Question

In the given figure, the capacitors $C_{1},\, C_{3},\, C_{4},\, C_{5}$ have a capacitance $4 \mu F$ each. If the capacitor $C_{2} \cdot$ has a capacitance $10\, \mu F$, then effective capacitance between $A$ and $B$ will be

• A. $2\, \mu F$
• B. $4\, \mu F$
• C. $6\, \mu F$
• D. $8\, \mu F$

Answer 1

Answer: (B)

As $\frac{C_{4}}{C_{3}}=\frac{C_{1}}{C_{5}}$

This is a balanced Wheatstone bridge arrangement in which no charge flows through $C_{2}$, hence $C_{2}$ is not considered.

Capacitors $C_{1}$ and $C_{5}$ are in series.
$\frac{1}{C'}=\frac{1}{C_{1}}+\frac{1}{C_{5}}$
$\Rightarrow C'=\frac{C_{1} C_{5}}{C_{1}+C_{5}}$
$=\frac{4 \times 4}{4+4}=2\, \mu F$
Also, $C_{3}$ and $C_{4}$ are in series, hence
$C''=\frac{C_{3} C_{4}}{C_{3}+C_{4}}$
$=\frac{4 \times 4}{4+4}=2\, \mu F$
Now, $C'$ and $C''$ are in parallel, hence net capacitance between $A$ and $B$ is
$C_{A B}=2+2=4\, \mu F$