54CHAPTER 16 Finally, using Figure 1, ()()()111.00 F16.0 V16.0 CabQCVµµ=∆==)()(55.00 FabQV80.0 Cµµ=()()=∆, 88.00 F64.0 CbcQVµµ=∆=and()()44.00 F32.0 CbcQVµµ=∆=16.36The technician combines two of the capacitors in parallel making a capacitor of capacitance 200 Fµ. Then she does it again with two more of the capacitors. Then the two resulting 200 Fµcapacitors are connected in series to yield an equivalent capacitance of 100 Fµ. Because of the symmetry of the solution, every capacitor in the combination has the same voltage across it, ()()290.0 VabVV∆= ∆=245.=0 V16.37(a) From ()QC, V=∆()()32525.0 F50.0 V1.2510 C1.25 mCQµµ==×=()()and 340.0 F50.0 V2.0010 C2.00 mCµµ==×=40Q(b) When the two capacitors are connected in parallel, the equivalent capacitance is
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