Unformatted text preview: K 3 is q = K 3 V = 3 C 5 V Stepping back one more level, K 2 and the other capacitor are in series and must therefore have the same charge. Hence, q = 3 C 5 V . We can use this information to calculate the voltage across K 2 . V = q K 2 = 3 C 5 2 3 C V = 2 5 V Stepping back another level, K 1 and C are now in parallel and must therefore have the same voltage diﬀerence. Using this to ﬁnd the charge gives. q = K 1 V = C 2 2 5 V = CV 5 Finally, since K 1 is made up of two capacitors in series (one of which is C 2 ), the charge must be that same on both. The charge on C 2 is therefore q 2 = CV 5 = 2 × 105 C Problem 25.64 The capacitances of the four capacitors shown in ﬁgure 21 are given in terms of a certain quantity C . (a) If C = 50 μF , what is the equivalent capacitance between points A and B ? (b) Repeat for point A and D . 25...
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 Spring '08
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 Physics, Capacitance, Charge, Work, Capacitors, Inductor, Farad, equivalent capacitor

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