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CourseNotes.25 - K 3 is q = K 3 V = 3 C 5 V Stepping back...

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Figure 20: An Arrangement of Capacitors Capacitor 2 is not as easy however. We will find the charge on capacitor 2 in the usual manner, by working the circuit down to its simplest form and then working back out. Before we start simplifying, we will note that the upper portion of the circuit plays no role in the bottom portion and we will therefore ignore it all together. The idea behind simplifying a circuit is to look for pairs of capacitors that are wired either in series or in parallel. The first pair that we will simplify is C 2 and the other capacitor on the same wire as C 2 . Since they are in parallel, K 1 = 1 1 C + 1 C = C 2 where we are using K to denote an imaginary equivalent capacitor. We can now treat this equivalent capacitor as being in parallel with the one which is nearest to it. K 2 = C 2 + C = 3 C 2 And finally, this equivalent capacitor is in series with the fourth capacitor in the lower portion of the circuit. K 3 = 1 1 C + 2 3 C = 3 C 5 We are now ready to step our way back out to the answer. The charge on
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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 × 10-5 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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