Question

In the arrangement of capacitors shown in figure, each capacitor is of $9\mu F$, then the equivalent capacitance between the points $A$ and $B$ is

• A. $9\mu F$
• B. $18\mu F$
• C. $4.5\mu F$
• D. $15\mu F$

Answer 1

Answer: (D)

The equivalent circuit of the given network is as shown in figure.

Capacitors $C_1$ and $C_3$ are connected in parallel, the effective capacitance $C^{\prime }$ of these capacitors is given by
$C^{\prime }=C_1+C_3=9\mu F+9\mu F=18\mu F$
Now, $C$' and $C_2$ are connected in series, the effective capacitance is given by
$C^{\prime \prime }=\frac{C^{\prime }C}{C^{\prime }+C_2}$
$=\frac{\left(18\mu F\right)\left(9\mu F\right)}{\left(18\mu F\right)+\left(9\mu F\right)}=6\mu F$
Now, $C^{\prime \prime }$ and $C_4$ are connected in parallel. Hence the equivalent capacitance between $A$ and $B$ is
$C_{eq}=C^{\prime \prime }+C_4=6\mu F+9\mu F=15\mu F$