Question

Effective capacitance of parallel combination of two capacitors $C_1$ and $C_2$ is $10\mu \text{F}$. When these capacitors are individually connected to a voltage source of $1\text{V}$, the energy stored in the capacitor $C_2$ is $4$ times that of $C_1$. If these capacitors are connected in series, their effective capacitance will be:

• A. $4.2\mu \text{F}$
• B. $3.2\mu \text{F}$
• C. $1.6\mu \text{F}$
• D. $8.4\mu \text{F}$

Answer 1

The correct option is C $1.6\mu \text{F}$

In parallel combination, $C_{eq}=C_1+C_2=10\mu \text{F}---\left(1\right)$

When connected across $1V$ battery, Given that,

$\frac{U_1}{U_2}=\frac{\left(\frac{1}{2}{C}_1{V}^2\right)}{\left(\frac{1}{2}{C}_2{V}^2\right)}=\frac{1}{4}$

$\Rightarrow \frac{C_1}{C_2}=\frac{1}{4}---\left(2\right)$
From, $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow C_2=8\mu \text{F}\text{and}{C}_1=2\mu \text{F}$

Now $C_1$ and $C_2$ are connected in series combination,
$\therefore C_{\text{equivalent}}=\frac{C_1{C}_2}{C_1+C_2}=\frac{2\times 8}{2+8}$

$\Rightarrow C_{\text{equivalent}}=\frac{16}{10}=1.6\mu \text{F}$

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