In the arrangement of the capacitors as shown in figure, each capacitor is of 6 μ F, then equivalent capacity between points A and B is (a) 12 μ F (b) 6 μ F (c) 4 μ F (d) 10 μ F

Answer: (d). A is on the left loop, B on the top-right loop, so C₄ is directly between A and B: 6. C₁ and C₃ both join the middle node to B: 6 + 6 = 12. In series with C₂: (6× 12)/(18) = 4. CAB = 6 + 4 = 10 So the answer is (d) 10 μ F.

In the arrangement of the capacitors as shown in figure, each… | MHT CET 2021 (20 Sep, Shift 2)

Q67MHT CET 2021 · 20 SEP · SHIFT 2Medium

Combination of Capacitors

In the arrangement of the capacitors as shown in figure, each capacitor is of 6 $\mu$ F, then equivalent capacity between points A and B is

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Answer: (d)

A is on the left loop, B on the top-right loop, so $C_4$ is directly between A and B: $6\ \mu\text{F}$ .
$C_1$ and $C_3$ both join the middle node to B: $6 + 6 = 12\ \mu\text{F}$ .
In series with $C_2$ : $\frac{6\times 12}{18} = 4\ \mu\text{F}$ .
$C_{AB} = 6 + 4 = 10\ \mu\text{F}$
So the answer is (d) 10 $\mu$ F.

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Combination of Capacitors question from MHT CET 2021 (20 Sep, Shift 2)

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