When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is $3.75 μ F$ . The capacity of each capacitor is

5 μ F

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6 μ F

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7 μ F

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8 μ F

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Solution

The correct option is A $5 μ F$
Let the capacitance be C
when connected parallel, $C_{1} = C + C + C = 3 C$
when other was connected in series
So $\frac{1}{C_{f}} = \frac{1}{3.75} = \frac{1}{3 C} + \frac{1}{C}$
$C = 5 μ F$

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When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is 3.75μF. The capacity of each capacitor is

The correct option is A 5μFLet the capacitance be Cwhen connected parallel, C1=C+C+C=3Cwhen other was connected in series So 1Cf=13.75=13C+1CC=5μF

When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is $3.75 μ F$ . The capacity of each capacitor is

The correct option is A $5 μ F$
Let the capacitance be C
when connected parallel, $C_{1} = C + C + C = 3 C$
when other was connected in series
So $\frac{1}{C_{f}} = \frac{1}{3.75} = \frac{1}{3 C} + \frac{1}{C}$
$C = 5 μ F$