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

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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.
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