Question

Q.5. The equivalent capacitance of two capacitors is
3 uF when connected in series and 16 uF when
connected in parallel. Find the capacitance of
individual capacitors.

Answer 1

Answer:

Explanation:

C=(C1*C2)/(C1+C2)=0.03F

When in Parallel

C=C1+C2=0.16F

C1*C2=0.03 x 0.16 = 0.0048F

Now, a quadratic will be formed

C1^2–0.16C1+0.0048=0

C1=0.12,0.04

Hence values of Capacitances are 0.12F & 0.04F.