Question

Three capacitors of capacitance 6 μF each are given. The minimum and maximum capacitance, which may be obtained are?
A. 6 μF,18 μF B. 3 μF,12 μF C. 2 μF,12 μF D. 2 μF,18 μF

Answer 1

Given that, three capacitors each of capacitance 6μF.

We have to find the minimum and maximum capacitance which may obtained from their combination.

For series: (minimum capacitance)

1/Cs = 1/C1 + 1/C2 + 1/C3

1/Cs = 1/6 + 1/6 + 1/6

1/Cs = (1 + 1 + 1)/6

1/Cs = 3/6

1/Cs = 1/2

Cs = 2μF

For parallel: (maximum capacitance)

Cp = C1 + C2 + C3

Cp = 6 + 6 + 6

Cp = 18μF

Therefore, the minimum and maximum capacitance, which may be obtained are 2μF and 18μF.

Option (D) 2μF, 18μF

Additional Information

Series Grouping

Consider two capacitors connected in series as shown:

----------| C1 |-------------| C2 |----------

(q is same and v is different)

Here,

q = CV

V1 = q/C1

V2 = q/C2

V = V1 + V2

q/Cs = q/C1 + q/C2

→ 1/Cs = 1/C1 + 1/C2

Parallel Grouping

Consider two capacitors connected in parallel as shown:

--------------| C1 |--------------

--------------| C2 |--------------

(V is same and q is different)

q = q1 + q2

CpV = C1V + C2V

→ Cp = C1 + C2

Answer 2

Answer:

• The answer is a option B

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