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Four identical capacitors are connected as shown in the diagram. When a battery of 6V is connected between A and B, the charge stored is found to be 1.5 $μ$ C. The value of C₁ is

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2.5 $μ$ F

15 $μ$ F

1.5 $μ$ F

0.1 $μ$ F

answer is D.

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Detailed Solution

the circuit can be re--drawn as

The capacitance across A and B

$= \frac{C_{1}}{2} + C_{1} + C_{1} = \frac{5}{2} C_{1}$

Given

$\frac{5}{2} C_{1} = \frac{1.5}{6} (c = q / v) C_{1} = 0.1 μF$

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