Question

Two capacitors of unknown capacitances C1 and C2 are connect first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.145 J and 0.25 J respectively, determine the values of C1 and C2. Also calculate the charge on each capacitor in parallel combination.

Answer 1

Hi,

Here is your answer,

Energy stored in a capacitor

                                 E = 1/2CV²

In series combination 

    0.045 = 1/2 C₁C₂/2C₁ + C₂ (100)²

→ C₁C₂/C₁ + C₂ = 0.09 × 10⁻⁴  -------------> (1)

In parallel combination 

       0.25 = (C₁ + C₂) (100)²

→ C₁ + C₂ = 0.5 × 10⁻⁴  ----------------------> (2)

On simplifying (1) and (2)

    C₁C₂ = 0.045 × 10⁻⁸
 
 (C₁ - C₂)²= ( C₁ + C₂)² - 4C₁C₂

            = (0.5 × 10⁻⁴) - 4 × 0.045 × 10⁻⁸

            = 0.25 × 10⁻⁸ - 0.180 × 10⁻⁸

(C₁ - C₂)² = 0.07 × 10⁻⁸

(C₁ - C₂)² = 2.6 × 10⁻⁵ = 0.25 × 10⁻⁴ -------------------> (3)

 Now, 

From (2) and (3)

C₁ = 0.38 × 10⁻⁴ F and C₂ = 0.12 × 10⁻⁴ F

Charges on capacitors C₁ and C₂ in parallel combination 

Q₁ = C₁V = (0.38 × 10⁻⁴ × 100) = 0.38 × 10⁻² C

Q₂ = C₂V = (0.12 ×10⁻⁴ × 100) = 0.12 × 10⁻² C


Hope it helps you !

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Answer 2

Hi,

Here is your answer,

Energy stored in a capacitor

E = 1/2CV²

In series combination

0.045 = 1/2 C₁C₂/2C₁ + C₂ (100)²

→ C₁C₂/C₁ + C₂ = 0.09 × 10⁻⁴ -------------> (1)

In parallel combination

0.25 = (C₁ + C₂) (100)²

→ C₁ + C₂ = 0.5 × 10⁻⁴ ----------------------> (2)

On simplifying (1) and (2)

C₁C₂ = 0.045 × 10⁻⁸

(C₁ - C₂)²= ( C₁ + C₂)² - 4C₁C₂

= (0.5 × 10⁻⁴) - 4 × 0.045 × 10⁻⁸

= 0.25 × 10⁻⁸ - 0.180 × 10⁻⁸

(C₁ - C₂)² = 0.07 × 10⁻⁸

(C₁ - C₂)² = 2.6 × 10⁻⁵ = 0.25 × 10⁻⁴ -------------------> (3)

Now,

From (2) and (3)

C₁ = 0.38 × 10⁻⁴ F and C₂ = 0.12 × 10⁻⁴ F

Charges on capacitors C₁ and C₂ in parallel combination

Q₁ = C₁V = (0.38 × 10⁻⁴ × 100) = 0.38 × 10⁻² C

Q₂ = C₂V = (0.12 ×10⁻⁴ × 100) = 0.12 × 10⁻² C

Hope it helps you !

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