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Question

If each capacitor is of capacitance  4 μF. The equivalent cpacitance (in μ F) between terminals A & C  is x. Then the value of  4x is .


Solution

For above arrangements following points can be noted:
(1) The charge recombines at C identical to the way it splits at A.
(2) Potentials at E and F are same. So, capacitor in between them can be replaced by conducting wire.
(3) Potentials at H and G are same. So, capacitor in between them can be replaced by conducting wire.
(4) Connections at B and D can be removed.


Above circuit can be rearranged as 



On simplification we get 


As capacitors of capicatance C and C2 are in parallel (circled in above figure) effective capacitance due to them is  C+C2=3C2

Now capacitors of capacitance CC, 3C2 are in series. So, effective capacitance due to them is 113C/2+1C+1C
=113C/2+2C

=3C2×C23C2+C2=3C242C=3C8

So, circuit can be further simplified as

From above circuit , effective capacitance between A and C is 


CAC=3C8+3C8+C=7C4
Given C=4 μF
x=7×44
x=7
4x=28

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