1. When two capacitors of capacitance C ₁ and C ₂ are connected in series the net capacitance is 12/7 m F and when connected in parallel its value is 7 m F. Calculate the value of C ₁ and C ₂ .

we have

in series combination

1/C₁ + 1/C₂ =  12/7 mF

or

C₁.C₂ / (C₁ + C₂) = 7/12 mF  .................(1)

and

in parallel combination

C₁ + C₂ =  7 mF  

or

C₁ = 7 - C₂ ........................(2)

by using (2) equation (1) becomes

[(7 - C₂).C₂ ] / 7  = 7/12

or

7C₂ - C₂²  = 49/12

or

12C₂² - 84C₂ + 49 = 0  

thus, by solving we get

C₂ =  6.36 mF or 0.64 mF

thus, as

C₁ =  7 - C₂

so,

C₁ = 0.64 mF or 6.36 mF