2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 1/2Master ItConsider the circuit shown in the figure below, where C1 = 5.00 µF, C2 = 8.00 µF, and ΔV = 21.0 V. CapacitorC1 is first charged by closing switch S1. Switch S1 is then opened, and the charged capacitor is connected tothe uncharged capacitor by closing S2.(a) Calculate the initial charge acquired by (b) Calculate the final charge on each capacitor.Part 1 of 5 ConceptualizeThe two switches are never closed at the same time. Capacitor is never connected directly across thebattery. The voltage across will be smaller than the given value for because the charge originallyplaced on is now shared with Part 2 of 5 CategorizeWe must visualize the series of operations. The definition of capacitance applied repeatedly will give us theanswers for the initial and final charges acquired by the capacitors.Part 3 of 5 Analyze(a) When is closed, the charge on is given byPart 4 of 5 Analyze(b) When is opened and is closed, the total charge remains constant and is shared by the twocapacitors. When the circuit is in this configuration, we denote the new charges on capacitors and by and respectively. The relationship between the charge when is closed and the tworesulting charges when is closed, is given byThe potential differences across the two capacitors are equal.Substituting into the above equation the expression for and the given values, we haveC1.C2C2ΔVC1C2.S1C1Q1 = C1ΔV = 5 µF 21 V = 105 µC.S1S2C1C2Q1' Q2', Q1S1S2Q1' = 105 µC − Q2'.ΔV' = = Q1'C1Q2'C2Q1'2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 2/2Solving for we obtainPart 5 of 5 AnalyzeSubstituting the value we found above for into the expression from the previous step for we haveFinalizeWe traced the circuit diagram with closed to see that capacitor is charged with the full chargeresulting from When we trace the circuit diagram again with open and closed, we see that thecharge originally resulting from is now shared between the capacitors. Equating the resulting potentialdifferences across the two parallel capacitors when is closed enables us to solve for the charge acrosseach capacitor in this configuration. = . 105 µC − Q2' 5 µFQ2' 8 µFQ2',Q2' = 64.6 µC.Q2' Q1',Q1' = 105 µC − Q2' = 105 µC − 64.6 µC = 40.4 µC.S1C1ΔV.
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