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Question

Some capacitors each of capacitance 30 μF are connected as shown in the figure. Calculate equivalent capacitance between terminals A and B (in μF).



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Solution

Clearly none of the capacitors are in series or parallel connection. Let us consider charge distribution.


Applying KVL in loop ADC:

Q2C+Q3CQ1C=0

Q1=Q2+Q3 ...(1)

Now, charge in FB = charge flowing in AC= Q1 [By symmetry]

Similalrly, charge in CE= charge in DE = Q2Q3

Applying KVL in loop DCFD:

Q3C+Q1Q2+2Q3C(Q2Q3)C=0

Q1Q2+3Q3=Q2Q3

Q12Q2+4Q3=0...(2)

Using (1) in (2) we get:

Q12Q2=4(Q2Q1)

5Q1=6Q2 ...(3)

Now, Q=CeqVAB

VAB=VAD+VDF+VFB

VAB=Q2C+Q2Q3C+Q1C

Using (1) and (3),

VAB=2Q2+Q1Q3C=3Q2C

Now,
Q = Q1+Q2, so

Ceq=Q1+Q23Q2C

From equation (1),

Ceq=6Q25+Q23Q2C

Ceq=11C15

As, C=30 μF,

Ceq=11×3015=22 μF
Why this question?
This question will help the students in developing a proper approach to deal with the questions where no series, parallel combination or bridges are observed.

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