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kanwal32

  • one year ago

equivalent capacitance of

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  1. kanwal32
    • one year ago
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    |dw:1434350864779:dw|

  2. kanwal32
    • one year ago
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    @Michele_Laino @IrishBoy123

  3. kanwal32
    • one year ago
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    @Ehsan18

  4. Ehsan18
    • one year ago
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    |dw:1434365430719:dw|

  5. Ehsan18
    • one year ago
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    the capacitors are labelled on the right side with alphabets

  6. kanwal32
    • one year ago
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    ok

  7. kanwal32
    • one year ago
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    how to find equivalent capacitance

  8. Ehsan18
    • one year ago
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    first A+C=C1 E+G=C2 then 1/D+1/C2=C3 C3+B=C4 1/C1+1/C4=C5 C5+F=Ce(Equivalent capacitance) just put the values yourself

  9. Ehsan18
    • one year ago
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    The answer is 10.1144

  10. kanwal32
    • one year ago
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    answer is 7.5 @Ehsan18 your answer is coming wrong

  11. kanwal32
    • one year ago
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    @IrishBoy123

  12. kanwal32
    • one year ago
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    @UnkleRhaukus hlp

  13. kanwal32
    • one year ago
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    @IrishBoy123

  14. UnkleRhaukus
    • one year ago
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    |dw:1434371149308:dw|

  15. UnkleRhaukus
    • one year ago
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    First. lets look at the section A–C

  16. UnkleRhaukus
    • one year ago
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    First. lets look at the section A–C |dw:1434371313237:dw|

  17. UnkleRhaukus
    • one year ago
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    the capacitance in this section can be found with the formula for parallel capacitance \[\frac1{C_\text{qe}}=\frac1C_1+\frac1{C_2}\] ie \[ C_\text{eq}= \left(\frac1{12}+\frac1{23}\right )^{-1} = 276/35\approx7.89 \]

  18. UnkleRhaukus
    • one year ago
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    so we can rewrite the diagram... |dw:1434371666596:dw|

  19. UnkleRhaukus
    • one year ago
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    similarly, with the section D–B... |dw:1434371905223:dw|

  20. UnkleRhaukus
    • one year ago
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    |dw:1434372308311:dw|

  21. UnkleRhaukus
    • one year ago
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    No two capacitors in this circuit are either directly in series or directly in parallel, you might need to use simultaneous equations from here

  22. Ehsan18
    • one year ago
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    @UnkleRhaukus you didn't add the two capacitors in parallel correctly it is Ce=C1+C2

  23. radar
    • one year ago
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    @Ehsan18 You're right, The total capacitance value of two capacitors connected in parallel is simply the sum of each capacitor value. It is the series configuration that is similar to resistors in parallel.

  24. radar
    • one year ago
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    |dw:1434387900902:dw| Does this help?

  25. radar
    • one year ago
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    Remove the 13 uF (mentally) You now have a pair of series capacitors one with a value of 35/6 uF and the other with a value of 10/6 uF connected in parallel. Now add those two values.

  26. radar
    • one year ago
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    I probably should have said an effective value of 35/6 and an effective value of 10/6 micro-farads. Any way when you add them you will get the approved solution of 7.5 uF.

  27. UnkleRhaukus
    • one year ago
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    you are quite right. i added those capacitors wrong, —sorry

  28. UnkleRhaukus
    • one year ago
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    @radar i have followed your method and got the same result, and it now makes sense to me why the 13µF cap can be effectively removed, on both branches either side of the 13, the ratio of capacitors is equal 2/10 = 7/35 = 1/5; the charge flows will be equal, and so there will be no potential difference between C and D. - thank you

  29. kanwal32
    • one year ago
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    |dw:1434454183746:dw|

  30. radar
    • one year ago
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    @UnkleRhaukus I actually used your diagram, you obviously understood the configuration, just a momentary slip on effective capacitance when capacitors are connected in parallel. I have followed many of your responses in the past, I know this was just a "momentary slip" and you do an excellent job of explaining your methods. Keep up the good work.

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