RC Circuits

From handout:

3.
a) 12-kOhm: .333 mA, 15 kOhm: .333 mA, 3 kOhm: 0 A
b) 50 microCoulombs

4. 2A

5.
a) 0.4 seconds
b) zero (capacitor and resistor are in series!)
c) 60 microCoulombs

6. 72 microCoulombs

From the book:
51.
a) 4.129 x 10^-5C
b) through .5, 6, and 5 ohm resistors: I = .635 A, through 8 ohm = .212 A, through 4 ohm = .424 A.

52.
a) C = 2.33 x 10^-9F

54.
a) 8V
b) 16 V
c) 8 V
d) 5.76 microcoulombs

Capacitors in Series/Parallel

40. minimum C = 1.36 x 10^-9F connected in series, maximum C = 1.95 x 10^-8F connected in parallel.

41. 300 pF connected in parallel.

42. C1 + (C2C3/(C2+C3)), Q on C1 = 562.5 microcoulombs, Q on C2 = Q on C3 = 375 microcoulombs

44
(a) For 0.4 microfarad capacitor, Q = 2 microcoulombs, V = 5 V
For 0.5 microfarad capacitor, Q = 2 microcoulombs, V = 4 V
(b) Both capacitors have 9V across them, For 0.4 microfarad capacitor, Q = 3.6 microcoulombs, 0.5 microfarad capacitor, Q = 4.5 microcoulombs

48. 5.128 pF

Kirchoff's Rules

1981B4.

a) 3A
b) 108 W
c) 6V

1989B3.

a) i. 40 W ii. 20 W iii. 60 W
b) i. 20 V ii. 10 V iii. 30 V
c) 15 V
d) 7.5 ohms

1983B3

a) 5 ohms
b) i. 4/3 A ii. 2/3 A
c) At point B: 10 V, at point C: -10 V, at point D: -2V
d) 40 W

1982B4

a) clock in parallel with the battery, radio in series with a resistor, and together in parallel with the battery.
b) 600 ohms
c) P = .45 W, energy = 27 J for a minute

Batteries & Internal Resistance

From Handout:

2.
a) 16 ohm current = .106 mA
b) battery current = 6.37 mA
c) New Req = 1413.8 ohms, terminal voltage = 8.987 V

3.
a) 2 A
b) 8V

From Textbook:
18 a) 8.406 V, b) 8.491 V
20. r = .4068 ohms
21. .06 ohms
24. current = .227 A

Circuit Problem Solutions

59. 99.8 Ohms
60. 6.762 Ohms
61. 4.601 Ohms
62. I = .083 A, P = .833 W
63. P = 2.222 W
64. R = 24.94 Ohms
65. current through 20 Ohm resistor = .75 A, current through 9 Ohm resistor = 2.11 A

Parallel and Combination Circuit HW

Equivalent Resistance HW:
1.
a) 2/3 R
b) 2 R
c) 15/2 R

2.
a) Req = 6 Ohms
b) Req = 5 Ohms

Book HW:

Questions:
2. Parallel lights - more difficult to connect, but all lights will stay on even if one goes out.
series lights - easy to connect lights together, but if one goes out, they all do!
4. For the bulbs in series, the current is the same. By P = I²R, the bulb with the largest resistance will have the highest power usage, and will therefore be the brightest.

If the bulbs are connected in parallel, the voltage is the same for each. The power equation becomes P = V²/R, which means the larger power will be used by the bulb with less resistance.

Problems.
2.
a) 360 Ohms
b) 8.89 Ohms
4. Maximum 2800 Ohms, minimum 261.4 Ohms
7. 4550 Ohms
13. 105.2 Ohms
24. I = 0.409 A.

Ohm's Law and Series Circuits HW

From Handout:

2. 1200 Ohms
3. 10 V, 20 Ohms
4.
b)60 Ohms,
c)0.2 A
d) 15 Ohm - 3 V, 21 Ohm - 4.2 V, 24 Ohm - 4.8 V
e) .6 W, .84 W, .96 W
f) .6W + .84 W _ ,96 W = 2.4 W OR (.2 A * 12 V) = 2.4 W
g) thickest filament - smallest resistance, since it has a larger cross sectional area. This must be the 15 Ohm resistor.
h) The brightest bulb uses the most power - this is the 24 Ohm bulb.

From book:
p. 551:
5. 35.5 Ohms
7. 2.1 x 10²¹ electrons
9. 24 Ohms, 360 W
26. Since P = V²/R, and V is the same for both, the resistance must be less for the higher power. Assuming a wall voltage of 120 V, the resistances are 24 Ohms and 12 Ohms for the 600 W and 1200 W settings.
31.
a) 8.57 Ohms, 1.05 W
b) 4 times the power. The bulb would probably not last for long!

p. 581
10. 27 Ohms