.46: The current gets split evenly between all the parallel bulbs. A single bulb will P 90 W 20 A draw I    0.75 A  Number of bulbs   26.7. So you can attach V 120 V 0.75 A 26 bulbs safely. V 120 V 26.47: a) I    6.0 A  P  IV  (6.0 A) (120 V)  720 W. R 20  b) At T  280 C, R  R0 (1  αT )  20  (1  (2.8  10 3 (C) 1 (257C))  34.4 . V 120 V I   3.49 A  P  (3.49 A) (120 V)  419 W. R 34.4 A
18. 26.48: a) 1  1 1   RR  Req  R3    R   R3   1 2    R1  R   1 R2   2   RR  R12 If Req  R1  R3  R1   1 2   R  R  R R .  1 2  1 2 1  1 1  R ( R  R2 ) b) Req   R  R    3 1  1 2 R3   R1  R2  R3 If Req  R1  R1 ( R1  R2  R3 )  R3 ( R1  R2 )  R3  R1 ( R1  R2 ) / R2 . 26.49: a) We wanted a total resistance of 400 Ω and power of 2.4 W from a combination of individual resistors of 400 Ω and 1.2 W power - rating. b) The current is given by: I  P / R  2.4 W / 400   0.077 A. In each leg half the current flows, so the power in each resistor in each resistor in each combination is the same: P  ( I / 2) 2 R  (0.039 A) 2 (400 )  0.6 W.
19. 26.50: a) First realize that the Cu and Ni cables are in parallel. 1 1 1   RCable RNi RCu L RNi  ρNi L / A  ρNi πa 2 L RCu  ρCu L/A  ρCu π (b  a 2 ) 2 1 πa 2 π (b 2  a 2 ) So:   Rcable ρ Ni L ρCu L π  a 2 b2  a 2    ρ  ρ   L  Ni Cu  π  (0.050 m) 2 (0.100 m) 2  (0.050 m)2    7.8  10 8 m   20m  1.72  10 8 m  RCable  13.6  10 6   13.6 μ L L b) R  ρeff   eff A πb 2 πb 2 R π (0.10 m) 2 (13.6  10 6 ) ρeff   L 20 m  2.14  10 8 m
20. 26.51: Let R  1.00 , the resistance of one wire. Each half of the wire has Rh  R 2. The equivalent resistance is Rh  Rh 2  Rh  5 Rh 2  5 (0.500 ) 1.25  2 26.52: a) The equivalent resistance of the two bulbs is 1.0 . So the current is: V 8.0 V I   4.4 A  the current through each bulb is 2.2 A. Rtotal 1.0   0.80  Vbulb  ε  Ir  8.0 V  (4.4 A) (0.80 )  4.4 V  Pbulb  IV  (2.2 A) (4.4 V)  9.9 W b) If one bulb burns out, then V 8.0 V I   2.9 A  P  I 2 R  (2.9 A) 2 (2.0 )  16.3 W, Rtotal 2.0   0.80  so the remaining bulb is brighter than before. 26.53: The maximum allowed power is when the total current is the maximum allowed value of I  P / R  36 W / 2.4   3.9 A. Then half the current flows through the parallel resistors and the maximum power is: 3 2 3 Pmax  ( I / 2) 2 R  ( I / 2) 2 R  I 2 R  I R  (3.9 A) 2 (2.4 )  54 W. 2 2