Question

Three capacitors of capacitance 6uFeach
are available. The minimum and maximum
capacitance, which may be obtained are
(A) 6uF, 18 F
(B) 34F, 12uF
(C) 21F, 12uF
(D) 24F, 189F​

Answer 1

Answer:

OPTION A -6uF,18F

Explanation:

the minimum capacitance can be obtained by connecting all capacitors in series. Hence this can be calculated as follows:-

         

C 1 =   61 +61 +  

6

1

​  

=  

2

1

​  

 

         C=2μF

Also the maximum capacitance can be obtained by connecting all capacitors in parallel. Hence this can be calculated as follows:-

         C=6+6+6=18μF

So, the correct answer is option (A).

Answer 2

Question:

Three capacitors of capacitance 6 μ F each are available. The minimum and maximum capacitance which may be obtained are :

(A) 6 μ F, 18 μ F

(B) 34 μ F, 12 μ F

(C) 21 μ F, 12 μ F

(D) 24 μ F , 189 μ F

(E) 2 μ F, 18 μ F

Answer:

$\bigstar{\bold{Option\:E:2\: \mu\: F, 18\: \mu \:F}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Three capacitors of capacitance 6 μ F

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Minimum capacitance
• Maximum capacitance

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ The mimimum capacitance is obtained when the capacitors are connected in series.

→ The effective capacitance when capacitors are connected in series is given by

  $\dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$

 where C₁ = C₂ = C₃ = 6 μ F

→ Substituting the datas we get,

   $\dfrac{1}{C}=\dfrac{1}{6} +\dfrac{1}{6} +\dfrac{1}{6}=\dfrac{3}{6}$

  1/C = 1/2

     C = 2 μ F

→ Hence the mimimum value of capacitance is 2 μ F

$\boxed{\bold{Minimum\:capacitance=2\: \mu \: F}}$

→ The maximum capacitance is obtained when capacitors are connected in parallel.

→ The effective capacitance when capacitors are connected in parallel is given by

 C = C₁ + C₂ + C₃

→ Substituting the datas we get,

  C = 6 + 6 + 6

  C = 18 μ F

$\boxed{\bold{Maximum\:capacitance=18\: \mu \:F}}$

→ Hence option E is correct.

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The effective capacitance when capacitors are connected in series is given by

$\dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+....+\dfrac{1}{C_n}$

→ The effective capacitance when capacitors are connected in parallel is given by

$C = C_1+C_2+C_3+....+C_n$