Question

Twelve capacitors each of capacitance 1 mu F are connected as shown.If effective capacitance between corners a and b is N mu F then value of 10N is equal to?​

Answer 1

In this kind of complex circuits, we use two types of tools to simplify the circuit namely:

• Plane of symmetry (here along body diagonal)
• Points with same potential are considered as same point.

Using the above tools, the simplified circuits look as (refer to diagram):

Now, Equivalent Capacitance will be :

$\dfrac{1}{C} = \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{3}$

$\implies\dfrac{1}{C} = \dfrac{2}{3} + \dfrac{1}{6}$

$\implies\dfrac{1}{C} = \dfrac{4 + 1}{6}$

$\implies C = \dfrac{6}{5} \: \mu F$

So, N = 6/5.

Now, value of :

$10 \: N = 10 \times \dfrac{6}{5} = 12 \mu F$

So, value of 10N is 12 micro-F

Answer 2

Explanation:

The question statement is incomplete as it does not give any information about the location of capacitors in the circuit, the image is missing.

However in general, the rule for combinational circuit of capacitors is that capacitances add up in parallel  

Ceq = Ca + Cb

Now if they are in series circuit, then net reduces reduces.

Its formula is;

1/Ceq = 1/Ca + 1/Cb