2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 1/4Master ItFour capacitors are connected as shown in the figure below. (Let C = 19.0 µF.)(a) Find the equivalent capacitance between points a and b.(b) Calculate the charge on each capacitor, taking ΔVab = 20.5 V.Part 1 of 9 ConceptualizeFrom outside the circuit in the figure, the section of circuit between a and b would behave like a singlecapacitor. When voltage is applied between points a and b, each of the capacitors will have some smallervoltage across it. We will find that the capacitor outside the loop stores the entire charge that passes points aand b, and the other capacitors store less charge.Part 2 of 9 CategorizeWe will identify a set of capacitors in series or in parallel, replace them with a single equivalent capacitor,draw the new circuit, and repeat these steps.Part 3 of 9 Analyze(a) We simplify the circuit in the figure in three steps shown in diagrams (i), (ii), and (iii) below. First, thecapacitor C and the 3.00 µF capacitor in series are equivalent to a capacitance ofCeq1 = = 2.59 µF.11/ 19 µF + 1/3.00 µF2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 2/4Part 4 of 9 AnalyzeNext, combines in parallel with 6.00 µF, creating an equivalent capacitance ofPart 5 of 9 AnalyzeFinally, and 20.0 µF are in series, with capacitance equivalent to the following.Part 6 of 9 Analyze(b) We find the charge on each capacitor and the voltage across each capacitor by working backward throughsolution diagrams (iii)–(i), alternately applying and to each capacitor, real orequivalent. For equivalent capacitor we havePart 7 of 9 AnalyzeThus, if the potential at a is higher than the potential at b, flows between the wires and the plates tocharge the capacitors in each diagram. In diagram (ii), we haveCeq1Ceq2 = 2.59 µF + 6.00 µF = 8.59 µF.Ceq2Ceq3 = = 6.01 µF11/ 8.59 µF + 1/20.0 µFQ = CΔV ΔV = Q/CCeq3,Qab = Ceq3 ΔVab = 6.01 µF 20.5 V = 123 µC.Qab2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 3/4Part 8 of 9 AnalyzeFor the 20.0 µF capacitor in diagrams (ii), (i) and the original circuit, we use diagram (ii) and the fact that forcapacitors in series, each stores the same charge as its equivalent capacitor. Then we haveand for the potential difference across the 20.0 µF capacitor, which we call we havePart 9 of 9 AnalyzeSince the circuit in diagram (ii) is equivalent to the circuit in diagram (i), we use the voltage to findthe charge across We call this charge Similarly, we find the charge across the 6.00 µF capacitor in the lower branch of the original circuit. We callthis capacitor The same current flows through both capacitors in the upper parallel branch in the original circuit in theoriginal figure. So, for the current in capacitor C and the 3.00 µF capacitor, with charges denoted by and respectively, and since either of these charges is equal to we have the following.FinalizeFirst, we recognized parallel and series connections in the original circuit and reduced this circuit step by stepinto equivalent circuits with equivalent capacitors. The final equivalent circuit consists of a single equivalentcapacitor across which a voltage is applied. We used the formulas for parallel and series capacitance tocalculate the capacitance of the equivalent capacitors in each step. Then, working backward through theequivalent circuit diagrams, we applied to find the charge and voltage related to each capacitance.We applied the relationships that parallel capacitors have the same voltage and capacitors in series have thesame charge.ΔVac = = = 14.3 V.QabCeq2 123 µC 8.59 µFQcb = Qab = 123 µC,C20,ΔVcb = = = 6.16 V.QcbC20 123 µC 20 µFΔVacCeq1. Qac.Qac = Ceq2ΔVac = 8.59 µF 14.3 V = 123 µCQ6.Q6 = 6 µF 14.3 V = 86 µCQCQ3, Qac − Q6,QC = Q3 = 37.2 µCQ = CΔV2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl
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