# Find The Minimum Number Of Capacitors Of 2UF Each Required To Obtain A Capacitance Of 5UF.

If two capacitors are in series and two capacitors are parallel, then the equivalent capacitance will be 5 μf.

If AB and BC capacitors are in series. Then the equivalent capacitance {X} will be:

$$\Rightarrow \frac{2\mu F}{2} = 1 \mu F. \\{In\;series\; \frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}}$$

If DE and GF capacitors are in parallel. Then the equivalent capacitance {Y} will be:

$$\Rightarrow 2 + 2 = 4\mu F.$$

In parallel Ceq = C1 + C2

If X and Y capacitors are in parallel, then:

Ceq = 4 + 1 = 5 microfarad.

Therefore, the minimum number of capacitors is 4.

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