Question

Find the equivalent capacitance between $\text{A}$ and $\text{B}$ of the circuit shown in the figure.


Given:
$C_1=5\mu \text{F};C_2=20\mu \text{F};C_3=10\mu \text{F};C_4=40\mu \text{F};C_5=30\mu \text{F}$

• A. $12\mu \text{F}$
• B. $24\mu \text{F}$
• C. $6\mu \text{F}$
• D. $21\mu \text{F}$

Answer 1

The correct option is A $12\mu \text{F}$


First, let's check for balanced Wheatstone bridge condition.

$\frac{C_1}{C_3}=\frac{5}{10}=\frac{1}{2}$

And $\frac{C_2}{C_4}=\frac{20}{40}=\frac{1}{2}$

$\because \frac{C_1}{C_3}=\frac{C_2}{C_4}$

So, this bridge is balanced.

$\therefore V_P=V_Q$

That means $C_5$ is ineffective and circuit reduces to

Now the effective capacitance of upper and lower branch is given by

$C_{upper}=\frac{C_1{C}_2}{C_1+C_2}=\frac{5\times 20}{5+20}=4\mu \text{F}$

$C_{lower}=\frac{C_{3}C{}_4}{C_3+C_4}=\frac{10\times 40}{10+40}=8\mu \text{F}$

So, net equivalent capacitance will be

$C_{AB}=C_{upper}+C_{lower}=12\mu \text{F}$

Hence, option $\left(a\right)$ is correct.

Key concept-Balanced Wheatstone bridge network of capacitors.