Question

The effective capacitance of two capacitors of 3 microfarrad and 16 microfarrad when they are connected in series and parallel respectively calculate the capacitance of each capacitor c1= 12microfarrad and c2= 4microfarrad​

Answer 1

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Answer 2

Question:

The effective capacitance of two capacitors are 3 microfarad and 16 microfarad when they are conected in series and paralllel respectively. Find the capacitance of each capacitor.

Answer:

The value of the capacitors are 12 μF or 4 μF each.

Explanation:

We know that effective capacitance when capacitors are connected in series is given by,

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

Substitute the data,

$\sf{\dfrac{1}{3} =\dfrac{C_2+C_1}{C_1\times C_2} }$

$\sf{\dfrac{C_1C_2}{C_1+C_2}=3}-----(1)$

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

C = C₁ + C₂

Substitute the data,

C₁ + C₂ = 16 ---(2)

Substitute this in the first equation

C₁ C₂/16 = 3

C₁C₂ = 16 × 3

C₁ C₂ = 48

C₁ = 48/C₂

Substitute this in second eqution,

48/C₂ + C₂ = 16

48 + (C₂)² = 16 C₂

(C₂)² -16 C₂ + 48 = 0

(C₂)² - 12 C₂ - 4C₂ + 48 = 0

C₂ (C₂ - 12) - 4 (C₂ - 12) = 0

(C₂ - 12) (C₂ - 4) = 0

Either

C₂ = 12 μF or C₂ = 4 μF

If,

C₂ = 12 μF, C₁ = 16  - 12 = 4 μF

If,

C₂ = 4 μF, C₁ = 16 - 4 = 12 μF

Hence the value of the capacitors are 12 μF or 4 μF each.