P26_018

# P26_018 - 18. (a) First, the equivalent capacitance of the...

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18. (a) First, the equivalent capacitance of the two 4 . 0 µ F capacitors connected in series is given by 4 . 0 µ F / 2=2 . 0 µ F. This combination is then connected in parallel with two other 2 . 0- µ Fca - pacitors (one on each side), resulting in an equivalent capacitance C = 3(2 . 0 µ F) = 6 . 0 µ F. This is now seen to be in series with another combination, which consists of the two 3 . 0- µ F capacitors connected in parallel (which are themselves equivalent to C 0 = 2(3 . 0 µ F) = 6 . 0 µ F). Thus, the equivalent capacitance of the circuit is C eq = CC 0 C + C 0 = (6 . 0 µ F)(6 . 0 µ F) 6 . 0 µ F+6 . 0 µ F =3 . 0 µ F . (b) Let V = 20 V be the potential di±erence supplied by the battery. Then q = C eq V = (3 . 0 µ F)(20 V) = 6 . 0 × 10 5 C. (c) The potential di±erence across C 1 is given by V 1 = CV C + C 0 = (6 . 0 µ F)(20 V) 6 . 0 µ
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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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