Question

The equivalent capacitance between $A$ and $B$ in the circuit given below, is :

• A. $2.4\mu F$
• B. $4.9\mu F$
• C. $3.6\mu F$
• D. $5.4\mu F$

Answer 1

Answer: (A)

The given circuit is as shown below:

Here capacitance $C_2,C_3$ and $C_4$ are connected in parallel. Therefore,
$C^{\prime }=C_2+C_3+C_4=5\mu F+5\mu F+2\mu F=12\mu F$
Also, capacitance $C_5$ and $C_6$ are in parallel. Therefore,
$C^{\prime \prime }=C_5+C_6=2\mu F+4\mu F=6\mu F$

Therefore, equivalent capacitance of the circuit is
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C^{\prime }}+\frac{1}{C^{\prime \prime }}=\frac{1}{6\mu F}+\frac{1}{12\mu F}+\frac{1}{6\mu F}$
$=\frac{2+1+2}{12\mu F}=\frac{5}{12\mu F}$
$\Rightarrow C_{\text{eq }}=\frac{12}{5}\mu F=2.4\mu F$