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,
$C'=C_{3}+C_{4}=1+1=2\, \mu F$
Now, capacitors $C_{2}$ and $C'$ are in series,
therefore their resultant capacitance,
$\frac{1}{C''}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}$ Capacitors $C_{6}$
and $C_{7}$ are in series, therefore their resultant capacitance,
$\frac{1}{C'''}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1$
$C'''=1\, \mu F$, Now, $C'''$ and $C_{7}$ are in
parallel, therefore their resultant capacitance,
$C''''=1+1=2\, \mu F$, Now, $C''''$ 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, $C''$ 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 $1\, \mu F$
$\therefore \frac{1}{1}=\frac{1}{2}+\frac{1}{C}$
$\therefore \frac{1}{C}=\frac{1}{1}-\frac{1}{2}=\frac{1}{2}$
$C=2\, \mu F$