Question

the capacitance between the point A and B of the combination is ​

Answer 1

Answer:

As the given circuit arrangement is balanced whetstone bridge of capacitor across terminal A and B

• 3uf×2uf=6uf×1uf (condition)

Hence the capacitor 5uf can be removed from whole circuit

• Then it become simple series and parallel combinations across the given terminals
• the effective capacitance above terminal AB becomes 3/4uf (series combination between 3uf and 1uf)
• similarly for effective capacitance below terminal AB becomes 3/2uf (series combination between 2uf and 6uf)
• then finally complete effective capacitance for the given complete circuit is 9/4uf ( as the above calculated capacitance are in parallel 3/2uf +3/4uf)

Answer 2

The capacitance between points A and B is (1) 9/4 μF.

Given: A capacitive circuit.

To Find: The capacitance between points A and B.

Solution:

• The Wheatstone bridge works on the principle of null deflection, i.e. if the ratio of the capacitance between two points is equal then the capacitance of the capacitor connected between the two nodes becomes zero.
• In a series connection, the equivalent capacitance is given by;

        1 / C = 1/C1 + 1/C2 + 1/C3 + ...up to n capacitors.

• Ina  parallel connection, the equivalent capacitance is given by;

        C = C1 + C2 + C3 + ...up to n capacitors.

Coming to the numerical, in the given circuit, we can say that;

The ratio of the capacitance = 1 / 2 and 3 / 6

Since the ratio of capacitance is equal so, the concept of the Wheatstone bridge can be applied here, and so the capacitance of 5 μF will be zero.

So, the circuit becomes simple. The 1 μF and 3 μF capacitors are in series. Also, the 2 μF and 6 μF capacitors are in series. These are in parallel connection to each other.

So, the equivalent series capacitance is given by the second bullet point.

    1 / C1 = 1 + 1/3 = 4/3

 ⇒  C1 = 3/4 μF

     1/C2 = 1/2 + 1/6 = 4/6

 ⇒  C2 = 6/4 μF

Now, for the parallel connection, we need to take the sum of C1 and C2.

       C = C1 + C2

  ⇒  C = 3/4 + 6/4

  ⇒  C = 9 / 4 μF

Hence, the capacitance between points A and B is (1) 9/4 μF.

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