P26_020 - F + 3 . µ F + 4 . µ F = 8 . 4 V . Thus, q 1 = C...

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20. (a) In this situation, capacitors 1 and 3 are in series, which means their charges are necessarily the same: q 1 = q 3 = C 1 C 3 V C 1 + C 3 = (1 . 0 µ F)(3 . 0 µ F)(12 V) 1 . 0 µ F+3 . 0 µ F =9 . 0 µ C . Also, capacitors 2 and 4 are in series: q 2 = q 4 = C 2 C 4 V C 2 + C 4 = (2 . 0 µ F)(4 . 0 µ F)(12 V) 2 . 0 µ F+4 . 0 µ F =16 µ C . (b) With switch 2 also closed, the potential diference V 1 across C 1 must equal the potential diference across C 2 and is V 1 = C 3 + C 4 C 1 + C 2 + C 3 + C 4 V = (3 . 0 µ F+4 . 0 µ F)(12 V) 1 . 0 µ F+2 . 0 µ
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Unformatted text preview: F + 3 . µ F + 4 . µ F = 8 . 4 V . Thus, q 1 = C 1 V 1 = (1 . µ F)(8 . 4 V) = 8 . 4 µ C, q 2 = C 2 V 1 = (2 . µ F)(8 . 4 V) = 17 µ C, q 3 = C 3 ( V − V 1 ) = (3 . µ F)(12 V − 8 . 4 V) = 11 µ C, and q 4 = C 4 ( V − V 1 ) = (4 . µ F)(12 V − 8 . 4 V) = 14 µ C....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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