Six capacitors of capacities 5, 5. 5, 5, 10 and X μ F are connected as shown in the network in the diagram.
- The value of X if the network is balanced, and
- The resultant capacitance between A and C.
Given balanced network of capacitors with
CAB = CBC = CAD = CBD = 5 µF, CDC = (10 + X) µF
The value of X = ?
Resultant capacitance between A and C (Ceq) = ?
a.Using formula, `C_s=(C_1C_2)/(C_1+C_2)` for branch DC of the given network we get
Now using formula
`C_(AB)/C_(BC)=C_(AD)/C_(DC)` (For balance condition)
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),
Also, 5 µF and 10 µF are in series in the branch ADC.
Now , `C_s^(')` and `C_s^('')` are in parallel
∴ CAC = CP = CS′ + CS′′ = 2.5 µF + 2.5 µF
= 5 µF
∴ The resultant capacitance between A and C is 5 µF.