Quiz 18 (ANSWERS)   Conceptual Questions (CQ): 7, 8, 9 Problems: 1, 2, 7, 8, 10, 13, 14, 16, 21, 26, 27, 30, 35, 38, 39, 41

TURN IN:  Conceptual Questions (CQ): 7, 8, 9 and Problems 2, 8, 10, 14,16, 26, 30, 38

Problem Discussions Follow:

2. See example 18.1.

8.  Find the current I using Ohm's Law. Then see example 18.1 to find the magnitude of the conventional charge (delta Q).  Then divide (delta Q) by the magnitude of the charge of the electron e = 1.6x10-19 C.

10. (a) Find R using Ohm's Law, section 18.3. Then find the new current using Ohm's Law when the Voltage becomes 15% less than the original voltage which you used to find R.  The new voltage is 15 % less than 240 (V). What does this mean? Think about a sale on shoes at Macy's. For example, how much do $80.00 shoes cost when they are marked down by 15 %,  i.e. are "15% off." Once you find the reduced  voltage, then find the reduced current from Ohm's Law. (b) Now instead of the voltage being reduced, the resistance is reduce by 15%.  In other words , the reduced resistance is 15% less than the resistance you computed in part (a). With this new value of R, find the new current using Ohm's Law when the voltage is 240 (V).

14. See example 18.5.  Find the resistance of the 10.0-long Al wire with radius 2.0 mm, remembering to convert to meters.  Then find the resistance of the 20.0-long Cu wire with radius 2.5 mm, again  converting to meters. Then find the ratio of these two resistances.

16. Use this logic: If the wire has one part that is 4 times the other, then the wire must be divided into 1/5  ths or 5 parts.  Starting with equation 18-3, you will find that you must solve this: x = 4*( L - x) , where L = is the wire length and x is the distance from the left end of the wire where the cut will be made.  Solve for x in terms of L to get where to cut the wire.  Once you find x will note that after cutting the wire, the longest section will have 4 times the resistance of the short section. Divide the total 10 ohm resistance into  5 parts. One part has the resistance of  the short  wire segment. The sum of the  other 4 parts is the  long wire segment's resistance.

26. V2/R = 3.3x10 3 (W) , where V is the magnitude of the voltage change across the electric oven's resistance.

30. V2/R =110 (W) , where V is the magnitude of the voltage change across the electric oven's resistance.    Note you can also write :  V2/R= I2*R. With this last relation you can find all of the requested quantities.

38. Power dissipate in the cord = 2*I2*R , where R is the resistance of each identical wire. Use equation 18-3 to find R for a single copper length of the wire given its length L and  and given its diameter, which gives you  area A.   The current I is given.