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Q.
For circuit the equivalent capacitance between $A$ and $B$ is (each capacitor is of $3 \mu F$ )
Electrostatic Potential and Capacitance
Solution:
$C _{ AE }=\frac{ C }{3}=\frac{3}{3}=1 \mu F$
$C _{ EF }=\frac{ C }{3}=\frac{3}{3}=1 \mu F$
$C _{ GF }=\frac{ C }{2}=\frac{3}{2}=1.5 \mu F$
$C _{ GB }=\frac{ C }{2}=\frac{3}{2}=1.5 \mu F$
All are joined in parallel
$C = C _{ AE }+ C _{ EF }+ C _{ FG }+ C _{ GB }$
$C=1+1+1.5+1.5$
$C =5\, \mu F$
