|QUIZ 19 CH. 19|
SELECTED DISCUSSIONS TO exercises/problems
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.|
|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.
(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
(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.