P26_071 - µ F equivalence from the top 3 capacitors has...

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71. We reduce the circuit, starting with C 1 and C 2 (in series) which are equivalent to 4 µ F. This is then parallel to C 3 and results in a total of 8 µ F, which is now in series with C 4 and can be further reduced. However, the ±nal step in the reduction is not necessary, as we observe that the 8
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Unformatted text preview: µ F equivalence from the top 3 capacitors has the same capacitance as C 4 and therefore the same voltage; since they are in series, that voltage is then 12 / 2 = 6 V....
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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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