Question

The equivalent capacitance of two capacitors is 6 μF when they are connected in series and 25 μF when they are connected in parallel. Find the capacitance of each capacitor

Answer 1

Answer:

See in image.

Explanation:

See in image.

Therefore C1 could be 10 or 15. Corresponding to that C2 can be 15 or 10.

Answer 2

Therefore the capacitance in each capacitor is C₁ = 0.167 μF and C₂ = 24.83 μF.

Given:

The equivalent capacitance of 2 capacitors when they are connected in series = 6 μF

The equivalent capacitance of 2 capacitors when they are connected in parallel = 25 μF

To Find:

The capacitance of each capacitor.

Solution:

The given question can be solved very easily as shown below.

Given that,

The equivalent capacitance of 2 capacitors when they are connected in series = 6 μF

The equivalent capacitance of 2 capacitors when they are connected in parallel = 25 μF

Let C₁ and C₂ be the capacitors and C be the equivalent capacitance.

Series Combination of the Capacitors:

⇒ ( 1/C ) = ( 1/C₁ ) + ( 1/C₂ )

In series combination, the equivalent capacitance ( C ) = 6 μF

⇒ ( 1/6 ) = ( 1/C₁ ) + ( 1/C₂ )

⇒ C₁ + C₂ = 6C₁C₂_______(i.)

The parallel combination of the Capacitors:

⇒ C = C₁ + C₂

In parallel combination, the equivalent capacitance ( C ) = 25 μF

⇒ C₁ + C₂ = 25_________(ii.)

From equations (i.) and (ii.),

⇒ 6C₁C₂ = C₁ + C₂ = 25

⇒ 6C₁C₂ = 25   [ ∵ C₂ = 25 - C₁ ]

⇒ 6C₁ ( 25 - C₁ ) = 25

⇒ 150C₁ - 6C₁² = 25

⇒ 6C₁² - 150C₁ + 25 = 0

On solving the above equation, we get,

⇒ C₁ = -24.33 or 1/6 ( Approximately )

As capacitance cannot be negative, C₁ = 1/6 or 0.167

From equation-(ii.),

⇒ C₂ = 25 - C₁ = 25 - ( 1/6 ) = 24.83

Therefore the capacitance in each capacitor is C₁ = 0.167 μF and C₂ = 24.83 μF.

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