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    Rubentheran's Avatar
    Rubentheran Posts: 35, Reputation: 1
    Junior Member
     
    #1

    May 16, 2011, 10:20 PM
    Capacitance
    5.a) Two capacitors , C1 = 3 F and C2=6 F , are connected in series and charged by connecting a battery of voltage V=10V in series with them. They are then disconnected from the battery, and the loose wires are connected again. What is the final charge on them.

    b) Repeat problem 5(a) , and after being disconnected from the battery , the capacitors are disconnected from each other. They are now reconnected as shown in Figure 3

    Can someone guide me a) and b) especially the b).I really don't know how to start. The answer for a) is weird because the given answer for a) is 0.

    the given answer for b) is Q'1 = 40/3 uC and Q'2= 80/3 uC
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    jcaron2's Avatar
    jcaron2 Posts: 986, Reputation: 204
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    #2

    May 17, 2011, 08:06 AM
    Rubentheran, regardless of voltage, capacitance value, etc. if the battery is removed and the wires hooked together (as in part A), the capacitors will be discharged (and quite violently so, since there's very little resistance to limit the current in this scenario). Ignoring the melted puddle of wire that would result :), the ultimate outcome would be that the capacitors would be totally discharged (i.e. they'd both have 0V across them). Just to elaborate, remember that charging a capacitor requires adding potential energy, like compressing a spring or lifting a heavy object into the air. Once there's a path for that potential to be released, it will do so entirely, just like letting go of the spring or dropping the heavy object.

    For part B, I'm not sure about the final configuration of the circuit, since you didn't post the schematic. However, I can tell you that in the steady state, when hooked to the battery as shown in the original figure, the capacitors will charge so that the total voltage across BOTH of them adds up to 10V. Their individual voltages will be proportional to the INVERSE of their capacitance. In other words, the 6F cap will end up with half the voltage of the 3F cap. Thus, the 6F cap will have 3.333V across it, while the 3F cap will have 6.667V across it. What happens after that depends on what the new circuit configuration is.
    Rubentheran's Avatar
    Rubentheran Posts: 35, Reputation: 1
    Junior Member
     
    #3

    May 18, 2011, 02:14 AM
    Thanks jcaron2.you really help me out

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