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, $C_S=\frac{C_1{C}_2}{C_1+C_2}=\mathrm{3...}(1)$
and $C_P=C_1+C_2=\mathrm{16...}(2)$
From (1) and (2), $C_1{C}_2=3(16)=\mathrm{48...}(3)$
From (2), (3), $C_1+48/C_1=16$
$\Rightarrow C_1^2-16C_1+48=0$
or $C_1^2-12C_1-4C_2+48=0$
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 $C_1=4,C_2=\frac{48}{4}=12\mu F$