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JEE Main 2016 · 10 Apr · Shift 1, Physics question 13

Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 μF is :

From A, capacitor C to a node X; from X, 8 μF in series with two 2 μF in parallel goes down to the B line; also from X, 1 μF to a node Y, from which 4 μF goes down to the B line and 6 μF in series with 12 μF also reaches it.
Choose one option
Answer and worked solution

Official answer: B

On the right, 6 μF and 12 μF in series make 4 μF, in parallel with 4 μF: 8 μF. That is in series with the 1 μF: 89\frac89 μF. On the left, 8 μF in series with the two 2 μF in parallel (4 μF) gives 83\frac83 μF. These two branches are in parallel: 89+249=329\frac89+\frac{24}9=\frac{32}9 μF. With C in series,
1C+932=1  ⟹  C=3223 μF.\frac1C+\frac9{32}=1\implies C=\frac{32}{23}\ \mu\text{F}.

Topics:Series and parallel combinations
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