hw6.8 - dt dQ 2 3 2 / R V I dt dQ b.) To find ,...

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HW 6.8 Given: V F i n d : 10 V ? / 1 dt dQ 6 1 R ? / 2 dt dQ ? 1 6 2 R I F C 4 1 ? 2 Q F C 8 2 ? 1 U ? . 1 Q ? . 2 Q Initially, all capacitors are uncharged. At t = 0, the switch is closed. (a) At what rate is charge being stored in C 1 and C 2 immediately afterward? (b) Calculate the following quantities after a long time. (c) Sometime later, the switch is opened. After another long time, how much charge remains stored in each of the capacitors ? Solutions a.) To find & we can use the following formula: dt dQ / 1 dt dQ / 2 you will need two voltage equations and one current equation since q=0 3 1 2 2 2 2 1 1 2 2 0 0 I I I V I R V I R I R V c 0 2 c V 2 2 R V I 0 1 1 2 2 I R R V R V 0 1 1 I R V V 0 1 I 0 2 1 2 3 R V I I I dt dQ I since current is charge/time (C/s) 2 2 1 / R V I
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Unformatted text preview: dt dQ 2 3 2 / R V I dt dQ b.) To find , & we can use the following formula: 1 I 2 Q 1 U since both capacitors are charged the current there is zero i.e. we have no complete loop 1 I 2 Q 2 2 1 CV U 2 1 1 2 1 V C U c.) To find & we can use the following formula: . 1 Q . 2 Q after the switch is opened all we have is the outside loop and since charge on 2 was zero it will remain zero and the energy stored on 1 while remain the same 2 2 2 1 2 1 C Q CV U 2 2 C Q CV CV V C Q 2 2 V C Q 1 . 1 . 2 Q...
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This note was uploaded on 01/26/2012 for the course PHYSICS 211 taught by Professor Selig during the Fall '10 term at University of Illinois, Urbana Champaign.

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hw6.8 - dt dQ 2 3 2 / R V I dt dQ b.) To find ,...

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