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JenniferSmart1

Capacitors...circuits?

  • one year ago
  • one year ago

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  1. JenniferSmart1
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    |dw:1362323572542:dw|

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

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

    • one year ago
  4. JenniferSmart1
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    @.Sam.

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

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

    • one year ago
  7. JenniferSmart1
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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|

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

    • one year ago
  9. UnkleRhaukus
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    This is physics

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

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

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

    • one year ago
  13. JenniferSmart1
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    yes

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

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

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

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

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

    • one year ago
  19. JenniferSmart1
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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|

    • one year ago
  20. JenniferSmart1
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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|

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

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

    • one year ago
  23. JenniferSmart1
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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 \(\frac 1 C =\frac 1{C_1}+\frac 1{C_2}\)

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

    • one year ago
  25. .Sam.
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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

    • one year ago
  26. JenniferSmart1
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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

    • one year ago
  27. JenniferSmart1
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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 \[\frac 1 C +C_3+C_4=C_{net}?\]

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

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

    • one year ago
  30. .Sam.
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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, \[C=C_1+C_2\]

    • one year ago
  31. JenniferSmart1
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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

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

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

    • one year ago
  34. .Sam.
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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, \[C_{T}=C+C_3 \\ \\ C_{T}=\frac{C_1 C_2}{C_1+C_2}+C_3\]

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

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

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

    • one year ago
  38. JenniferSmart1
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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?

    • one year ago
  39. JenniferSmart1
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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|

    • one year ago
  40. JenniferSmart1
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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

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

    • one year ago
  42. JenniferSmart1
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    oops I forgot something

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

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

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

    • one year ago
  46. .Sam.
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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 :)

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

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

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

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

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

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

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

    • one year ago
  54. .Sam.
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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

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

    • one year ago
  56. JenniferSmart1
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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?

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

    • one year ago
  58. .Sam.
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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

    • one year ago
  59. JenniferSmart1
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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?

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

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

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

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

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

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

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

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

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

    • one year ago
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