Question

how to solve this Q. please answer.​

Answer 1

The equivalent capacitance of two capacitors $\sf{C_1}$ and $\sf{C_2}$ connected in series is,

$\longrightarrow\sf{C_s=\dfrac{C_1C_2}{C_1+C_2}}$

The equivalent capacitance of two capacitors $\sf{C_1}$ and $\sf{C_2}$ connected in parallel is,

$\longrightarrow\sf{C_p=C_1+C_2}$

In the figure there are four capacitors, two each are connected in series. Hence their equivalent capacitance each will be,

$\longrightarrow\sf{C_1=\dfrac{C^2}{C+C}}$

$\longrightarrow\sf{C_1=\dfrac{C}{2}}$

These two capacitors each are connected in parallel with another capacitance. Hence their equivalent capacitance is,

$\longrightarrow\sf{C_{eq}=\dfrac{C}{2}+C+\dfrac{C}{2}}$

$\longrightarrow\underline{\underline{\sf{C_{eq}=2C}}}$

Hence (4) is the answer.

Answer 2

Answer:

option : 4

Explanation:

A to B has 3 parallel lines

let's assume these as 1,2 and 3 lines

in 1st line we see there are 2 capacitors in series

so, the eq capacitance is

1st equivalance capacitance is

1/eq c = 1/C + 1/C

1/eq c = 2/C

1st eq c = C/2

2nd eq c = C

3rd is similar to first one so,

eq c = C/2

in parallel equivalence is added

final eq c = C/2 + C/2 + C

= 2C