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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Apr 2008 23:07:42 IST
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All have a capacitance of 1 micro farad. Find Ceq across AB.
Am not getting the equivalent circuit right :(
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Will nip in at times to solve problems :)
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7 Apr 2008 23:12:26 IST
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3 C / 4
??
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it is not important where u stand, but in which direction u are moving |
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7 Apr 2008 23:14:12 IST
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I mean
3 microfarad / 4
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7 Apr 2008 23:16:11 IST
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is the answer is 10/3 micro farad
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
      
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
if i helped u plzzzzz rate me,,,,,,, |
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7 Apr 2008 23:17:02 IST
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see i have no means to draw the figure but let me try to ans these
i am adopting the following approach
let the capacitors be numbered so that from moving from left to right the numbering is
A, 1, 2, 3 , 4, 5 , 6 , B
on redrawing the circuit u get that 4 and 5 are short-circuited
equivalent circuit is that 1,2 and 3 are in parallel and are in series with 6th capacitor
then
equivalent capacitance = 3C/4
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all the best ... |
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7 Apr 2008 23:26:26 IST
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Let the pot of A be V1 and the pot of B be V4 ( if we go on naming 1, 2 etc we reach 4).
3 capacitors have P.D V1 - V2 . So they r in || and eq is 3C
2 capacitors have P.D V2 - V3 . So they r in || and eq is2C
1 capacitor have P.D V3 - V4 .
As we hav to find the eq cap btw A and B , across which the PD is V1 -V4 , the 2 cap of eq cap ( 2C ) is out of circuit.
So 3C and 1C is in series and result is 3C / 4.
Finally we put C = 1 micro F
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7 Apr 2008 23:30:32 IST
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The diagram wil be A goes to 3C ( pd V1 - V2 ) and 2C ( pd V2 -V3 ) in series and frm a point between them comes out 1C ( pd V2 - V 4 ) which ends in B.
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7 Apr 2008 23:35:41 IST
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Re:Solve...
Download the img:
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7 Apr 2008 23:36:36 IST
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sorry
guess i was busy painting wen ramyani already posted
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7 Apr 2008 23:41:01 IST
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circuit diagram by elasti is right
that is what i meant in my explanation
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all the best ... |
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