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

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F)(2.25 V) = 22.5C.57. The pairC3andC4are in parallel and consequently equivalent to 30F.Since thisnumerical value is identical to that of the others (with which it is in series, with thebattery), we observe that each has one-third the battery voltage across it.Hence, 3.0 V isacrossC4, producing a chargeq4=C4V4= (15F)(3.0 V) = 45C.58. (a) HereDis not attached to anything, so that the 6Cand 4Ccapacitors are in series(equivalent to 2.4C). This is then in parallel with the 2Ccapacitor, which produces anequivalence of 4.4C. Finally the 4.4Cis in series withCand we obtaineq4.40.820.82(40F)33F4.4CCCCCCwhere we have used the fact thatC= 40F.(b) Now,Bis the point that is not attached to anything, so that the 6Cand 2Ccapacitorsare now in series (equivalent to 1.5C), which is then in parallel with the 4Ccapacitor (andthus equivalent to 5.5C). The 5.5Cis then in series with theCcapacitor; consequently,eq5.50.8534F .5.5CCCCCC59. The pairC1andC2are in parallel, as are the pairC3andC4; they reduce to equivalentvalues 6.0F and 3.0F, respectively.These are now in series and reduce to 2.0F,across which we have the battery voltage. Consequently, the charge on the 2.0Fequivalence is (2.0F)(12 V) = 24C.This charge on the 3.0F equivalence (ofC3andC4) has a voltage ofV=qC=24C3F= 8.0 V.
CHAPTER 251032Finally, this voltage on capacitorC4produces a charge (2.0F)(8.0 V) = 16C.60. (a) Equation 25-22 yieldsUCV1212200107 0104 910212323FVJchch...(b) Our result from part (a) is much less than the required 150 mJ, so such a spark shouldnot have set off an explosion.61. Initially the capacitorsC1,C2, andC3form a series combination equivalent to a singlecapacitor, which we denoteC123. Solving the equation1223131231231231111C CC CC CCCCCC C C,we obtainC123= 2.40F.WithV =12.0 V, we then obtainq=C123V= 28.8C.In thefinal situation,C2andC4are in parallel and are thus effectively equivalent to2412.0FC.Similar to the previous computation, we use124243131234124312431111C CC CC CCCCCC C Cand findC1234= 3.00F.Therefore, the final charge isq=C1234V= 36.0C.(a) This represents a change (relative to the initial charge) ofq= 7.20C.(b) The capacitorC24which we imagined to replace the parallel pairC2andC4,is in serieswithC1andC3and thus also has the final chargeq=36.0C found above.The voltageacrossC24would be242436.0C3.00 V12.0FqVC.This is the same voltage across each of the parallel pairs. In particular,V4= 3.00 Vimplies thatq4=C4V4= 18.0C.

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Term
Spring
Professor
N/A
Tags
Capacitance, Charge, Energy, Potential difference, Electric charge, Dielectric

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