Question

Capacitance of two capacitors are C1 & C2. When they are connected in series the effective capacitance is 2.4F, when they are connected in parallel effective becomes 10F. Find C1 & C2.
2 points
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Answer 1

Answer:

$\bigstar{\bold{Capacitances=4F,6F}}$

Explanation:

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

• Capacitance of two capacitors are C₁ and C₂
• When connected in series, effective capacitance is 2.4 F
• When connected in parallel, effective capacitance becomes 10 F

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

• Value of C₁ and C₂

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

➞ Given that the effective capacitance when the capacitors are connected in series in 2.4 F

➞ Capacitance when capacitors are connected in series is given by,

    $\boxed{\sf \dfrac{1}{C} =\dfrac{1}{C_1} +\dfrac{1}{C_2}}$

➞ Hence,

    $\sf \dfrac{1}{C_1} +\dfrac{1}{C_2} =\dfrac{1}{2.4}$

➞ Cross multiplying,

    $\sf \dfrac{C_2+C_1}{C_1C_2}=\dfrac{1}{2.4}$

➞ Taking the reciprocal,

     $\sf \dfrac{C_1C_2}{C_1+C_2} =2.4----(1)$

➞ Now effective capacitance when two capacitors are connected in parallel is given by,

    $\boxed{\sf C=C_1+C_2}$

➞ Hence by given the effective capacitance when capacitors are connected in parallel is 10 F.

➞ Therefore,

    C₁ + C₂ = 10  --------(2)

     C₁ = 10  - C₂ ---------(3)

➞ Substitute equations 2 and 3 in equation 1

     $\sf \dfrac{(10\:F-C_2)\times C_2}{10} =2.4$

     C₂ (10 - C₂) = 24

     10 C₂ - (C₂)² = 24

     (C₂)² - 10 C₂ + 24 = 0

➞ Factorizing by splitting the middle term,

    (C₂)² - 6 C₂ - 4 C₂ + 24 = 0

    C₂ (C₂ - 6) - 4 (C₂ - 6) = 0

    (C₂ - 6) (C₂ - 4) = 0

➞ Either

    C₂ - 6 = 0

    C₂ = 6

    OR

    C₂ - 4 = 0

    C₂ = 4

➞ Case 1 :

    If C₂ is 6 F,

    then C₁ = 10 - C₂

    C₁ = 10 - 6 = 4 F

➞ Case 2 :

    If C₂ = 4 F,

    then C₁ = 10 - 4

   C₁ = 6 F

➞ Hence the capacitances of the capacitors are 6 F and 4 F.

    $\boxed{\bold{Capacitances=4F,6F}}$