Class examples for Chapter 25

Class examples for Chapter 25 - Class examples for Chapter...

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1 Class examples for Chapter 25 13. The charge initially on the charged capacitor is given by q = C 1 V 0 , where C 1 = 100 pF is the capacitance and V 0 = 50 V is the initial potential difference. After the battery is disconnected and the second capacitor wired in parallel to the first, the charge on the first capacitor is q 1 = C 1 V , where V = 35 V is the new potential difference. Since charge is conserved in the process, the charge on the second capacitor is q 2 = q – q 1 , where C 2 is the capacitance of the second capacitor. Substituting C 1 V 0 for q and C 1 V for q 1 , we obtain q 2 = C 1 ( V 0 V ). The potential difference across the second capacitor is also V , so the capacitance is   0 2 21 50V 35V 100pF 43pF. 35V VV q CC 16. We determine each capacitance from the slope of the appropriate line in the graph. Thus, C 1 = (12 C)/(2.0 V) = 6.0 F. Similarly, C 2 = 4.0 F and C 3 = 2.0 F. The total equivalent capacitance is given by 1 2 3 123 1 2 3 1 2 3 1 1 1 () C C C C C C C C C C  , or 1 2 3 123 1 2 3 (6.0 F)(4.0 F 2.0 F) 36 F 3.0 F 6.0 F 4.0 F 2.0 F 12 C C C C C C C   . This implies that the charge on capacitor 1 is 1 q (3.0 F)(6.0 V) = 18 C. The voltage across capacitor 1 is therefore V 1 = (18 C)/( 6.0 F) = 3.0 V. From the discussion in section 25-4, we conclude that the voltage across capacitor 2 must be 6.0 V – 3.0 V = 3.0 V. Consequently, the charge on capacitor 2 is (4.0 F)( 3.0 V) = 12 C.
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This document was uploaded on 11/04/2011 for the course PHY 108 at SUNY Buffalo.

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Class examples for Chapter 25 - Class examples for Chapter...

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