Answer :

Figure ‘a’ and ‘b’ can be solved using Y- Delta transformation while figure ‘c’ and ‘d’ can be solved using the concept of Balanced bridge circuit.

__Y- Delta or Star-Delta) Transformation:__

The Y-Delta transformation technique is used to simplify electrical circuits. It is an extension of Kirchoff’s Loop Rule.

As shown on the figure, the capacitance arranged in between 3 terminals of the first figure can be transformed into the form shown in the second figure. As we converts from the first form to the second one, the capacitance P,Q and R will be replaced by capacitance A, B and C.

The capacitance between terminals 1 and 2 in the second figure corresponding to that of in the first figure, can be written as,

Similarly between terminals 2 and 3 will be

Similarly between terminals 3 and 1 will be

With these values of B, C, and A , the first figure can be transformed into an easier second figure.

a)

In the figure ‘a’ , as the circuit is not balanced ∵ ), this must be changed into a simpler form using Y-Delta transformation.

Below we consider the capacitance in the ‘circled portion’, and by the transformation equations,

The capacitance equivalent to 1μF and 3μF is,

Similarly, corresponding to the capacitance 1μF and 4μF, the equivalent capacitance after transformation is,

Similarly, corresponding to the capacitance 3μF and 4μF, the equivalent capacitance after transformation is,

Hence the resultant figure can be drawn as shown,

All the values are in μF)

And it can be further simplified, by re-arranging parallel and series arrangements as shown in figure below.

The above arrangement of capacitances is a simple one, and can be done using the basic equations.

First, consider the two parallel arrangements at the bottom, the equivalent capacitance in the left one is,

Or,

Similarly for the bottom right arrangement,

Or,

Hence the effective capacitance, considering two series capacitance C_{eq1},C_{eq2}) connected in series with the 3/8 μF, is

Hence, the Effective capacitance between the terminals is 11/6)μF

b)

In figure ‘b’ we have to apply Y-Delta transformation at two portions, as circled in the picture below.

We apply Y- Delta transformation in each circled portion.

In the upper portion,

The capacitance equivalent to 1μF and 3μF is,

Similarly, corresponding to the capacitance 1μF and 4μF, the equivalent capacitance after transformation is,

Similarly, corresponding to the capacitance 3μF and 4μF, the equivalent capacitance after transformation is,

At the lower circled portion,

The same values will come , as the two portions are symmetrical with respect to the central horizontal line. Hence the arrangement becomes,

All the values are in μF)

By simplifying further, it becomes,

Hence Effective capacitance is,

Hence, the Effective capacitance between the terminals is 11/4)μF.

**c)**

This problem can be done by either Y-Delta transformation or by the concept of balanced bridge circuits. Here we choose the concept of balanced bridge circuits for simplicity.

In the below figure, the circled portion is a balance bridge since it obeys balancing condition which is,

Or,

And hence the 5μF capacitor will be ineffective as per the principle. Hence the arrangement will be reduced into,

Or, by combining the series capacitance together, it will be reduced into,

This is a simple parallel arrangement, and effective capacitance can be calculated as,

By substituting the values, we get

Hence, the Effective capacitance between the terminals is 8μF.

**d)**

This problem can be done by the concept of balanced bridge circuits. There are three balanced bridges present in the arrangement.

For which

Since,

and

They are balanced and hence the three 6 μF capacitance will be ineffective.

Hence the resultant arrangement will be,

It is further reduced, by combining series capacitors together, into,

This is a simple parallel arrangement, and effective capacitance can be calculated as,

Or,

Hence, the Effective capacitance between the terminals is 8μF.

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