Question

find the value of ce as shown in figure..​

Answer 1

Given:

A circuit containing capacitors has been provided in the diagram.

To find:

Equivalent capacitance in between terminals XY.

Calculation:

The whole circuit can be divided into three parts:

• Capacitors C1 and C2 above XY
• Capacitor C along XY
• Capacitors C1 and C2 below XY.

In the 1st part , they are in series , so net Capacitance in that part is :

$C_{n} = \dfrac{C1 \times C2 }{C1 + C2}$

In the 2nd part, net Capacitance is C.

In the 3rd part, net Capacitance is :

$C_{n} = \dfrac{C1 \times C2 }{C1 + C2}$

So, equivalentcapacitance of the circuit will be algebraic summation of the net capacitance of all the three parts :

$\therefore \: C_{e} = C_{n} + C + C_{n}$

$= > C_{e} = \dfrac{C1 \times C2 }{C1 + C2} + C + \dfrac{C1 \times C2 }{C1 + C2}$

$= > C_{e} = \dfrac{2(C1)( C2) }{C1 + C2} + C$

So, final answer is:

$\boxed{ \sf{ C_{e} = \dfrac{2(C1)( C2) }{C1 + C2} + C }}$