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Question

Find the equivalent capacitance between A and B in the circuit shown in the figure. Given C=5 μF


A
25 μC
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B
10 μC
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C
7.5 μC
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D
7 μC
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Solution

The correct option is D 7 μC
Let us imagine a battery of EMF E connected between A and B. Let's suppose that the battery supplies a charge Q which moves from teminal A to B.


From the symmetry of circuit, the charges appearing on diagonally opposite capacitors will be equal.

Applying the charge conservation at junction P we get, the charge(Q2) on capacitor C connected between P & R as,
Q1+(QQ1)+Q2=0
The three plates of the capacitor taken together forms an isolated system
Q2=2Q1Q
The distribution of charges on each capacitor is shown below;


The potential difference between A and B can be written as,
VAVB = (VAVP)+(VPVB)

Substituting for the potential difference across capacitors in terms of charge on them we get,
E = Q1C + QQ12C

Or, 2CE=2Q1+(QQ1)=Q1+Q ....(1)

Also, VAVB=(VAVP)+(VPVR)+(VRVB)
E=Q1C+2Q1QC+Q1C
CE = Q1+(2Q1Q)+Q1=4Q1Q ....(2)

From Eq.(1) and (2),

Q1=3CE5

Q=2CEQ1=7CE5

Now replacing the whole circuit by an equivalent capacitance we get,

Q=Ceq×E

Ceq = QE = 7CE5E = 7×C5

Ceq=7×55=7 μC
Why this question?

Tip––: This problem presents the situation of an unbalanced wheatstone bridge. In such problems always stick to the application of fundmental concepts like symmetric distribution of charge, charge conservation & KVL.

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