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Q.
The equivalent capacitance between $A$ and $B$ is (in $pF$)
KEAMKEAM 2010Electrostatic Potential and Capacitance
Solution:
Starting from the right end of the network, three $3 \,\mu F$ capacitors are connected in series. The equivalent capacitance of these three capacitors is
$\frac{1}{C_S} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3}$
$\Rightarrow C_S = 1\, \mu F$
$C_S$ and $2 \,\mu F$ are in parallel. The equivalent capacitance of these two capacitors is
$C_P = C_S + 2 \,\mu F = 1 \,\mu F + 2\, \mu F = 3\, \mu F$
Proceeding in this way, finally three 3 pF are in series. Therefore, the equivalent capacitance between A and B is
$ \frac{1}{C_{AB}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3}$
$\Rightarrow C_{AB} = 1\, \mu F$
