Question

If the equivalent capacitance between $A$ and $B$ is $1\mu F,$ then the value of $C$ will be

• A. $2\mu F$
• B. $4\mu F$
• C. $3\mu F$
• D. $6\mu F$

Answer 1

Answer: (A)

Capacitors $C_3$ and $C_4$ are in parallel,
therefore their resultant capacitance,
$C^{\prime }=C_3+C_4=1+1=2\mu F$
Now, capacitors $C_2$ and $C^{\prime }$ are in series,
therefore their resultant capacitance,
$\frac{1}{C^{\prime \prime }}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}$ Capacitors $C_6$
and $C_7$ are in series, therefore their resultant capacitance,
$\frac{1}{C^{\prime \prime \prime }}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1$
$C^{\prime \prime \prime }=1\mu F$, Now, $C^{\prime \prime \prime }$ and $C_7$ are in
parallel, therefore their resultant capacitance,
$C^{\prime \prime \prime \prime }=1+1=2\mu F$, Now, $C^{\prime \prime \prime \prime }$ and $C_5$ are in series.
Therefore, their resultant capacitance,
$\frac{1}{(C{)}^5}=\frac{1}{2}+\frac{1}{2}$
$(C{)}^5=1\mu F$,
Now, $C^{\prime \prime }$ and $(C{)}^5$, are in parallel.
Therefore, their resultant capacitance,
$(C{)}^6=1+1=2\mu F$ Now, $C_1$ and $(C{)}^{6,}$ are in series and their resultant capacitance is given $1\mu F$
$\therefore \frac{1}{1}=\frac{1}{2}+\frac{1}{C}$
$\therefore \frac{1}{C}=\frac{1}{1}-\frac{1}{2}=\frac{1}{2}$
$C=2\mu F$