Question

The effective capacitance of the network between terminals is
$($Given, $C=6\mu \text{F})$

• A. $6\mu \text{F}$
• B. $20\mu \text{F}$
• C. $3\mu \text{F}$
• D. $10\mu \text{F}$

Answer 1

The correct option is A $6\mu \text{F}$


It can be observed clearly that in the given circuit,

$\frac{C_P}{C_S}=\frac{C_Q}{C_T}=\frac{C}{C}=1$

So, this means the given circuit is balanced Wheatstone bridge.

Now let's redraw the above circuit, considering balanced Wheatstone bridge


Since, both upper and lower branch is in series, so effective capacitance of both part will be

$C_{eq}^{\prime }=\frac{C\times C}{C+C}=\frac{C}{2}$

Since both $C/2$ capacitors are in parallel , So effective capacitance will be

$C_{eq}=\frac{C}{2}+\frac{C}{2}=C=6\mu \text{F}$

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