Answer:
[tex]C_1=7.23pF\ and\ C_2=2.19pF[/tex]
Explanation:
Let the two capacitance are [tex]C_1\ and\ C_2[/tex]
It is given that when capacitors are connected in parallel their equvilaent capacitance is 9.42 pF
So [tex]C_1+ C_2=9.2[/tex]--------EQN 1
And when they are connected in series their equivalent capacitance is 1.68 pF
So [tex]\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{1.68}[/tex]
[tex]\frac{C_1+C_2}{C_1C_2}=\frac{1}{1.68}[/tex]
[tex]C_1C_2=1.68\times 9.42=15.8256pF[/tex]
[tex]C_1-C_2=\sqrt{(C_1+C_2)^2-4C_1C_2}=\sqrt{9.42^2-4\times 15.8256}=5.0432pF[/tex]-----EQN
On solving eqn 1 and eqn 2
[tex]C_1=7.23pF\ and\ C_2=2.19pF[/tex]
