Download Solutions - UMD Physics

P28.2
(a)
(b)
P28.4
(a)
V term  IR
becomes
10.0 V  I 5.60 
so
I 1.79 A .
Vterm    Ir
becomes
10.0 V     1.79 A  0.200 
so
  10.4 V .
Here   I R  r , so I

Rr

12.6 V
 2.48 A .
 5.00   0.080 0 
Then, V  IR   2.48 A  5.00   12.4 V .
(b)
Let I1 and I2 be the currents flowing through the battery and
the headlights, respectively.
Then,
I1  I2  35.0 A , and   I1r I2r  0
so
   I2  35.0 A   0.080 0   I2  5.00   12.6 V
giving
I2  1.93 A .
Thus,
V2  1.93 A  5.00   9.65 V .
FIG. P28.4
Rp 
P28.6
(a)
1
 4.12 
1 7.00   1 10.0 
Rs  R1  R2  R3  4.00  4.12 9.00  17.1 
(b)
V  IR
FIG. P28.6
34.0 V  I 17.1 
I 1.99 A
for 4.00  , 9.00  resistors.
Applying V  IR ,
1.99 A  4.12   8.18 V
8.18 V  I 7.00 
so
I 1.17 A
for 7.00  resistor
8.18 V  I 10.0 
so
I 0.818 A
for 10.0  resistor.
P28.9
If we turn the given diagram on its side, we find that it is the same as
figure (a). The 20.0  and 5.00  resistors are in series, so the
first reduction is shown in (b). In addition, since the 10.0  ,
5.00  , and 25.0  resistors are then in parallel, we can solve
for their equivalent resistance as:
Req 

1
1
10.0 
 5.001   25.10 

 2.94 
.
This is shown in figure (c), which in turn reduces to the circuit shown in
figure (d).
Next, we work backwards through the diagrams applying
I
V
R
and
V  IR alternately to every resistor, real and equivalent. The
12.94  resistor is connected across 25.0 V, so the current
through the battery in every diagram is
I
V
25.0 V

 1.93 A
R
12.94 
.
In figure (c), this 1.93 A goes through the 2.94  equivalent resistor to
give a potential difference of:
V  IR   1.93 A  2.94   5.68 V
.
From figure (b), we see that this potential difference is the same across
V ab , the 10  resistor, and the 5.00  resistor.
Vab  5.68 V
(b)
Therefore,
.
(a)
Since the current through the 20.0  resistor is also the current through
the 25.0  line ab,
I
V ab 5.68 V

 0.227 A  227 m A
R ab
25.0 
.
FIG. P28.9
P28.15
1 
 1
Rp  

 3.00 1.00
1
 0.750 
R s   2.00  0.750  4.00   6.75 
Ibattery 
V 18.0 V

 2.67 A
R s 6.75 
P2   2.67 A   2.00 
2
P  I2R :
P2  14.2 W
P4   2.67 A 
in 2.00 
 4.00 A   28.4 W
V 2   2.67 A  2.00   5.33 V ,
V 4   2.67 A  4.00   10.67 V
2
in 4.00 
V p  18.0 V  V 2  V 4  2.00 V   V 3  V1 
P3 
P1 
P28.20
 V 3 
R3
2

 2.00 V  2
3.00 
 V1    2.00 V  2 
R1
1.00 
 1.33 W
4.00 W
FIG. P28.15
in 3.00 
in 1.00 
15.0   7.00 I1   2.00 5.00  0
5.00  7.00I1
so
I1  0.714 A
I3  I1  I2  2.00 A
0.714  I2  2.00
FIG. P28.20
so
   2.00 1.29  5.00 2.00  0
I2  1.29 A
  12.6 V
P28.21
We name currents I1 , I2 , and I3 as shown.
From Kirchhoff’s current rule, I3  I1  I2 .
Applying Kirchhoff’s voltage rule to the loop containing I2 and I3 ,
12.0 V   4.00 I3   6.00 I2  4.00 V  0
FIG. P28.21
8.00   4.00 I3   6.00 I2
Applying Kirchhoff’s voltage rule to the loop containing I1 and I2 ,
  6.00 I2  4.00 V   8.00 I1  0
 8.00 I1  4.00   6.00 I2 .
Solving the above linear system, we proceed to the pair of simultaneous
equations:
8  4I1  4I2  6I2

8I1  4  6I2
or
8  4I1  10I2

I2  1.33I1  0.667
and to the single equation 8  4I1  13.3I1  6.67
I1 
14.7 V
 0.846 A . Then
17.3 
I2  1.33 0.846 A   0.667
and
I3  I1  I2
give I1  846 m A ,I2  462 m A ,I3  1.31 A .
All currents are in the directions indicated by the arrows in the circuit diagram.
P28.23
We use the results of Problem 28.21.
(a)
By the 4.00-V battery:
U   V  It  4.00 V  0.462 A  120 s  222 J .
By the 12.0-V battery:
12.0 V 1.31 A  120 s
1.88 kJ .
(b)
(c)
By the 8.00- resistor:
I2R t  0.846 A   8.00  120 s  687 J .
By the 5.00- resistor:
 0.462 A  2  5.00  120 s 
128 J .
By the 1.00- resistor:
 0.462 A  2 1.00  120 s 
25.6 J .
By the 3.00- resistor:
1.31 A  2  3.00  120 s 
616 J .
By the 1.00- resistor:
1.31 A  2 1.00  120 s 
205 J .
2
222 J 1.88 kJ 1.66 kJ from chemical to electrical.
687 J 128 J 25.6 J 616 J 205 J 1.66 kJ from electrical to internal.
P28.32
(a)
I t  I0etRC
I0 
Q
5.10  106 C

 1.96 A
RC 1300  2.00  109 F




9.00  106 s
  61.6 m A
I t    1.96 A  exp 
 1300  2.00  109 F 




(b)


8.00  106 s
  0.235 C
q t  Q etRC   5.10 C  exp 
  1 300  2.00  109 F 


(c)
The magnitude of the maximum current is I0  1.96 A .


P28.39



et RC 
12.0
et 12.0 s
4.00  106
(a)
  RC  4.00  106  3.00  106 F  12.0 s
(b)
I

R
FIG. P28.39
q  C  1 et RC   3.00  106  12.0 1 et 12.0 
q  36.0 C 1 et 12.0 
P28.41


V  Igrg  I Ig R p
Rp 
, or
I 3.00 A et 12.0
Igrg
 I I 
g

Ig  60.0 
 I I 
g
I  0.500 m A
Therefore, to have I 0.100 A  100 m A when g
:
FIG. P28.41
Rp 
 0.500 m A  60.0 
99.5 m A
 0.302 
.