Your are provided with 8 $μ$ F capacitors. Show with the help of a diagram how you will arrange minimum number of them to get a resultant capacitance of 20 $μ$ F.

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Solution

Let $C^{'}$ be equivalent resistance of the loop.
$\frac{1}{C^{'}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$
$⟹ C^{'} = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$
$⟹ C^{'} = 4$
Let $C$ be equivalent resistance between the end points.
$⟹ C = C_{1}^{'} + C_{3} + C_{4}$
$⟹ C = 4 + 8 + 8$
$⟹ C = 20 μ F$

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Your are provided with 8μF capacitors. Show with the help of a diagram how you will arrange minimum number of them to get a resultant capacitance of 20 μF.

Let C′ be equivalent resistance of the loop.1C′=1C1+1C2⟹C′=18+18=14⟹C′=4Let C be equivalent resistance between the end points.⟹C=C′1+C3+C4⟹C=4+8+8⟹C=20μF

Your are provided with 8 $μ$ F capacitors. Show with the help of a diagram how you will arrange minimum number of them to get a resultant capacitance of 20 $μ$ F.