Answers to Electricity Homework
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Electric Fields
1.a. Ew = 8x10-17J
b. Ew = Ek(gain) = 8x10-17J
c. v = 13.26x106m/s
2. v = 26.52x106m/s
3. pd = 25V
4. E = 30J
5. E (1995 PI Q11)
6. D (1992 PI Q13)
7. D (1994 PI Q11)
8. v= 29.63x106m/s (1999 PI Q33)
9.a.i. Ek = 8.03x10-16J (1998 PII Q7ai)
a.ii.V = 5021.9V (1998 PII Q7aii)
b. (1998 PII Q7b)
10.a.i. v = 41.9x106m/s (1992 PII Q7ai)
a.ii.v2 = v1/sqrt2 (1992 PII Q7aii)
b. N = 9.375x1016 (1992 PII Q7b)
Resistance Networks
11. R = 8.33kW
12. R = 6.27kW
13. R = 2W
14. R = 1.33W
15. R = 40W
16. R = 81.58W
17. C (2.4W)
18.a.
b.
c.
19. B (1999 PI Q11)
20. C (1995 PI Q12)
21. C (1994 PI Q12)
22. E
23. Rp = 7.5W (1992 PI Q34)
24. C
Ohm's Law
25. I = 0.05A
26. I = 0.44mA
27. I = 0.135A
28. VR = 2V
29. VR = 3V
30. VR1 = 6V, VR2 = 6V
31. D (1995 PI Q13)
32. C (1994 PI Q13)
33. E (1997 PI Q13)
34. D (1998 PI Q11)
35. Rtotal = 2W (1998 PI Q33)
Current, Voltage, Resistance and Power
36. A (1997 PI Q15)
37. C (1996 PI Q16)
38. D (1995 PI Q14)
Internal resistance and EMF
39. emf = 5V
40. Vtpd = 4V
41. 1V is dropped across the internal resistor.
42. The 1V is called the "lost" volt. (Vlost = emf - Vtpd)
43. r = 10W
44. emf = 16V
45. I = 0.25A
46. r = 5W
47. R = 12W
48. No current is drawn by a load in circuit A.
This means there will be no voltage drop across
the internal resistance. With no lost volts the
voltmeter reading will be equal to the emf.
49. Vtpd = 4V
50. E (r = 10W)
51. B
52. B (1994 PI Q14)
53. A (1996 PI Q13)
54. D (1996 PI Q14)
55. D (1999 PI Q14)
56. C (1998 PI Q12)
57. E (1993 PI Q14)
58.a. Vlost = 0.10V (1999 PII Q6a)
b. r = 2.5W (1999 PII Q6b)
c.i. R2 = 47.5W (1999 PII Q6ci)
c.ii. Vtpd decreases (1999 PII Q6cii)
59.a.i. Vtpd = 1.2V (1998 PII Q5ai)
a.ii.Vlost = 0.3V (1998 PII Q5aii)
60. (1995 PII Q6a)
61.a. (1997 PII Q5a)
b.i. R = 3W (1997 PII Q5bi)
b.ii. (1997 PII Q5bii)
c.i. Rtotal = 0.165W (1997 PII Q5ci)
c.ii.I = 72.7A (1997 PII Q5cii)
62.a.i. Q = 1800C (1994 PII Q6ai)
a.ii.E = 2160J (1994 PII Q6aii)
b.i. (1994 PII Q6bi)
b.ii.emf = 1.4V (1994 PII Q6bii)
r = 4W
63.a. (1995 PII Q7a)
bi. r = 2.5W (1995 PII Q7bi)
b.ii.emf = 17.1V (1995 PII Q7bii)
c. I = 6.8A (1995 PII Q7c)
64.i. Vtpd = 4V (1998 PII Q5bi)
ii. r = 0.5W (1998 PII Q5bii)
65.a. r = 1.5W (1997 PI Q35a)
b. I = 3.67A (1997 PI Q35b)
66.a. (1993 PII Q6a)
b.i. I = 5A (1993 PII Q6bi)
b.ii. V = 1.6V (1993 PII Q6bii)
b.iii.r = 0.08W (1993 PII Q6biii)
c. (1993 PII Q6c)
Potential Dividers
67. V(R2) = 4V
68. V(R2) = 1V
69. V(R2) = 3V
70. D (1998 PI Q13)
71. B
72. D (1996 PI Q12)
73.a. V = 3.6V (1999 PI Q34a)
b. (1999 PI Q34b)
74.a. Vmeter = 5V (1992 PI Q33a)
b. Vmeter = 2.5V (1992 PI Q33b)
75. Rx = 4W
76. Rx = 25.38W
77. Rx = 45W
78. Rx = 9W
79. Rx = 200W
80. B (1992 PI Q16)
81. B (1998 PI Q14)
82. B
83. D
84. D (1994 PI Q15)
85. B (1999 PI Q15)
86a. False
b. True
C(b only)
87. The wheatstone bridge is balanced if:
R1/R2 = R3/R4
R1/R2 = 80/50 = 1.6
R3/R4 = 240/150 = 1.6
=>Balanced wheatstone bridge.
88. Zero
89. V80 = 3V
90. V50 = 3V
91. V240 = 9V
92. V150 = 9V
93. Zero
94.a. (1993 PI Q33a)
b. None (1993 PI Q33b)
95. E
96.i. RA/RB = RC/RD (1994 PII Q5bi)
OR RA/RC = RB/RD
ii. (1994 PII Q5bii)
| Resistance of A/W | Resistance of B/W | Voltmeter reading/mV |
| 120 | 120 | 0 |
| 121 | 120 | -21 |
| 121 | 121 | 0 |
| 121 | 122 | +21 |
| 121 | 119 | -42 |
97.a. Rthermistor = 810W (1999 PII Q5a)
b.i. Mean = 851.6W (1999 PII Q5bi)
b.ii.Random Error = 1.2W (1999 PII Q5bii)
c. (1999 PII Q5c)
98a. Rt = 150W (1998 PI Q34a)
b. T = 24oC (1998 PI Q34a)
99.a.i. 0V (1996 PII Q5ai)
a.ii.Rplatinum = 0.5KW (1996 PII Q5aii)
b.i. Rplatinum = 420W (1996 PII Q5bi)
b.ii.V1K = 6.3V (1996 PII Q5bii)
b.iii.Voltmeter reading = 0.3V(1996 PII Q5biii)
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