Question

Two capacitors have an equivalent capacitance of $20\mu \text{F}$ when connected in parallel and $4.8\mu \text{F}$ when connected in series. The capacitance of these capacitors are

• A. $8\mu \text{F}$ & $12\mu \text{F}$
• B. $10\mu \text{F}$ & $16\mu \text{F}$
• C. $12\mu \text{F}$ & $18\mu \text{F}$
• D. $6\mu \text{F}$ & $8\mu \text{F}$

Answer 1

The correct option is A $8\mu \text{F}$ & $12\mu \text{F}$

Let the capacitance of capacitors be $C_1$ and $C_2$.

Given, $C_p=20\mu \text{F}$ and $C_s=4.8\mu \text{F}$

In parallel combination,

$C_1+C_2=20\mu \text{F}.......\left(1\right)$

And for series combination

$\frac{C_1{C}_2}{C_1+C_2}=4.8\mu \text{F}.......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{C_1{C}_2}{20}=4.8\Rightarrow C_1{C}_2=96$

$\therefore C_2=\frac{96}{C_1}$

Putting the value of $\left(C_2\right)$ in eq.$\left(1\right)$

$C_1+\frac{96}{C_1}=20$

$C_1^2-20C_1+96=0$

$C_1^2-12C_1-8C_1+96=0$

$(C_1-12)(C_1-8)=0$

$\therefore C_1=8\mu \text{F}$ or $12\mu \text{F}$

This implies either
$(C_1=8\mu \text{F}$ & $C_2=12\mu \text{F})$ or $(C_1=12\mu \text{F}$ & $C_2=8\mu \text{F})$

Hence, option $\left(a\right)$ is correct.