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Q. The equivalent capacitance between points $A$ and $B$ in the given figure, isPhysics Question Image

Chhattisgarh PMTChhattisgarh PMT 2010

Solution:

$C_{1}$ and $C_{2}$ are in series
$\therefore \frac{1}{C'}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{3}+\frac{1}{6} $
$\Rightarrow $ $C'=2 \mu F$
Similarly, $C_{3}$ and $C_{4}$ are in series
$\frac{1}{C''}=\frac{1}{C_{3}}+\frac{1}{C_{4}}=\frac{1}{2}+\frac{1}{2}$
Now, $C'$ and $C''$ are in parallel
$\therefore $ $C_{\text {comb }}=C'+C''$
$=2 \mu F+1 \mu F=3 \mu F$

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