Question

Two capacitors when connected in series have a capacitance of $3\, \mu F$, and when connected in parallel have a capacitance of $16\, \mu F$. Their individual capacities are :

• A. $1\, \mu F, 2\,\mu F$
• B. $6\, \mu F, 2\,\mu F$
• C. $12\, \mu F,\, 4\, \mu F$
• D. $3\, \mu F,\, 16\, \mu F$

Answer 1

Answer: (C)

Let their individual capacities are $C _{1}$ and $C _{2}$.
Here, $C _{ S }=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}=3 ...(1)$
and $C_{P}=C_{1}+C_{2}=16 ...(2)$
From (1) and (2), $C _{1} C _{2}=3(16)=48 ...(3)$
From (2), (3), $C _{1}+48 / C _{1}=16$
$\Rightarrow C _{1}^{2}-16 C _{1}+48=0$
or $C _{1}^{2}-12 C _{1}-4 C _{2}+48=0$
or $C_{1}\left(C_{1}-12\right)-4\left(C_{1}-12\right)=0$
or $\left( C _{1}-12\right)\left( C _{1}-4\right)=0$
$C _{1}=12,4\, \mu F$
From (3), for $C _{1}=12, C _{2}=\frac{48}{12}=4\, \mu F$
and for $C _{1}=4,\, C _{2}=\frac{48}{4}=12\, \mu F$