QUIZ 19 CH. 19  
QUIZ 19  


SELECTED DISCUSSIONS TO exercises/problems
BELOW.
YOU ARE RESPONSIBLE FOR ALL PROBLEMS EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES. 

CH. 19 MULTIPLE CHOICE TBA  Exercise/problems 2 , 5, 9, 10, 15, 21, 26, 27, 40, 42, 46, 49, 52, 57, 64, 65, 66  
TURN IN: 2 ( try 5), 10 (try 9), 15, 21, 26, 27, 40 (try 42), 52 (try 49 and 46), 57, 65 (try 64 and 66)  
* DISCUSSIONS PROVIDED.  
DISCUSSIONS  
2. Q= I*t.  
10. R = (Cu resistivity)*L/A and R '
= ( Al resistivity)*L/A'. R/R' = (Cu resistivity)*A'/[(Al resistivity)*A] = (Cu resistivity)*d'^{2}/[( Al resistivity)*d^{2}] = 1. Find d given d' = 3.26 mm. 

15. (a) I = 18.5 (A) = change in
V/R , where R = (resistivity)*L/A at 20.0 degrees C.
Find resistivity. (b) 17.2 (A )= change in V/R' , where R' = (resistivity')*L/A, where resistivity' = resistivity*[1 + alpha*change in temp]; change in temp = (92.0  20.0) degrees C and resistivity was found in part a. Find alpha. 

21. I = change in V/R , where R varies between and 1 k ohm and 500 kohm. Set R to the lowest value (when you are wet) and I = 5.0 mA to find change in potential .  
26. 6.00 (V) = I*r + I*R = I*(r + R). Find r given R and I.  
27. Open circuit (switch open) means I = 0. Then the voltmeter would read the voltage difference across the battery only since there is no voltage drop across the resistors r and R. That battery voltage = 3.08 (V). When switch is closed, use battery voltage to solve: battery voltage = I*r + I*R for r and R. Use 2.97 (V) = battery voltage  I*r. Find r from the latter equation and solve for R in the former equation. Note: I = 1.65 (A) when switch is closed.  
40. (a) P = 540 (W) = V^{2}/R (b)
I = V/R, where V is the magnitude of the voltage drop across R. (c)
P*time*7.4 cents/KWh = cost. Note: time = 1 h. Convert P = 540
(W) to (kW) = 1000 (W). (d) P = V^{2}/R . We assume R is constant though we know it'll change as the temperature varies. 

49. (a) See Test #1, Problem
3(Sp 12) : 1/Rp = 1/40.0 + 1/90.0 so Rp = (40)*(90)/(40 + 90) which you
should simplify. (b) I = 120 (V)/Rp (c) I_{90} = 120 (V)/90 and I_{40} = 120 (V)/40. 

52. (a) Moving left to right on the bottom
branch of circuit: 1/R_{p1} = 1/3.00 + 1/6.00
and 1/R_{p2} = 1/12.0 + 1/4.00. Find R_{p1} + R_{p2} . (b) Now, 60.0 (V) = I* ( R_{p1} + R_{p2} ) . Find I. Then find I* R_{p1} and I* R_{p2} . From these latter two values you can find each of 4 currents: I_{3} = I* R_{p1} /3, I_{6} = I* R_{p1} /6, I_{12} = I* R_{p2} /12 and I_{4} = I* R_{p2} /4 . 

57. See example 11. At top Node: I = I_{30} + I_{20}, where I flows through the 10.0 (V)battery upward in the branch, and I_{30} and I_{20} flow downward in the branches for the 30 ohm and 20.0 ohm resistor/5.00 (V) battery, respectively. You need two other equations from which to find I, I_{30} & I_{20}. They are, using the law that the sum of the voltage drops around an internal loop of the circuit is zero: clockwise in the left loop, 10.00 (V)  I_{30}*(30 ohm) = 0 and counterclockwise in the right loop, 5.00 (V) + I_{20}*(20 ohm)  I_{30}*(30 ohm) = 0. Note the positive term I_{20}*(20 ohm) in the last equation due to the fact we are going counter clockwise in the right loop and thus against current I_{20} . 

65. This is a very important problem which
I will be sure to grade. It is an RC circuit from which we'll do a lab
later this coming month. (a) This is a charging processQ on the capacitor's positive plate grows according to the second of equations 19.17 . Q = Qm*(1  e^{t/RC}). Note: Q_{m} = maximum charge at the end of the process  when current I = 0. Q_{m} = C*V, where V = 60.0 (V). Note RC is easily computed from the given info. Plug in various times into the formula and you will notice Q increases with time t. (b) Plug in various times t into the first of equations 19.17: I = Io* e^{t/RC}. You will notice I decreases with time t. (c) Make a sketch. We 'll examine detailed graphs in upcoming RC lab. 