20_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

20_Ch 16 College - 54 CHAPTER 16 Finally using Figure 1 Q1 = C1 V ab = 1.00 F 16.0 V = 16.0 C Q5 = 5.00 F V ab = 80.0 C and 16.36 Q8 = 8.00 F V)bc

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54 CHAPTER 16 Finally, using Figure 1, ()( ) ( ) 11 1.00 F 16.0 V 16.0 C ab QC V µ =∆ = = ) () ( 5 5.00 F ab QV 80.0 C = ( ) =∆ , 8 8.00 F 64.0 C bc = and ( ) 4 4.00 F 32.0 C bc = 16.36 The technician combines two of the capacitors in parallel making a capacitor of capacitance 200 F . Then she does it again with two more of the capacitors. Then the two resulting 200 F capacitors are connected in series to yield an equivalent capacitance of 100 F . Because of the symmetry of the solution, every capacitor in the combination has the same voltage across it, 29 0 . 0 V ab VV ∆=∆ = 24 5 . = 0 V ±² 16.37 (a) From ( ) QC , V ( ) 3 25 25.0 F 50.0 V 1.25 10 C 1.25 mC Q µµ == × = ( ) and 3 40.0 F 50.0 V 2.00 2.00 mC × = 40 Q (b) When the two capacitors are connected in parallel, the equivalent capacitance is
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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