Six capacitors of capacities 5, 5. 5, 5, 10 and X μ F are connected as shown in the network in the diagram.

Find :

- The value of X if the network is balanced, and
- The resultant capacitance between A and C.

#### Solution

Given balanced network of capacitors with

C_{AB} = C_{BC} = C_{AD} = C_{BD} = 5 µF, C_{DC} = (10 + X) µF

The value of X = ?

Resultant capacitance between A and C (C_{eq}) = ?

a.Using formula, `C_s=(C_1C_2)/(C_1+C_2)` for branch DC of the given network we get

`C_s=(10xxX)/(10+X)` ..........(i)

Now using formula

`C_(AB)/C_(BC)=C_(AD)/C_(DC)` (For balance condition)

`∴ 5/5=5/((10xxX)/(10+X))`

`∴ 1=(10+X)/(2X)`

`∴ 2X=10+X`

`∴ x=10muF`

The value of X is 10 µF.

b. As the bridge is balanced, no current flows through branch BD. Hence the network can be reduced as follows:

Here, 5 µF and 5 µF are in series in the branch ABC.

Using formula (i),

`C_s=(5xx5)/(5+5)=2.5muF`

Also, 5 µF and 10 µF are in series in the branch ADC.

`C_s^('')=(5xx10)/(10+10)=50/20=2.5 muF`

Now , `C_s^(')` and `C_s^('')` are in parallel

∴ CAC = CP = C_{S}′ + C_{S}′′ = 2.5 µF + 2.5 µF

= 5 µF

∴ The resultant capacitance between A and C is 5 µF.