ANSWERS

Quiz  6 Chapter 25  problems:
Quiz 6
Ch. 25- 4, 6, 8, 10, 21, 22, 29, 30, 35, 36, 38, 39, 42, 58, 92, 93

Simulation: Click here for the motion of an electron in a uniform electric field.

Click here for a  great derivation of the resistivity in terms of the mean free collision time, the electron mass m, the electronic charge magnitude e. http://www.nvaphysics.com/4BAU09/quizzes/4B_Au_09_quiz5.htm
4. The heating element of a toaster is a resistor that gets so hot, it glows. It radiates like a   black body object, and thus you can estimate it's  temperature by the color of its reddish glow--we will cover this in Physics 4C.  Use Ohm's Law to find R.
error log #1 : See correction to #8--you must include the distance since R is resistance per unit meter--be careful. 
error log #2: Corrected  #42 (b) I = V/R , where V = 6.0 (V).  Note: R has the same value as in part (a); current I will be doubled if V is doubled.
error log #3: Corrected #22 For example in (a) , the area is (2)*(4) cm 2 and the length is 1.0 cm.
6. 
(a) SEE THE PREVIOUS PROBLEM. 
(b) CHARGE = CURRENT*TIME. REMEMBER TO COVERT TIME TO SECONDS..
8. Birds can stand on high voltage lines without being electrocuted-- a process by which your heart stops when  a large enough current  passes through it. That can only happen if the voltage difference across your body is large enough to support such a current . The distance between  a bird's feet is small enough to avoid electrocution. Remember  
| voltage difference | = I*R*distance, where R = (resistance per meter) = 0.000025 ohm/meter.
10.
(a) Find the resistance R from the given information : |voltage difference| = 240 (V) and current I = 6.50 (A). Use  Ohm's Law. Then , for the R value, find the new current when the voltage has dropped by an amount equal to 15 % of 240 (V). 
(b) Let the resistance drop by an amount  equal to 15 % of R computed in  (a). Use Ohm's Law to find the current I when
| voltage difference| = 240 (V)
21. R = (resistivity)*length/area. When you find the ratio, the resisitivity will cancel out. For example,
R1/R2 = 1 = (area 2*length 1)/(area 1*length 2). Use the formula for the area of a circle to find the ratio of diameters.
22. Use R = (resistivity)*length/area. This is easy; for the current  of each direction, examine the area.  For example in (a) , the area is (2)*(4) cm 2 and the length is 1.0 cm. Please convert to meters. Use the resistivity of carbon. Use the same  method for (b) and (c).
29.
(a) RNET = R1 +R2 = (resistivity 1)*(length 1)/(area 1) + (resistivity 2)*(length 2)/(area 2). Note that area 1 = area 2  and length 1 = length 2. The only difference between substance 1 and substance 2 is the resistivity.

(b) I =  85x10 -3 (V)/ RNET .
(c) This is the fun part. Across substance 1, | voltage difference | = I*R1 .   Across substance 2,
| voltage difference | = I*R2 .  
30.
(a) This is for current flow perpendicular to  cylinder's central axis.  Following the hint, the differential resistance  for the differential shell of radius  r' is dR = (resistivity)*(dr')/A, where A = surface area of  the  curvy part of a cylindrical shell of radius r'.  A = 2*pi*r'*L, where L is the length of the hollow cylinder .  Use your calculator  to compute a precise value of pi.
Integrate  (resistivity)*(dr')/2*pi*r'*L  between the two given limits .
(b) Plug in the numbers.
(c) For current along the axis, R = (resistivity)*(length )/(area), where area = area  of a ring of thickness
r2 - r1 .  Note:  area = ( area of circle of radius r2 ) -  ( area of circle of radius r1 ).
35.
(a) See section 25-5. Think of a battery of voltage  V in series with a resistor R and RL.  For example see figure 26-7 (b). Take V = 120 (V),  take the 60 W-rated resistor to be R and take the 100 W-rated resistor to be RL.  

Here V = V+ - V- , where V + = voltage of battery's positive terminal and V- = voltage of battery's negative terminal.

R = resistance of the power line leading to the load resistance.
RL = load resistance. 
Again, see figure 26-7(b) and make your mental substitutions.

You want the rate of heat production in  resistor  R, where R =  resistance in the line leading  to load resistor RL.  This quantity is   the  difference  between the  rate of heat production in the load resistor RL and  the total power delivered to the circuit P = I*V,  where I is the circuit current  and V is the battery voltage. This difference must  equal the rate of heat production in the resistor  R. Think about it for a few minutes. 

Note:  P = (rate of heat production in resistor R) + (rate of heat production in resistor RL ). You want to evaluate (rate of heat production in resistor R) = P  - (rate of heat production in resistor RL).

Note that (rate of heat production in resistor RL) = I2*RL . = I*I*  RL.  Note that for the entire circuit,
I = V/(R + RL). Thus I*I*  RL  = I*[V/((R + RL)]*RL  . = I*V*[RL/(R + RL)] = P*[RL/(R + RL)].

Thus you want: P -  P*[RL/(R + RL)]. = P*[R/(R + RL)] .   Note that (R + RL) = V/I. Thus,
 P*[R/(R + RL)]   = P*I*R/V.  You should be able to manipulate this with substitutions to show that
 P*I*R/V = P2R/V2.

(b)  Study the expression to reach your conclusion.

36. P = I*V = I2*R = V2/R.  Thus, P = V2/R.  V = 120 (V)  and  P = 850 (W) or 1250 (W). Evaluate R in each case. Which is bigger? 
38. P = I*V = I*I*R = I2*R in general, where R is the bulb resistance.  In Europe, V = 240 (V) and in the US,  V = 120 (V). 
In  Europe:  P = I2*R, where  I = V/R and V = 240(V).
In US:  P'  = I'2*R, where I' = V'/R and V = 120 (V). 
Find the ratio P/P'. It turns out the bulbs are brighter in Europe. By how much? 
39. kWh, spelled kilowatt hour, is a unit of energy (J), since kilo watt has units J/s and 1 h =  1 hour = 3600 seconds. Before you begin this problem, please covert 1 kW-h into joules. Note 1kW = 1000 W. Find the energy (in J)  consumed by the 550 W toaster for a period of 360 seconds ( = 6 minutes) .  Then find the cost by multiplying by 
9.0 cents/ kWh, which should be  converted to cents/J. 
42.  
(a) I = V/R, where V = 3.0 (V). Find R. Note: 1 (mA)  is 10-3 (A).  P = I*V = V2/R = I2*R; there are three ways to compute it.  
(b) I = V/R , where V = 6.0 (V).  Note: R has the same value as in part (a); current I will be doubled if V is doubled. Both  I and V are doubled together producing an effect increasing the power consumption by a factor of 4.  P = I*V = V2/R = I2*R; again, there are three ways to compute it.  Find the ratio of the power for this part with the power P of part (a). Show that power consumption, in this case the rate of heat production in the circuit, is increased by a factor of 4. 
58.  Start with I = V/R and find R from the given information for I and V. Then use R = (resistivity)*L/A to get the resistivity if you want.
(a) See above comments.
(b) See above comments.
(c) J = I/A.
(d) Ex = (resistivity)*J
(e) J = n*e*vd.  Find n.  Can you state what material this is? 
(f) Find the *mean free time* between collisions as discussed in  class. Use the published electron mass m and charge magnitude e. This part not required but is assigned as a thought problem. 
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error log #1 : See correction to #8--you must include the distance since R is resistance per unit meter--be careful. 
error log #2: Corrected  #42 (b) I = V/R , where V = 6.0 (V).  Note: R has the same value as in part (a); current I will be doubled if V is doubled.
error log #3: Corrected #22 For example in (a) , the area is (2)*(4) cm 2 and the length is 1.0 cm.

More hints soon. Check back later.