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ch25_p19 - 19(a and(b We note that the charge on C3 is q3 =...

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19. (a) and (b) We note that the charge on C 3 is q 3 = 12 µ C – 8.0 µ C = 4.0 µ C. Since the charge on C 4 is q 4 = 8.0 µ C, then the voltage across it is q 4 / C 4 = 2.0 V. Consequently, the voltage V 3 across C 3 is 2.0 V ¡ C 3 = q 3 / V 3 = 2.0 µ F. Now C 3 and C 4 are in parallel and are thus equivalent to 6 µ F capacitor which would then be in series with C 2 ; thus, Eq 25-20 leads to an equivalence of
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