RC Cola Isn't Just A Soft Drink....

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 - the NEW resistors? More on Page 7!

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

72. (a) When the capacitors are connected in parallel, we find the equivalent capacitance from
Cparallel = C1 + C2 = 0.40 µF + 0.60 µF = 1.00 µF.
The stored energy is
Uparallel = 0.5*Cparallel*V^2 = (1.00 x 10–6 F)(45 V)^2 = 1.0 x 10^–3 J.
(b) When the capacitors are connected in series, we find the equivalent capacitance from
1/Cseries = (1/C1) + (1/C2) = [1/(0.40 µF)] + [1/(0.60 µF)], which gives Cseries = 0.24 µF.
The stored energy is
Useries = 0.5*CseriesV^2 = 0.5*(0.24 x 10–6 F)(45 V)^2 = 2.4 ´ 10^–4 J.
(c) We find the charges from
Q = CeqV;
Qparallel = CparallelV = (1.00 µF)(45 V) = 45 µC.
Qseries = CseriesV = (0.24 µF)(45 V) = 11 µC.

There was some internal resistance to the revolution - Batteries & Internal Resistance

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
29.
For the conservation of current at point c, we have
Iin = Iout ;
I1 = I2 + I3 .
When we add the internal resistance terms for the two
loops indicated on the diagram, we have
loop 1: V1 – I2R2 – I1R1 – I1r1 = 0;
+ 9.0 V – I2(15 W) – I1(22 W) – I1(1.2 W) = 0;
loop 2: V3 + I2R2 + I3r3 = 0;
+ 6.0 V + I2(15 W) + I2(1.2 W) = 0.
When we solve these equations, we get
I1 = 0.60 A, I2 = – 0.33 A, I3 = 0.93 A.

Oh, and here's something interesting I found:

February break clue #2:

February Break Assignment - Clue #1

Clue #1

Solutions will be posted tomorrow. Here are the solutions to Friday's work:

Problem 3.
a) This is a standard problem where you have to find Req. The trick here is to realize that the 3 ohm and 6 ohm resistors are in parallel since their ends are connected directly together. You should obtain that Req = 6 ohms for the entire circuit. The battery current is therefore 1 ampere, and since the 4 ohm resistor is in series with the battery, it also has a current of 1 ampere.

b) To show that the junction rule holds, you must demonstrate that the current into the junction (which is the current through the 3 ohm and 6 ohm resistors) is equal to the current out (the current going through the 4 ohm resistor). This emans the only thing missing is the current through the 3 ohm and 6 ohm resistors.

You can use Ohm's law to find that the voltage drop across the 4 ohm resistor is (4 ohm)*(1 A) = 4 V, which means the voltage left to drop across the 3 ohm and 6 in parallel is 2 V. Since both resistors are in parallel, they BOTH must have 2 volts across them. This allows you to calculate the current through them by Ohm's law, 2/3 A and 1/3 A respectively. Since 2/3 A + 1/3 A (current in) = 1 A (current out) we show that the node rule holds.

c) The only thing you must show is that for the large loop, the sum of potential differences adds to zero. Going clockwise from the negative terminal of the battery:

6 V - (2/3 A)(3 ohms) - (1 A)(4 ohms) = 0.

The rule holds!


1989B3.

a)
i. 40 W (direct application of P = IV)
ii. 20 W (calculation of power for a constant force, P = F*v)
iii. 60 W (the battery is supplying ALL of the power to the circuit, so it must be equal to the sum of the power used by the other circuit elements.)

b)
i. 20 V (Ohm's law)
ii. 10 V (You know the power is 20 W, and the current through the motor is 2 A. Using P = IV, the voltage must be 10 V)
iii. 30 V (Similar to part (a), the battery is the source of all potential difference in the circuit, so it must be the sum of the potential differences for the motor and resistor.)


c) 15 V (You can get this by calculating the new power required to lift the mass and using P = IV, or by using the fact that it says the voltage is directly proportional to the speed of the motor. )

d) 7.5 ohms (15 V potential difference across the motor, and the battery potential difference is still 30 V. Thus the resistor has 15 V across it, with the same 2 A current. This makes the resistance 7.5 ohms by Ohm's law.)

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

Also, here are a couple more circuit problems to make sure you're where you need to be. The answers are included. Click to see them full size (unless your eyes are really THAT good....)

Dr. StrangeCircuit, or How I Learned To Stop Worrying And Love Finding Req.

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

Series? Parallel? Neither? This question offends me.

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.

MC Solutions and Results

Hi everyone,

Here is a PDF of the multiple choice solutions and the class results sorted from the worst to the best questions for our class.

Please go through and start looking at the questions so you come in Tuesday with some ideas for what did/didn't go well.

See you all Tuesday,

EMW