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JenniferSmart1

  • one year ago

Capacitors...circuits?

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  1. JenniferSmart1
    • one year ago
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    |dw:1362323572542:dw|

  2. JenniferSmart1
    • one year ago
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    @UnkleRhaukus My first time doing a circuit problem Let's see how much I know....

  3. JenniferSmart1
    • one year ago
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    should I "add" \(C_1 \;\text{and}\; C_2\) first?

  4. JenniferSmart1
    • one year ago
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    @.Sam.

  5. JenniferSmart1
    • one year ago
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    |dw:1362324461114:dw| these are in parallel...no they're in series correct?

  6. JenniferSmart1
    • one year ago
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    |dw:1362324513695:dw| is my charge distribution correct so far?

  7. JenniferSmart1
    • one year ago
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    is it true that the part circled doesn't have a charge? or a net charge of zero?|dw:1362324758141:dw|

  8. JenniferSmart1
    • one year ago
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    does that mean \(Q_1 \ne Q_2\)

  9. UnkleRhaukus
    • one year ago
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    This is physics

  10. UnkleRhaukus
    • one year ago
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    http://openstudy.com/study#/groups/Physics

  11. JenniferSmart1
    • one year ago
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    I'll move my question

  12. .Sam.
    • one year ago
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    |dw:1362325488170:dw| Series

  13. JenniferSmart1
    • one year ago
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    yes

  14. JenniferSmart1
    • one year ago
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    so \(C_1=C_2\)

  15. .Sam.
    • one year ago
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    I don't think you can say that they're equal

  16. JenniferSmart1
    • one year ago
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    are the charges equal? \[Q_1=Q_2\] |dw:1362325598105:dw|

  17. JenniferSmart1
    • one year ago
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    or should they be opposite in sign?

  18. .Sam.
    • one year ago
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    The sign is correct

  19. JenniferSmart1
    • one year ago
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    is it true that these two plates don't have a net charge? There is something special about these two plates, but I don't remember what it is. |dw:1362325688912:dw|

  20. JenniferSmart1
    • one year ago
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    \[C_1=\frac{Q_1}{V_1}\] \[C_2=\frac{Q_2}{V_2}\] so far \(Q_1=Q_2\) the Potential.....is that the potential across these plates? |dw:1362325884681:dw|

  21. JenniferSmart1
    • one year ago
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    |dw:1362325984158:dw| The potential is probably different between \[C_1 & C_2\]

  22. .Sam.
    • one year ago
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    The charges must be the same, otherwise they won't come up with a formula for seires: \[V=V_1+V_2\\\ \\ \text{From} Q=CV \\ \\ \frac{Q}{C}=\frac{Q}{C_1}+\frac{Q}{C_2} \\ \\ \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2} \]

  23. JenniferSmart1
    • one year ago
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    makes sense. Does this mean that the Potential difference between (parallel plates 1) and (parallel plates 2) is \(\frac 1 C\) where \(\frac 1 C =\frac 1{C_1}+\frac 1{C_2}\)

  24. JenniferSmart1
    • one year ago
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    I mean it makes sense mathematically, I'm trying to conceptually understand it

  25. .Sam.
    • one year ago
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    That C you're calculating is actually here|dw:1362326867484:dw| Then, You have to combine with C3 with C, which is in parallel, Just add them and you'll get the total capacitance for the top part without C4, as for C4 just use the formula 1/C... again

  26. JenniferSmart1
    • one year ago
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    I have access to the answer. I'm trying to recreate an example by Prof. Walter Lewin....unfortunately I've been very unsuccessful so far :( https://www.youtube.com/watch?v=UFK-kof1WLQ&list=PLOggKvajo_owEtF3edwu7BYxTOFD3-CpP

  27. JenniferSmart1
    • one year ago
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    so my goal is to have C4 and C3 and (C_2 and C_1) in parallel....and the only way to do it is to combine the series C_2 and C_1 We've come to the conclusion that C_1 and C_2=\[\frac 1 C\] and Now I'm adding \[\frac 1 C +C_3+C_4=C_{net}?\]

  28. JenniferSmart1
    • one year ago
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    Let's see....

  29. JenniferSmart1
    • one year ago
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    are \[\frac 1 C+C_3=\text{to some other C}\] since they're within the same enclosed circuit

  30. .Sam.
    • one year ago
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    Don't do them together, that's the biggest mistake you can do in circuit problems. For "All the capacitors" in series, remember that Voltages always add up, so: \[V_{Total}=V_1+V_2+V_3\\\ \\ \text{From} Q=CV \\ \\ V=\frac{Q}{C} \\ \\ \frac{Q}{C}=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3} \\ \\ \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\] For "All the capacitors" in parallel, Voltages in Parallel are equal, \[V=V_1=V_2\] So, \[C=C_1+C_2\]

  31. JenniferSmart1
    • one year ago
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    but why are \[C_1, C_2 \;\text{and}\; C_3\] in series? I agree that C_1 and C_2 are, but wouldn't the addition of C_1 and C_2 be in parallel to C_3

  32. JenniferSmart1
    • one year ago
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    oh I see...you're just giving me a general statement...sorry :S

  33. JenniferSmart1
    • one year ago
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    so yes, for "all the capacitors in series" the Voltages add up

  34. .Sam.
    • one year ago
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    Yes so for C1 and C2 we have \[ \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}\] \[C=\frac{C_1 C_2}{C_1+C_2}\] Now we have parallel, |dw:1362328161892:dw| Then, \[C_{T}=C+C_3 \\ \\ C_{T}=\frac{C_1 C_2}{C_1+C_2}+C_3\]

  35. .Sam.
    • one year ago
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    |dw:1362328282074:dw|

  36. .Sam.
    • one year ago
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    Then we reduced the circuit into |dw:1362328321279:dw| Series, \[C_{Total}=\frac{1}{C_T}+\frac{1}{C_4} \\ \\ C_{Total}=\frac{1}{\frac{C_1 C_2}{C_1+C_2}+C_3}+\frac{1}{C_4}\]

  37. .Sam.
    • one year ago
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    Whops, that should be \[\frac{1}{C_{Total}}=\frac{1}{C_T}+\frac{1}{C_4} \\ \\ \frac{1}{C_{Total}}=\frac{1}{\frac{C_1 C_2}{C_1+C_2}+C_3}+\frac{1}{C_4}\]

  38. JenniferSmart1
    • one year ago
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    ok I've been able to follow you up to the point where \[C_T=C+C_3\] \[C_T=\frac{C_1C_2}{C_2+C_1}+C_3\] that means that... |dw:1362328737355:dw| does that mean that \(C_T\) and \(C_4\) are in parallel? does that mean that \(V=V_T+V_4\) does not apply anymore because it's not in series anymore? or is that always true...whether we're working with capacitors in parallel or in series?

  39. JenniferSmart1
    • one year ago
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    wait....are C_T and C_4 in parallel? If we line them up it looks like they're in series |dw:1362329073958:dw|

  40. JenniferSmart1
    • one year ago
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    oh that's why we're able to say \[\frac 1 {C_{Total}}=\frac 1{C_T}+\frac 1{C_4}\] \[\frac 1 {C_{Total}}=\frac{1}{\frac{C_1C_2}{C_2+C_1}+C_3}\] Ok I think I'm following

  41. .Sam.
    • one year ago
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    That's series, there's no obvious branching lines.

  42. JenniferSmart1
    • one year ago
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    oops I forgot something

  43. JenniferSmart1
    • one year ago
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    \[\frac 1 {C_{Total}}=\frac{1}{\frac{C_1C_2}{C_2+C_1}+C_3}+\frac{1}{C_4}\]

  44. JenniferSmart1
    • one year ago
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    did we just achieve the goal of this problem? Is that it?

  45. JenniferSmart1
    • one year ago
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    why do you have C_Total instead of 1/C_total?

  46. .Sam.
    • one year ago
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    The final answer should be\[\frac{1}{C_{Total}}=\frac{1}{\frac{C_1 C_2}{C_1+C_2}+C_3}+\frac{1}{C_4}\] But then, I usually work out the numbers first then substitute into another equation that way is not so confusing :)

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  47. JenniferSmart1
    • one year ago
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    oh I see....

  48. JenniferSmart1
    • one year ago
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    one more question |dw:1362329932879:dw| would this be a parallel series?

  49. .Sam.
    • one year ago
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    You need to know where's the power coming from..

  50. JenniferSmart1
    • one year ago
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    |dw:1362330036313:dw|

  51. JenniferSmart1
    • one year ago
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    that's the power right?

  52. .Sam.
    • one year ago
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    Series, you can ignore that middle line|dw:1362330096032:dw|

  53. JenniferSmart1
    • one year ago
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    can you draw capacitors in parallel for me

  54. .Sam.
    • one year ago
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    Lets say there's another capacitor there |dw:1362330161795:dw| Then, You can tell C_T and C_4 is in series, but both C_T and C_4 is in parallel with C_5 So u work out the series for C_T and C_4 first, then parallel for C_5

  55. JenniferSmart1
    • one year ago
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    oh so does parallel mean parallel to the power supply?

  56. JenniferSmart1
    • one year ago
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    |dw:1362330342861:dw| C_5 and C_6 would be capacitors in parallel , whereas C_T and C_3 are capacitors in series correct?

  57. JenniferSmart1
    • one year ago
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    I mean C_T and C_4....typo :S

  58. .Sam.
    • one year ago
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    no, just look at the branching first, C_T , C_6, and C_4 is in series because they're in the same line, Then C_5 is the only one parallel to all of the other 3 capacitors, you're actually relating to other capacitors

  59. JenniferSmart1
    • one year ago
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    oh I see.... so for capacitors in parallel \[V=V_1+V_2\] still holds, correct? but the Q's are not equal?

  60. .Sam.
    • one year ago
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    Parallel capacitors is \[V=V_1=V_2\]

  61. JenniferSmart1
    • one year ago
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    oh interesting Let's see \[V=\frac Q C\] \[\frac Q C=\frac{Q_1}{C_1}=\frac{Q_2}{C_2}\]

  62. JenniferSmart1
    • one year ago
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    Q's the same again?

  63. .Sam.
    • one year ago
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    For Parallel, Q is not the same, but voltage is the same, so, example: \[Q=Q_1+Q_2+Q_3\] \[CV=C_1V+C_2V+C_3V\] \[C=C_1+C_2+C_3\]

  64. .Sam.
    • one year ago
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    That's in parallel, you have to remember.

  65. .Sam.
    • one year ago
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    Series= Charges equal Parallel= Voltage equal

  66. JenniferSmart1
    • one year ago
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    Yes! Makes sense :)

  67. JenniferSmart1
    • one year ago
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    You are awesome! Thank you soo much for teaching me capacitors in parallel and series...Thank You @.Sam.

  68. .Sam.
    • one year ago
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    hahaha, welcome :)

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