54CHAPTER 16 Finally, using Figure 1, ()( )( )111.00 F16.0 V16.0 CabQC Vµ=∆ ==)()(55.00 FabQV80.0 C=()=∆, 88.00 F64.0 Cbc=and()44.00 F32.0 Cbc=16.36The 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 Fcapacitors 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, 290.0VabVV∆=∆=245.=0V±²16.37(a) From ( )QC, V()32525.0 F50.0 V1.2510 C1.25 mCQµµ==×=()and 340.0 F50.0 V2.002.00 mC×=40Q(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.