**C. Circuits
(Chapter 17) Homework Part I**

**50 Points**

**17.1 Batteries and Direct Current**

**1, 2, 7, 9**

**17.2 Current and Drift Velocity**

**11, 12, 13, 14, 19**

** **

13: I = q/t solve for t

14: a. I = q/t b. E = DU = qDV

19: a. q = It. Then convert q into number of protons. Remember that each proton has a charge of 1.6 x 10-19C.

b. P = W/t = (# of protons x KE (in eV’s) x 1.6 x 10-19J/ev) / t

**17.3 Resistance and Ohm’s Law**

**20, 21, 22, 23, 25, 32, 37, 40, 42, 46, 48**

21: Realize that a perfect ohmic resistor means that it is NOT affected by temp. In the real world, however, there are no perfect ohmic resistors.

25: a. R = rL/A b.
A = pd^{2}/4. Therefore, half the diameter means 1/4 the
area. So, R increases by 4 times and
therefore I will decrease by 1/4.

37: R = R_{o}
(1+aDT). Remember that R = rL/A. You’ll need table 17.1 p. 550.

40: water boils at
100^{o}C

42: R = rL/A
solve for L. A = pd^{2}/4

48: first find the
initial current (I_{o}) then,
use I = V/R_{o}(1+aDT) to
determine second current (I). Subtract
I – I_{o} to determine DI.

**17.4 Electric Power**

**49, 50, 51, 52, 53, 56, 58, 60, 65,70**

Mr. Young Questions about Traveling
and Electricity (Yes, you must answer
these too) Don’t peek at the answers until you have tried them!

Young_{1} = Europe has 240 V coming out of their
electric sockets and we have 120 V, who needs more current to run
appliances? USA or EU?

Young_{2} = If you want to plug your 1200 W hair dryer
into an outlet in Europe, you know that you need to buy a “converter” before
you go “over the pond”. Does this
converter effectively increase or decrease the resistance in your hair
dryer? (Hint: P = V^{2}/R).

Young_{3} = Consider
the question posed by Young_{2}. What multiple would the resistance change in
Europe for your hair dryer? (Hint: USA has 120 V and Europe has 240 V – Europe
supplies 2 times the voltage as the US.
Hint #2: P = V^{2}/R). If you still need help, let us assume you
have resistance (R ) in the USA. What
would the effective resistance be in Europe when you hook up your
converter: 2R? 4 R?
1/2 R? 1/4 R?…..etc.

50: P = V^{2}/R

51: P = I^{2}R

53: P = V^{2}/R. Higher bulb has smaller resistance so it is
thicker.

65: a. E = Pt = Ivt

70: a. I = P/V b. Q = cmDT where c = specific heat of water =
4190 J/kg^{o}C. Water has a
mass of 3.785 kg/gal

Answers:

2: d

12: 0.25 A

14: a. 0.30 C b. 0.90 J

20: a

22: The one with the smaller slope

32: 75 V

40: 0.429W

42: 105 m

46: a. yes b. 2.0W

48: a decrease of 0.27 A

50: b

52: Power output would quadruple and it would overheat (see young questions)

56: 144 W

58: 1.2 W

60: a. 0.40 A b. 2.4 W

70: a. at least a 30 amp breaker b. 3.1 hours

Young_{1} = USA draws more current. Remember P = VI. So, if V is low (as it is in the USA, then I must compensate so
that the 1200 watt hair dryer can operate)

Young_{2} = The
effective resistance must increase.
Realize that V = IR. So, in the
US with 120 V, you need a resistance R and you get a current I. But, if V increases, your hair dryer can
only take a current I running through its wires. Therefore, you must increase the Resistance when the new 240
volts is applied.

Young_{3}= P = V^{2}/R. So, if V doubles, to 2V, (2V)^{2} = 4V^{2}. That means that R must increase by a factor
of 4. WOW, how cool! I ©
math, don’t you?