ch25_p16 - µ C. The voltage across capacitor 1 is...

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16. We determine each capacitance from the slope of the appropriate line in the graph. Thus, C 1 = (12 µ C)/(2.0 V) = 6.0 µ F. Similarly, C 2 = 4.0 µ F and C 3 = 2.0 µ F. The total equivalent capacitance is C 123 = ( ( C 1 ) 1 + ( C 3 + C 2 ) 1 ) 1 = 3.0 µ F. This implies that the charge on capacitor 1 is (3.0 µ F)(6.0 V) = 18
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Unformatted text preview: µ C. The voltage across capacitor 1 is therefore (18 µ C)/( 6.0 µ 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 µ F)( 3.0 V) = 12 µ C....
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.

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