Question

when two capacitors are connected in series, the effective capacitance is 2.4 micro farad and when connected in parallel, the effective capacitance is 10 micro farad. calculate the individual capacitances.

Answer 1

When two capacitors are connected in series then total capacitance i.e C is
1/C= 1/C₁+1/C₂
where,
C₁ is the capacitance of 1^st capacitor
C₂ is the capacitance of 2^nd capacitor
Given,
C = 2.4 micro farad
1/2.4 = 1/C₁+1/C₂
C₁× C₂ = 2.4 C₁ + 2.4 C₂............(1)

Now if capacitors are connected in parallel,then total capacitance is i.e. C
C= C₁+C₂
given, C = 10 micro farad
10 = C₁+ C₂...................(2)
Now, from 1st and 2nd equation,
C₁×C₂ = 24
i.e C₁ = 24/C₂ ..............(3)
now substitute value of C₁ in equation 2^nd,
We get,
C₁² -10 C₁ +24 = 0
⇨ C₁ = 6 or 4
also from 3rd equation, C₂ = 4 or 6