Question

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

Answer 1

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 $\frac{C}{2}$ are in parallel (circled in above figure) effective capacitance due to them is $C+\frac{C}{2}=\frac{3C}{2}$

Now capacitors of capacitance $C$, $C$, $\frac{3C}{2}$ are in series. So, effective capacitance due to them is $\frac{1}{\frac{1}{3C/2}+\frac{1}{C}+\frac{1}{C}}$
$=\frac{1}{\frac{1}{3C/2}+\frac{2}{C}}$

$=\frac{\frac{3C}{2}\times \frac{C}{2}}{\frac{3C}{2}+\frac{C}{2}}=\frac{\frac{3C^2}{4}}{2C}=\frac{3C}{8}$

So, circuit can be further simplified as

From above circuit , effective capacitance between A and C is


$C_{AC}=\frac{3C}{8}+\frac{3C}{8}+C=\frac{7C}{4}$
Given $C=4\mu F$
$x=\frac{7\times 4}{4}$
$x=7$
$4x=28$