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)

$15 \,\mu F$

Answer 2

undefined 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 $C'$ of these capacitors is given by $C'=C_{1}+C_{3}=9\,\mu F+9\,\mu F=18\,\mu F$ Now, $C$' and $C_{2}$ are connected in series, the effective capacitance is given by $C''=\frac{C'C}{C'+C_{2}}$ $=\frac{\left(18\,\mu F\right)\left(9\,\mu F\right)}{\left(18\,\mu F\right)+\left(9\,\mu F\right)}=6\,\mu F$ Now, $C''$ and $C_{4}$ are connected in parallel. Hence the equivalent capacitance between $A$ and $B$ is $C_{eq}=C''+C_{4}=6\,\mu F+9\,\mu F=15\,\mu F$