College Physics for AP® Courses

# Chapter 21

1.

(a) $2.75 kΩ2.75 kΩ size 12{2 "." "75"" k" %OMEGA } {}$

(b) $27.5Ω27.5Ω size 12{"27" "." 5 %OMEGA } {}$

3.

(a) $786Ω786Ω size 12{"786" %OMEGA } {}$

(b) $20.3Ω20.3Ω size 12{"20" "." 3 %OMEGA } {}$

5.

$29 . 6 W 29 . 6 W size 12{"29" "." 6" W"} {}$

7.

(a) 0.74 A

(b) 0.742 A

9.

(a) 60.8 W

(b) 3.18 kW

11.

(a) Rs=R1+R2⇒Rs≈R1R1>>R2Rs=R1+R2⇒Rs≈R1R1>>R2alignl { stack { size 12{R rSub { size 8{s} } =R rSub { size 8{1} } +R rSub { size 8{2} } } {} # drarrow R rSub { size 8{s} } »R rSub { size 8{1} } left (R rSub { size 8{1} } ">>"R rSub { size 8{2} } right ) {} } } {}

(b) $1Rp=1R1+1R2=R1+R2R1R21Rp=1R1+1R2=R1+R2R1R2 size 12{ { {1} over {R rSub { size 8{p} } } } = { {1} over {R rSub { size 8{1} } } } + { {1} over {R rSub { size 8{2} } } } = { {R rSub { size 8{1} } +R rSub { size 8{2} } } over {R rSub { size 8{1} } R rSub { size 8{2} } } } } {}$,

so that

R p = R 1 R 2 R 1 + R 2 ≈ R 1 R 2 R 1 = R 2 R 1 >> R 2 . R p = R 1 R 2 R 1 + R 2 ≈ R 1 R 2 R 1 = R 2 R 1 >> R 2 . alignl { stack { size 12{R rSub { size 8{p} } = { {R rSub { size 8{1} } R rSub { size 8{2} } } over {R rSub { size 8{1} } +R rSub { size 8{2} } } } » { {R rSub { size 8{1} } R rSub { size 8{2} } } over {R rSub { size 8{1} } } } } {} # =R rSub { size 8{2} } left (R rSub { size 8{1} } ">>"R rSub { size 8{2} } right ) "." {} } } {}

13.

(a) $-400 kΩ-400 kΩ size 12{ +- "400 k" %OMEGA } {}$

(b) Resistance cannot be negative.

(c) Series resistance is said to be less than one of the resistors, but it must be greater than any of the resistors.

14.

2.00 V

16.

2.9994 V

18.

$0 . 375 Ω 0 . 375 Ω size 12{0 "." "375" %OMEGA } {}$

21.

(a) 0.658 A

(b) 0.997 W

(c) 0.997 W; yes

23.

(a) 200 A

(b) 10.0 V

(c) 2.00 kW

(d) $0.1000Ω; 80.0 A, 4.0 V, 320 W0.1000Ω; 80.0 A, 4.0 V, 320 W size 12{0 "." "1000" %OMEGA ;" 80" "." "0 A, 4" "." "0 V, 320 W"} {}$

25.

(a) $0.400 Ω0.400 Ω size 12{0 "." "400" %OMEGA } {}$

(b) No, there is only one independent equation, so only $rr size 12{r} {}$ can be found.

29.

(a) –0.120 V

(b) $-1.41×10−2Ω-1.41×10−2Ω size 12{ +- 1 "." "41"´"10" rSup { size 8{-2} } %OMEGA } {}$

(c) Negative terminal voltage; negative load resistance.

(d) The assumption that such a cell could provide 8.50 A is inconsistent with its internal resistance.

31.

$− I 2 R 2 + emf 1 − I 2 r 1 + I 3 R 3 + I 3 r 2 - emf 2 = 0 − I 2 R 2 + emf 1 − I 2 r 1 + I 3 R 3 + I 3 r 2 - emf 2 = 0 size 12{ {underline {-I rSub { size 8{2} } R rSub { size 8{3} } + "emf" rSub { size 8{1} } - ital " I" rSub { size 8{2} } r rSub { size 8{1} } + ital " I" rSub { size 8{3} } r rSub { size 8{3} } + ital " I" rSub { size 8{3} } r rSub { size 8{2} } +- "emf" rSub { size 8{2} } =" 0"}} } {}$

35.

$I 3 = I 1 + I 2 I 3 = I 1 + I 2 size 12{I rSub { size 8{3} } = ital " I" rSub { size 8{1} } + ital " I" rSub { size 8{2} } } {}$

37.

$emf 2 - I 2 r 2 - I 2 R 2 + I 1 R 5 + I 1 r 1 - emf 1 + I 1 R 1 = 0 emf 2 - I 2 r 2 - I 2 R 2 + I 1 R 5 + I 1 r 1 - emf 1 + I 1 R 1 = 0 size 12{ {underline { "emf" rSub { size 8{2} } +- ital " I" rSub { size 8{2} } r rSub { size 8{2} } +- ital " I" rSub { size 8{2} } R rSub { size 8{2} } + ital " I" rSub { size 8{1} } R rSub { size 8{5} } +I rSub { size 8{1} } r rSub { size 8{1} } +- "emf" rSub { size 8{1} } + ital " I" rSub { size 8{1} } R rSub { size 8{1} } = 0}} } {}$

39.

(a) $I1=4.75 AI1=4.75 A size 12{I rSub { size 8{1} } =4 cdot "75 A"} {}$

(b) $I2 =-3.5 AI2 =-3.5 A size 12{I rSub { size 8{"2 "} } = +- 3 "." "5 A"} {}$${}$

(c) $I3=8.25 AI3=8.25 A size 12{I rSub { size 8{3} } =8 "." "25"" A"} {}$

41.

(a) No, you would get inconsistent equations to solve.

(b) $I1≠I2+I3I1≠I2+I3 size 12{I rSub { size 8{1} } <> I rSub { size 8{2} } +I rSub { size 8{3} } } {}$. The assumed currents violate the junction rule.

42.

$30 μA 30 μA size 12{"30" μA} {}$

44.

$1 . 98 k Ω 1 . 98 k Ω size 12{1 "." "98 k" %OMEGA } {}$

46.

$1 . 25 × 10 - 4 Ω 1 . 25 × 10 - 4 Ω size 12{1 "." "25"´"10" rSup { size 8{ +- 4} } %OMEGA } {}$

48.

(a) $3.00 MΩ3.00 MΩ size 12{3 "." "00 M" %OMEGA } {}$

(b) $2.99 kΩ2.99 kΩ size 12{2 "." "99 k" %OMEGA } {}$

50.

(a) 1.58 mA

(b) 1.5848 V (need four digits to see the difference)

(c) 0.99990 (need five digits to see the difference from unity)

52.

$15 . 0 μA 15 . 0 μA size 12{"15" "." 0 mA} {}$

54.

(a)

(b) $10.02 Ω10.02 Ω size 12{10 "." "02 " %OMEGA } {}$

(c) 0.9980, or a $2.0×10–12.0×10–1$ percent decrease

(d) 1.002, or a $2.0×10–12.0×10–1$ percent increase

(e) Not significant.

56.

(a) $−66.7Ω−66.7Ω size 12{-"66" "." 7 %OMEGA } {}$

(b) You can’t have negative resistance.

(c) It is unreasonable that $IGIG size 12{I rSub { size 8{G} } } {}$ is greater than $ItotItot size 12{I rSub { size 8{"tot"} } } {}$ (see Figure 21.33). You cannot achieve a full-scale deflection using a current less than the sensitivity of the galvanometer.

57.

24.0 V

59.

$1 . 56 k Ω 1 . 56 k Ω size 12{1 "." "56 k" %OMEGA } {}$

61.

(a) 2.00 V

(b) $9.68 Ω9.68 Ω size 12{9 "." "68 " %OMEGA } {}$

62.
$Range = 5 . 00 Ω to 5 . 00 k Ω Range = 5 . 00 Ω to 5 . 00 k Ω size 12{"Range=5" "." "00 " %OMEGA " to "5 "." "00"" k" %OMEGA } {}$
63.

$range 4 . 00 to 30 . 0 M Ω range 4 . 00 to 30 . 0 M Ω size 12{"range 4" "." "00 to 30" "." "0 M" %OMEGA } {}$

65.

(a) $2.50 μF2.50 μF size 12{2 "." "50 "mF} {}$

(b) 2.00 s

67.

86.5%

69.

(a) $1.25 kΩ1.25 kΩ size 12{1 "." "25 k" %OMEGA } {}$

(b) 30.0 ms

71.

(a) 20.0 s

(b) 120 s

(c) 16.0 ms

73.

$1 . 73 × 10 − 2 s 1 . 73 × 10 − 2 s size 12{1 "." "73" times "10" rSup { size 8{ - 2} } " s"} {}$

74.

$3 . 33 × 10 − 3 Ω 3 . 33 × 10 − 3 Ω size 12{3 "." "33"´"10" rSup { size 8{-3} } %OMEGA } {}$

76.

(a) 4.99 s

(b) $3.87ºC3.87ºC size 12{3 "." "87"°C} {}$

(c) $31.1 kΩ31.1 kΩ size 12{"31" "." "1 k" %OMEGA } {}$

(d) No

1.

(a), (b)

3.

(b)

5.

(a) 4-Ω resistor; (b) combination of 20-Ω, 20-Ω, and 10-Ω resistors; (c) 20 W in each 20-Ω resistor, 40 W in 10-Ω resistor, 64 W in 4-Ω resistor, total 144W total in resistors, output power is 144 W, yes they are equal (law of conservation of energy); (d) 4 Ω and 3 Ω for part (a) and no change for part (b); (e) no effect, it will remain the same.

7.

0.25 Ω, 0.50 Ω, no change

9.

a. (c)

b. (c)

c. (d)

d. (d)

11.
1. I1 + I3 = I2
2. E1 - I1R1 - I2R2 - I1r1 = 0; - E2 + I1R1 - I3R3 - I3r2 = 0
3. I1 = 8/15 A, I2 = 7/15 A and I3 = -1/15 A
4. I1 = 2/5 A, I2 = 3/5 A and I3 = 1/5 A
5. PE1 = 18/5 W and PR1 = 24/25 W, PR2 = 54/25 W, PR3 = 12/25 W. Yes, PE1 = PR1+ PR2 + PR3
6. R3, losses in the circuit
13.

(a) 20 mA, Figure 21.41, 5.5 s; (b) 24 mA, Figure 21.32, 2 s