18_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

18_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

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Unformatted text preview: 52 CHAPTER 16 16.33 a 15.0 mF 6.00 mF 3.00 mF c b a 20.0 mF 2.50 mF 6.00 mF Figure 1 c b 20.0 mF Figure 2 c a 8.50 mF b 20.0 mF Figure 3 (a) The equivalent capacitance of the upper branch between points a and c in Figure 1 is Cs = ( 15.0 µ F ) ( 3.00 µ F ) 15.0 µ F + 3.00 µ F = 2.50 µ F Then, using Figure 2, the total capacitance between points a and c is Cac = 2 .50 µ F+6.00 µ F=8.50 µ F From Figure 3, the total capacitance is −1 1 1 + Ceq = = 5.96 µ F 8.50 µ F 20.0 µ F (b) Qab = Qac = Qcb = ( ∆V ) ab Ceq = ( 15.0 V ) ( 5.96 µ F ) = 89.5 µ C Thus, the charge on the 20.0 µ C is Q20 = Qcb = 89.5 µ C 89.5 µ C = 10.53 V 20.0 µ F ( ∆V )ac = ( ∆V )ab − ( ∆V )bc = 15.0 V − Then, Q6 = ( ∆V ) ac ( 6.00 µ F ) = 63.2 µ C and Q15 = Q3 = ( ∆V ) ac ( 2 .50 µ F ) = 26.3 µ C ...
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