Question

Find the equivalent capacitances of the combinations shown in figure between the indicated points.

Answer 1

Applying star-delta conversion in the part indicated in the diagram.

The Capacitance of the `C_1` is given by

`C_1^' = (C_2C_3)/(C_1+C_2+C_3)`

`C_2^' = (C_1C_3)/(C_1+C_2+C_3)`

`C_3^' = (C_1C_2)/(C_1+C_2+C_3)`

⇒ `C_1'^ = (3 xx 4)/(1+3+4) = 12/8  "uF"`

⇒ `C_2^' = (1 xx 3)/(1+3+4) = 3/8  "uF"`

⇒ `C_3^' = (1 xx 4)/(1+3+4) = 4/8  "uF"`

Thus, the equivalent circuit can be drawn as :

Therefore, the equivalent capacitance is given by 

`C_(eq) = 3/8 + [((3+1/2) xx (3/2+1))/((3+1/2)+(3/2+1))]` = `3/8 + 35/24 = (9+35)/24 = 11/6  "uF"`

`1+1 = 2  "uF"`

(c)
It is a balanced bridge.
Therefore, the capacitor of capacitance 5 μF can be removed.

Cef = `4/3+8/3+4 = 8  "uF"`

The capacitance of the two branches are : 

`C_1 = (2 xx 4)/(2+4) = 4/3 uF`

`C_2 = (4 xx 8)/(4+8) = 8/3 uF`

∴ Equivalent capacitance = `4/3 + 8/3 + 4 = 8  "uF"`

It can be observed that the bridges are balanced.
Therefore, the capacitors of capacitance 6 μF between the branches can be removed

The capacitances of the four branches are :

`C_1 = (2 xx 4)/(2+4) = 4/3 uF`

`C_2 = (4 xx 8)/(4+8) = 8/3 uF`

`C_3 = (4 xx 8)/(4+8) = 8/3 uF`

`C_4 = (2 xx 4)/(2+4) = 4/3 uF`

∴ Equivalent capacitance 

= `4/3 + 8/3 + 8/3 + 4/3`

= ` 8  "uF"`