19_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

19_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

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Unformatted text preview: Electrical Energy and Capacitance 16.34 (a) The combination reduces to an equivalent capacitance of 12 .0 µ F in stages as shown below. 24.0 mF 36.0 V 4.00 mF 36.0 V 2.00 8.00 mF mF 4.00 mF Figure 1 2.00 mF 6.00 mF 36.0 V Figure 2 (b) From Figure 2, 12.00 mF Figure 3 Q4 = ( 4.00 µ F ) ( 36.0 V ) = 144 µ C Q2 = ( 2.00 µ F ) ( 36.0 V ) = 72.0 µ C and Q6 = ( 6.00 µ F ) ( 36.0 V ) = 216 µ C Then, from Figure 1, Q24 = Q8 = Q6 = 216 µ C 16.35 a 1.00 mF 24.0 V a 6.00 mF 5.00 mF 24.0 V b 8.00 mF b Figure 2 The circuit may be reduced in steps as shown above. Qac = ( 4.00 µ F ) ( 24.0 V ) = 96.0 µ C Then, in Figure 2, ( ∆V ) ab = and c c c Using the Figure 3, 4.00 mF 24.0 V 12.0 mF 4.00 mF Figure 1 a Qac 96.0 µ C = = 16.0 V Cab 6.00 µ F ( ∆V )bc = ( ∆V ) ac − ( ∆V )ab = 24.0 V − 16.0 V = 8.00 V Figure 3 53 ...
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