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20_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

# 20_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

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54 CHAPTER 16 Finally, using Figure 1, ( ) ( )( ) 1 1 1.00 F 16.0 V 16.0 C ab Q C V µ µ = = = ) ( )( 5 5.00 F ab Q V 80.0 C µ µ = ( )( ) = , 8 8.00 F 64.0 C bc Q V µ µ = = and ( )( ) 4 4.00 F 32.0 C bc Q V µ µ = = 16.36 The technician combines two of the capacitors in parallel making a capacitor of capacitance 200 F µ . Then she does it again with two more of the capacitors. Then the two resulting 200 F µ capacitors are connected in series to yield an equivalent capacitance of 100 F µ . Because of the symmetry of the solution, every capacitor in the combination has the same voltage across it, ( ) ( ) 2 90.0 V ab V V = ∆ = 2 45. = 0 V 16.37 (a) From ( ) Q C , V = ( )( ) 3 25 25.0 F 50.0 V 1.25 10 C 1.25 mC Q µ µ = = × = ( )( ) and 3 40.0 F 50.0 V 2.00 10 C 2.00 mC µ µ = = × = 40 Q (b) When the two capacitors are connected in parallel, the equivalent capacitance is
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