P28_086 - . 67 V. Thus the nal charge on C is q f = V C ....

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86 .Wh en S is open for a long time, the charge on C is q i = E 2 C .W h e n S is closed for a long time, the current i in R 1 and R 2 is i =( E 2 −E 1 ) / ( R 1 + R 2 )=(3 . 0V 1 . 0V) / (0 . 20 Ω + 0 . 40 Ω) = 3 . 33 A. The voltage diFerence V across the capacitor is then V = E 2 iR 2 =3 . 0V
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Unformatted text preview: . 67 V. Thus the nal charge on C is q f = V C . So the change in the charge on the capacitor is q = q f q i = ( V E 2 ) C = (1 . 67 V 3 . 0 V)(10 ) = 13 C....
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