chapter 25 homework solutions - Physics and Astronomy at TAMU

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CHAPTER 25 HOMEWORK SOLUTIONS. 25.8.IDENTIFY: /. I Q t. = . Positive charge flowing in one direction is equivalent to negative charge flowing in.
CHAPTER 25 HOMEWORK SOLUTIONS 25.8.IDENTIFY:

I  Q / t . Positive charge flowing in one direction is equivalent to negative charge flowing in

the opposite direction, so the two currents due to Cl and Na + are in the same direction and add. SET UP: Na + and Cl  each have magnitude of charge q  e EXECUTE:

I

(a) Qtotal  (nCl  nNa )e  (3.92 1016  2.68 1016 )(1.60 1019 C)  0.0106 C. Then

Qtotal 0.0106 C   0.0106A  10.6 mA. t 1.00 s

(b) Current flows, by convention, in the direction of positive charge. Thus, current flows with Na  toward the negative electrode. EVALUATE: The Cl  ions have negative charge and move in the direction opposite to the conventional current direction.

25.27.

IDENTIFY: (a) SET UP: EXECUTE: (b) SET UP:

Apply R  R0 1   T  T0   to calculate the resistance at the second temperature.

  0.0004  C  (Table 25.1). Let T0 be 0.0C and T be 11.5C. 1

R0 

R 100.0    99.54  1   T  T0  1+ 0.0004  C 1 11.5 C 

  0.0005  C 



1



(Table 25.2). Let T0  0.0C and T  25.8C.





1 R  R0 1   T  T0    0.0160  1+ 0.0005  C   25.8 C    0.0158    EVALUATE: Nichrome, like most metallic conductors, has a positive  and its resistance increases with temperature. For carbon,  is negative and its resistance decreases as T increases.

EXECUTE:

25.32.

IDENTIFY: When current passes through a battery in the direction from the  terminal toward the + terminal, the terminal voltage Vab of the battery is Vab  E  Ir . Also, Vab  IR, the potential across the circuit resistor. SET UP: E  24.0 V . I  4.00 A. E  Vab 24.0 V  21.2 V EXECUTE: (a) Vab  E  Ir gives r    0.700 . I 4.00 A V 21.2 V  5.30 . (b) Vab  IR  0 so R  ab  I 4.00 A EVALUATE: The voltage drop across the internal resistance of the battery causes the terminal voltage of the battery to be less than its emf. The total resistance in the circuit is R  r  6.00  . 24.0 V I  4.00 A, which agrees with the value specified in the problem. 6.00 

25.36.

IDENTIFY: The sum of the potential changes around the circuit loop is zero. Potential decreases by IR when going through a resistor in the direction of the current and increases by E when passing through an emf in the direction from the  to + terminal. SET UP: The current is counterclockwise, because the 16 V battery determines the direction of current flow. EXECUTE: 16.0 V  8.0 V  I (1.6   5.0   1.4   9.0 )  0

16.0 V  8.0 V  0.47 A 1.6   5.0   1.4   9.0  (b) Vb  16.0 V  I (1.6 )  Va , so Va  Vb  Vab  16.0 V  (1.6 )(0.47 A)  15.2 V. (c) Vc  8.0 V  I (1.4   5.0 )  Va so Vac  (5.0 )(0.47 A)  (1.4 )(0.47 A)  8.0 V  11.0 V. (d) The graph is sketched in Figure 25.36. EVALUATE: Vcb  (0.47 A)(9.0 )  4.2 V. The potential at point b is 15.2 V below the potential at point a and the potential at point c is 11.0 V below the potential at point a, so the potential of point c is 15.2 V  11.0 V  4.2 V above the potential of point b. I

Figure 25.36

25.37.

IDENTIFY: The voltmeter reads the potential difference Vab between the terminals of the battery. SET UP: open circuit I  0. The circuit is sketched in Figure 25.37a.

EXECUTE:

Vab  E  3.08 V

Figure 25.37a SET UP:

switch closed The circuit is sketched in Figure 35.37b. EXECUTE: Vab  E  Ir  2.97 V E  2.97 V r I 3.08 V  2.97 V  0.067  r 1.65 A Figure 25.37b

Vab 2.97 V   1.80 . I 1.65 A EVALUATE: When current flows through the battery there is a voltage drop across its internal resistance and its terminal voltage V is less than its emf.

And Vab  IR so R 

25.45.

IDENTIFY: A “100-W” European bulb dissipates 100 W when used across 220 V. (a) SET UP: Take the ratio of the power in the US to the power in Europe, as in the alternative method for problem 25.44, using P = V2/R. 2 2 2 2 PUS VUS / R  VUS   120 V   120 V   2  EXECUTE:    = 29.8 W.  . This gives PUS  (100 W)  PE VE / R  VE   220 V   220 V  (b) SET UP: Use P = IV to find the current. EXECUTE: I = P/V = (29.8 W)/(120 V) = 0.248 A EVALUATE: The bulb draws considerably less power in the U.S., so it would be much dimmer than in Europe.

25.55.

IDENTIFY: SET UP:

V2  VI . V  IR . R The heater consumes 540 W when V  120 V . Energy  Pt. P  I 2R 

V2 V 2 (120 V) 2   26.7  so R  R P 540 W P 540 W (b) P  VI so I    4.50 A V 120 V V 2 (110 V) 2   453 W . P is smaller by a factor of (c) Assuming that R remains 26.7  , P  R 26.7  (110 /120) 2 . EVALUATE: (d) With the lower line voltage the current will decrease and the operating temperature will decrease. R will be less than 26.7  and the power consumed will be greater than the value calculated in part (c). EXECUTE:

25.63.

(a) P 

(a) IDENTIFY: Apply Eq. (25.10) to calculate the resistance of each thin disk and then integrate over the truncated cone to find the total resistance. SET UP: EXECUTE: The radius of a truncated cone a distance y above the bottom is given by r  r2   y / h  r1  r2   r2  y 

with    r1  r2  / h Figure 25.63

Consider a thin slice a distance y above the bottom. The slice has thickness dy and radius r. The resistance of the slice is  dy  dy  dy dR   2  2 A r   r2   y  The total resistance of the cone if obtained by integrating over these thin slices: h

 h dy  1   1 1 1  R   dR      r2  y          r h r  0  r2   y 2      0 2  2 But r2  h  r1  1 1    h  r1  r2   h        r2 r1    r1  r2  r1r2   r1r2 (b) EVALUATE: Let r1  r2  r. Then R   h /  r 2   L / A where A   r 2 and L  h. This agrees with Eq. (25.10). R

25.68.

 

IDENTIFY: Consider the potential changes around the circuit. For a complete loop the sum of the potential changes is zero. SET UP: There is a potential drop of IR when you pass through a resistor in the direction of the current. 8.0 V  4.0 V  0.167 A . Vd  8.00 V  I (0.50   8.00 )  Va , so EXECUTE: (a) I  24.0 

Vad  8.00 V  (0.167 A) (8.50 )  6.58 V. (b) The terminal voltage is Vbc  Vb  Vc . Vc  4.00 V  I (0.50 )  Vb and

Vbc   4.00 V  (0.167 A) (0.50 )   4.08 V.

(c) Adding another battery at point d in the opposite sense to the 8.0 V battery produces a 10.3 V  8.0 V  4.0 V  0.257 A . Then counterclockwise current with magnitude I  24.5 

Vc  4.00 V  I (0.50 )  Vb and Vbc  4.00 V  (0.257 A) (0.50 )  3.87 V. EVALUATE: When current enters the battery at its negative terminal, as in part (c), the terminal voltage is less than its emf. When current enters the battery at the positive terminal, as in part (b), the terminal voltage is greater than its emf.

25.85.

IDENTIFY:

Apply R 

EXECUTE:

(a) dR 

L

to find the resistance of a thin slice of the rod and integrate to find the total A R. V  IR . Also find R ( x ), the resistance of a length x of the rod. SET UP: E ( x)   ( x) J

R

0 A

 dx A

L

 exp [ x

L] dx 

0



0 A

 0 exp[  x L] dx A

[ L exp[ x L]]0L 

so

0 L A

(1  e 1 ) and I 

upper limit of x rather than L in the integration, R ( x )  (b) E ( x)   ( x) J 

0 L A

V0 V0 A  . With an R  0 L(1  e 1 )

1  e  . x/ L

I  0e  x L V e x L  0 1 . A L 1  e 

   0 L  V0 A (e x / L  e1 ) x/ L   (c) V  V0  IR ( x) . V  V0   1 e V    0   1 (1  e 1 )  0 L[1  e ]   A  (d) Graphs of resistivity, electric field and potential from x  0 to L are given in Figure 25.85. Each quantity is given in terms of the indicated unit. EVALUATE: The current is the same at all points in the rod. Where the resistivity is larger the electric field must be larger, in order to produce the same current density.

Figure 25.85