ch25-p062

# ch25-p062 - Therefore, with 2.00 V across C 2 we find q 2 =...

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bottom right capacitor. The bottom right capacitor has, as a result, a potential across it equal to V = q C = 60 µ C 10 µ F = 6.00 V which leaves 10.0 V 6.00 V = 4.00 V across the group of capacitors in the upper right portion of the circuit. Inspection of the arrangement (and capacitance values) of that group reveals that this 4.00 V must be equally divided by C 2 and the capacitor directly below it (in series with it).
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Unformatted text preview: Therefore, with 2.00 V across C 2 we find q 2 = C 2 V 2 = (10.0 F)(2.00 V) = 20.0 C . 62. (a) The potential across C 1 is 10 V, so the charge on it is q 1 = C 1 V 1 = (10.0 F)(10.0 V) = 100 C. (b) Reducing the right portion of the circuit produces an equivalence equal to 6.00 F, with 10.0 V across it. Thus, a charge of 60.0 C is on it -- and consequently also on the...
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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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