Question

The equivalent capacity across M and N in the following figure is

Answer 1

I can't see the figure ? where is figure?

Answer 2

Given that,

Each capacitor is 3 μF.

We know that,

In series :

If the capacitor connected in series then  the resultant of capacitor will be

$\dfrac{1}{C_{eq}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}$

In parallel :

If the capacitor connected in parallel then  the resultant of capacitor will be

$C_{eq}=C_{1}+C_{2}+C_{3}$

We need to calculate the capacity across M and N

Using parallel formula

$C_{eq}=C_{1}+C_{2}+C_{3}$

Put the value into the formula

$C_{eq}=3\times10^{-6}+3\times10^{-6}+3\times10^{-6}$

$C_{eq}=9\times10^{-6}\ F$

$C_{eq}=9\ \mu F$

Hence, The capacity across M and N is 9μF.