F225 v 225 c 57 the pair c 3 and c 4 are in parallel

F)(2.25 V) = 22.5  C.
57. The pair C
3
and C
4
are in parallel and consequently equivalent to 30  F. Since this
numerical value is identical to that of the others (with which it is in series, with the
battery), we observe that each has one-third the battery voltage across it. Hence, 3.0 V is
across C
4
, producing a charge
q
4
= C
4
V
4
= (15  F)(3.0 V) = 45  C.
58. (a) Here D is not attached to anything, so that the 6 C and 4 C capacitors are in series
(equivalent to 2.4 C ). This is then in parallel with the 2 C capacitor, which produces an
equivalence of 4.4 C . Finally the 4.4 C is in series with C and we obtain
  
eq
4.4
0.82 0.82(40 F) 33 F
4.4
C C
C C
C C
     

where we have used the fact that C = 40  F.
(b) Now, B is the point that is not attached to anything, so that the 6 C and 2 C capacitors
are now in series (equivalent to 1.5 C ), which is then in parallel with the 4 C capacitor (and
thus equivalent to 5.5 C ). The 5.5 C is then in series with the C capacitor; consequently,
  
eq
5.5
0.85 34 F .
5.5
C C
C C
C C
   

59. The pair C
1
and C
2
are in parallel, as are the pair C
3
and C
4
; they reduce to equivalent
values 6.0  F and 3.0  F, respectively. These are now in series and reduce to 2.0  F,
across which we have the battery voltage. Consequently, the charge on the 2.0  F
equivalence is (2.0  F)(12 V) = 24  C. This charge on the 3.0  F equivalence (of C
3
and
C
4
) has a voltage of
V =
q
C
=
24 C
3 F


= 8.0 V.

CHAPTER 25 1032
Finally, this voltage on capacitor C
4
produces a charge (2.0  F)(8.0 V) = 16  C.
60. (a) Equation 25-22 yields
U CV      
 
1
2
1
2
200 10 7 0 10 4 9 10
2 12 3
2
3
F V J
c hc h
. . .
(b) Our result from part (a) is much less than the required 150 mJ, so such a spark should
not have set off an explosion.
61. Initially the capacitors C
1
, C
2
, and C
3
form a series combination equivalent to a single
capacitor, which we denote C
123
. Solving the equation
1 2 2 3 1 3
123 1 2 3 1 2 3
1 1 1 1 C C C C C C
C C C C C C C
 
    ,
we obtain C
123
= 2.40  F. With V = 12.0 V, we then obtain q = C
123
V = 28.8  C. In the
final situation, C
2
and C
4
are in parallel and are thus effectively equivalent to
24
12.0 F C   . Similar to the previous computation, we use
1 24 24 3 1 3
1234 1 24 3 1 24 3
1 1 1 1
C C C C C C
C C C C C C C
 
   
and find C
1234
= 3.00  F. Therefore, the final charge is q = C
1234
V = 36.0  C.
(a) This represents a change (relative to the initial charge) of  q = 7.20  C.
(b) The capacitor C
24
which we imagined to replace the parallel pair C
2
and C
4
,
is in series
with C
1
and C
3
and thus also has the final charge q =36.0  C found above. The voltage
across C
24
would be
24
24
36.0 C
3.00 V
12.0 F
q
V
C


  
.
This is the same voltage across each of the parallel pairs. In particular, V
4
= 3.00 V
implies that q
4
= C
4
V
4
= 18.0  C.