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.
$\frac{1}{C^{\prime }}=\frac{1}{C_1}+\frac{1}{C_5}$
$\Rightarrow C^{\prime }=\frac{C_1{C}_5}{C_1+C_5}$
$=\frac{4\times 4}{4+4}=2\mu F$
Also, $C_3$ and $C_4$ are in series, hence
$C^{\prime \prime }=\frac{C_3{C}_4}{C_3+C_4}$
$=\frac{4\times 4}{4+4}=2\mu F$
Now, $C^{\prime }$ and $C^{\prime \prime }$ are in parallel, hence net capacitance between $A$ and $B$ is
$C_{AB}=2+2=4\mu F$