An extended wheatstone bridge is as shown in the figure below. If $C_1=C_2=C_3=C_4=10\mu \text{F}$ and $C_5=C_6=C_7=C_8=5\mu \text{F}$ , then find the value of equivalent capacitance.

Applying the principle of wheatstone bridge, the given circuit reduces to the following


Let $C^{\prime }$ be the equivalent capacitance in the upper arm and $C^{\prime \prime }$ be the equivalent capacitance in the lower arm.

Thus, Equivalent capacitance of the given circuit $C_{eq}=C^{\prime }+C^{\prime \prime }$

Equivalent capacitance in the upper arm:

$\frac{1}{C^{\prime }}=\frac{1}{10}+\frac{1}{10}+\frac{1}{5}$

$\Rightarrow C^{\prime }=\frac{5}{2}\mu \text{F}$

Equivalent capacitance in the lower arm:

$\frac{1}{C^{\prime \prime }}=\frac{1}{10}+\frac{1}{10}+\frac{1}{5}$

$\Rightarrow C^{\prime \prime }=\frac{5}{2}\mu \text{F}$

Therefore, $C_{eq}=\frac{5}{2}+\frac{5}{2}=5\mu \text{F}$

Accepted answers : $5,5.0,5.00$