QUIZ 19  CH. 19

QUIZ 19

SAMPLE EXAM 1 SP'11 SAMPLE EXAM 1 SP'12

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)*d2] = 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 k-ohm. 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) = V2/R (b) I = V/R, where V is the magnitude of the voltage drop across R. (c) P*time*7.4 cents/KW-h = cost. Note: time = 1 h. Convert P =  540 (W) to (kW) = 1000 (W). (d) P =  V2/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) I90 = 120 (V)/90  and I40 = 120 (V)/40.

52. (a) Moving left to right on the bottom branch of circuit:  1/Rp1 = 1/3.00  + 1/6.00 and  1/Rp2 = 1/12.0  + 1/4.00. Find Rp1   + Rp2 . (b) Now, 60.0 (V) = I* ( Rp1   + Rp2 ) . Find I. Then find I* Rp1  and I* Rp2  . From these latter two values you can find  each of 4 currents: I3 = I* Rp1 /3, I6 = I* Rp1 /6, I12 = I* Rp2 /12 and  I4 = I* Rp2 /4 .

57. See example 11. At top Node:  I = I30 + I20,  where I flows through the 10.0 (V)-battery upward in the branch, and I30 and I20 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,  I30 & I20. 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) - I30*(30 ohm) = 0 and counterclockwise in the right loop, 5.00 (V) +  I20*(20 ohm) -  I30*(30 ohm) = 0. Note the positive term   I20*(20 ohm)  in the last equation due to the fact  we are going counter clockwise in the right loop and thus against current I20 .

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  process-Q on the capacitor's positive plate  grows according to the second of  equations 19.17 . Q = Qm*(1 - e-t/RC). Note: Qm = maximum charge at the end of the process - when current I = 0. Qm = 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.