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JenniferSmart1

  • one year ago

Capacitors in series and in parallel

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  1. JenniferSmart1
    • one year ago
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    |dw:1362354546520:dw| @experimentX

  2. JenniferSmart1
    • one year ago
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    no let me try it

  3. experimentX
    • one year ago
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    |dw:1362354809409:dw|

  4. JenniferSmart1
    • one year ago
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    |dw:1362354804183:dw|

  5. JenniferSmart1
    • one year ago
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    oh ok

  6. Xavier
    • one year ago
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    An ok way to remember it is that capacitors in series add like resistors in parallel. Other way around too

  7. JenniferSmart1
    • one year ago
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    ok since they're in parallel, the voltages are the same. \[V=V_1=V_2\] \[Q=Q_1+Q_2\] \[C=\frac{Q}{V}\] \[C_1=\frac{Q_1}{V_1}=\frac{Q_1}{V}\] \[C_2=\frac{Q_2}{V_2}=\frac{Q_2}{V}\] \[CV=C_1V+C_2V\] \[C\cancel{V}=C_1\cancel{V}+C_2\cancel V\] C=C_1+C_2

  8. JenniferSmart1
    • one year ago
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    but they have different capacitances.....hmmm?

  9. JenniferSmart1
    • one year ago
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    oh I add the capacitances to get C....the total capacitance of the capacitors in parallel \[C=6 \mu F\]

  10. experimentX
    • one year ago
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    yeah yeah ... parallel is like absolute capitalism!!

  11. JenniferSmart1
    • one year ago
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    |dw:1362355380229:dw| LOL!!!!!

  12. JenniferSmart1
    • one year ago
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    oh ok let's see|dw:1362355422603:dw| \[C=\frac{Q}{V}\] \[C_1=\frac{Q}{V_1}\] \[C_2=\frac{Q}{V_2}\] \[Q=Q_1=Q_2\] \[V=V_1+V_2\] \[\frac Q C=\frac{Q}{C_1}+\frac{Q}{C_2}\] \[\frac {\cancel Q} C=\frac{\cancel Q}{C_1}+\frac{\cancel Q}{C_2}\] \[\frac 1 C=\frac 1 {C_1}+\frac 1{C_2}\]

  13. JenniferSmart1
    • one year ago
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    \[\frac 1 C=\frac 1 6+\frac 2 6=\frac 3 6\] \[C=2\]

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