The equivalent capacitance between $A$ and $B$ in the circuit given below is.....$μF$
JEE MAIN 2018, Diffcult
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The simplified circuit of the circuit given in question as follows:
The equivalent capacitance between $C \& D$ capacitors of $2\, \mu \mathrm{F}, 5\, \mu \mathrm{F}$ and $5\, \mu \mathrm{F}$ are in parallel.
$\therefore \mathrm{C}_{\mathrm{CD}}=2+5+5=12 \,\mu \mathrm{F}$
($\because$ In parallel grouping $C_{e q}=C_{1}+C_{2}+\ldots .+C_{n})$
Similarly equivalent capacitance between $\mathrm{E}$ $\& \mathrm{BC}_{\mathrm{EB}}$
$=4+2=6\, \mu \mathrm{F}$
Now equivalent capacitance between $A \& B$
$\frac{1}{C_{e q}}=\frac{1}{6}+\frac{1}{12}+\frac{1}{6}=\frac{5}{12}$
$\Rightarrow \mathrm{C}_{\mathrm{eq}}=\frac{12}{5}=24\, \mu \mathrm{F}$
($\because $ In series group-ing, $\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots \ldots + \frac{1}{{{C_n}}}$
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