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Homework Help: Capacitor with Key

  1. Dec 20, 2008 #1
    1. The problem statement, all variables and given/known data
    It says,
    The potantial difference between K-L = V , when the key is closed...
    And asks whats the potantial difference between K-L, when the key is open ?

    You see 2 drawings,
    1 above the question,
    1 below the question,

    I have drawed these,
    these shows us the different waypaths, electric would follow as being related to the key.
    But i couldn't bring this question to the solution... Can you show me ?
    Exactly i dont know what i have to do too...
    I did, what i could do.

    Attached Files:

  2. jcsd
  3. Dec 20, 2008 #2


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    Hi FiskiranZeka! :smile:

    (are all the capacitances the same?)

    When the key is open, from K to N has only one capacitor

    but when the key is closed, it has two capacitors in parallel …

    what is the combined capacitance? :wink:
  4. Dec 20, 2008 #3
    ( yeah all Capacitors are Same ( C ) )

    Key Closed -->
    v(KL) = V, so -> v(KN) = 2V, so -> v(MK)=4V, so -> v(MN)=12Volt...
    C Combined = C+C --> 2C ; 2C + C --> 2/3 C ; 2/3 C + C --> 5/3 C

    Key Open -->
    The batteries are same :) So still must be 12 Volt,
    C Combined = C + C --> 1/2 C ; 1/2 C + C --> 3/2 C ; 3/2 C + C --> 3/5 C
    q = V.C and ? How to continue ?
    Must find V (KL)
  5. Dec 21, 2008 #4
    << Not solved yet.
  6. Dec 21, 2008 #5


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    Hi FiskiranZeka! :smile:

    I'm honestly not following what you're doing …

    remember the capacitor rule: in parallel, you add the capacitances, and in series, you add the inverses of the capacitances.
  7. Dec 21, 2008 #6
    Yes, i know that...
    I found the Combined Capacitence for both positions ( Key open / Key closed )
    Look at above...
    And then ? What should i do now ?

    OFF Topic;
    In the country i live, education is different then you are used to see...
    Those things aren't my homework :) Here we have to do these to learn the unit...
    Pratically or teorically, education system doesn't prove/teach us many things in physic.
    And they also don't want from us to make an electric motor or something imporant/usable in real life... So, they ask us simple questions like above and we have to learn how to solve these questions and this is enough for us to do.
    Now gotta learn how to solve this...
    Don't expect from me to do it, 'cos in school we don't even do it in the lesson.
    But we are expected to do it in exam.
    Last edited: Dec 21, 2008
  8. Dec 21, 2008 #7
    hey, I 'm trying to solve it but still, this is an DC or AC ?
    If DC, it remains V.
    If AC, a lots trouble now ( u must use triangle or Y formation to find the total C). I'll post the formula ASAP.
  9. Dec 21, 2008 #8
    Thanks for your interrest and,
    I dont know what is DC and what is AC...
    Also there is no info in the question if its AC or DC...

    There is a solution in book but honestly i dont understand it;
    'Cos book shows me different electric waypaths ( Cycles / Contrivances ) as being compared to the 2 electric waypaths i added to the question image...( You see 1 electric waypath for closed key ( at the top of paper ), 1 electric waypath for open key ( at the bottom of paper ) Exactly those waypaths are what i'm thinking and what seems true for me )... Anyway, i couldn't understand books solution but i'll took important points from there, so you may help me on my own solution...

    Book says;
    In both positions ( Key On / Off ) Voltage of Cycle / Contrivance is 3 Volt...
    And Correct answer is V2 = 3/5 V

    By the way, what is AC / DC ?
    I have seen AC / DC button on an Electric Lamp...
    Was switching from Home electric to Battery Electric.

    * If AC/ DC is related with stable Voltage and Changable Voltage,
    Then i think in this question the battery is same ( Steady / Stable ),
    Voltage shouldn't change when we open the key or we close.

    *Sorry for my bad english skills.
    Last edited: Dec 21, 2008
  10. Dec 21, 2008 #9


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    HI FiskiranZeka! :smile:
    Yes, that's right. :smile:
    Sorry, I don't understand any of this. :confused:
    Yes, in the key closed position you have to prove V(MN) = 3V.

    Try that first, using the rules:

    for capacitors in parallel, Q is the same
    for capacitors in series, the Qs add to (total V)(total C) :smile:

    (DC is direct current, and AC is alternating current. The voltages in these problems are always DC :wink:)
  11. Dec 22, 2008 #10
    I'm sorry about what i posted, it's totally wrong.
    And here's why the Voltage of circle is 3:
    These capacitors is the same, so all have the same Q and thus, same V then use the rule u've just said.
    V(KL)=2V and V(KL)=V(KM)-> V(KM)=2V->V(MN)=2V+V=3V
  12. Dec 22, 2008 #11
    I will look at all these.. and check again, but im away for a while ( not at home )
    So can't check them right now... Thank you for your help...
    Will check all answers and try to resolve when im back at home.
  13. Dec 23, 2008 #12
    Didn't understand totally,
    let me practice and see;
    please check my calculations,

    Look at new image;

    New Image;

    I'm trying to solve the main question, through Image 2 and Image 3...
    First tell me, are both images drawed true ?

    Paralels has the same voltage;
    The question says; Voltage of KL = V,
    So, look at IMAGE 2;
    v (KL) = V --> v ( KN ) = V too
    ( MN ) has 2 wires, underwire and upperwire;
    v ( MN_Underwire ) = v ( KN ) + v ( MK ) { Because KN and MK are Serial }
    v ( MN_Underwire ) = V + v ( MK )
    q= V.C -->
    v ( KN ) = V and C ( KN ) = 2C !!! --> q ( KN ) = V. 2C
    q ( KN ) = q ( MK ) --> V . 2C = v ( MK ) . C --> v ( MK ) = 2V !!!
    v ( MN_Underwire ) = V + 2V = 3V

    Voltage of Underwire = Voltage of Upperwire = Total Voltage ;
    Voltage of System = 3V


    Now i go further to solve the problem;
    IMAGE 3; (key open)
    generator is same; as it is DC ...
    So; System has still 3 Volt...
    C_combined = 3/5 C
    V2 = ?

    q= V.C -->
    q System = q (MK) { Because , Serial Q's are equal }
    q (MK) = q (MK_Downwire) + q (MK_Upperwire)
    q (MK_Upperwire) = Vupper . 0,5C
    q (MK_Downwire) = Vdown . C
    Vdown = Vupper ('cos paralel )
    --> q (MK) = q (MK_Downwire) + q (MK_Upperwire)
    --> q (MK) = Vdown.C + Vdown. 0,5C --> q (MK) = 1,5C . Vdown
    q System = q (MK)
    3V . 3/5 C = Vdown . 1,5C
    Vdown = Vup = 6/5 V
    q = V.C --> q (ML) = q (KL) = q Upper
    v ( ML ) . C = v (KL) . C = 6/5 V . 1/2 C
    v ( KL ) = 3/5 V
    V2 = 3/5V = Solution

    Huhh... Finally... Thanks... Check the maths all right ?:P
    Any mental mistake ?
  14. Dec 24, 2008 #13
    Any mistake ?
  15. Dec 24, 2008 #14


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    Hi FiskiranZeka! :smile:
    Yes, that's fine …

    except that there's no need to say v ( MN_Underwire ) …

    just say v(MN) (because the voltage difference between two points is independent of the path!)
    hmm … a bit long …

    you could say that C(MK) = 3/2 times C(KN),

    so V(MK) = 2/3 times C(KN), so V(MK) = 2/5 V(KN)

    so V(LK) = 1/2 times V(MK) = 1/5 V(KN) = 3/5 V …

    oh … but isn't V(KL) = -V(LK)? :smile:
  16. Dec 25, 2008 #15
    understood this...
    Didn't understand above;
    Below must be true ? why not ?
    V(MK)= 2/3 times V(KN) ... so, V(MK) = 6/5V and V(KN)=9/5 V total = 3V
    V(LK) = 1/2 times V(MK) , V(LK)=3/5V

    and yes, V(LK) = -V(KL) right. But as it seems Turks don't care about saying V(KL) or saying V(LK) :) 'cos the book says 3/5 V is true... That means they dont care for the direction of electric... But what you said is right, correct answer must be -3/5 V

    * I think you made a mistake at calculating...
    I solved it from the way you showed me and found different numbers then you have finden for voltage, but even i found different numbers; those numbers bringed me to the true answer.
  17. Dec 25, 2008 #16
  18. Dec 26, 2008 #17


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    merry fishmas!

    Hi FiskiranZeka! :smile:

    (i didn't reply yesterday. 'cos it was xmas, so i took the day off to do important things like eat and drink and watch doctor who, and wallace and gromit :wink:)
    oops! :redface:

    should have been V(MK) = 2/3 times V(KN), so V(MK) = 2/5 V(KN) :redface:
    Yes, the way you did it is correct also :smile:

    but my reason for preferring my way is that I used the rule that in capacitors in series, V is inversely proportional to C (so you can "balance" the voltages, a bit like balancing weights in mechanics) … so I didn't need to spend time finding Q …

    you did find Q, which is ok, but it takes longer! :wink:
  19. Dec 26, 2008 #18
    No, i made a new solution (2nd solution); i didnt find Q,
    C and V are inversely proportional.... I used this rule to solve the problem as you teached me.

    My 2nd Solution is made by your way; Using the C and V inversation :))
    Look at it carefully ;

    C(MK) = 3/2 times C(KN)
    V(MK) = 2/3 times V(KN),
    so, V(MK) = 6/5V and V(KN)=9/5 V total = 3V
    V(LK) = 1/2 times V(MK) , V(LK)=3/5V

    I didn't write the maths ( calculated at mind ), if you don't understand i can write the maths too so you see how i found those numbers...
    Last edited: Dec 26, 2008
  20. Dec 26, 2008 #19


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    ah, i didn't realise you calculated it in your head …

    there were so many q's i assumed you used them!

    is everything ok now? :smile:
  21. Dec 26, 2008 #20
    Yes, everything well,
    i learn both ways;

    In my first solution surely there were too many Q's because i solved the problem through Q's...( Not at mind )

    But the point you missed is, that after you showed me an alternative and a shorter way that, " C's and V's are opposite related " ; i made a new solution that you may missed ( you may not saw ), that new solution is the one that i calculated at mind...
    its like k + 2/3 k = 5/3k , 5/3k =150, k = 90 :P

    Happy x mas happy holiday thank you...
    Here in Turkey there is no holiday for X mas :)
    We don't celebrate it.

    // Problem Solved...
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