Question

Each capacitor in the figure has a capacitance of $10\mu \text{F}$. The emf of the battery is $100\text{V}$. Find the ratio of the energy stored in $b$ to $a$.

• A. $0.25$
• B. $4$
• C. $25$
• D. $0.4$

Answer 1

The correct option is A $0.25$

Since, capacitors $b$ & $c$ are in parallel, so their effective capacitance will be

$C^{\prime }=(10+10)\mu \text{F}=20\mu \text{F}$

The circuit reduces in the following arrangement.


So, the net Capacitance,

$\frac{1}{C_{eq}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{10}=\frac{1}{4}$

$\Rightarrow C_{eq}=4\mu \text{F}$

In series connection, charge will be the same which is

$Q=C_{eq}V=4\times 100=400\mu \text{C}$

Now, energy stored in $a$ will be

$E_a=\frac{Q^2}{2C_a}=\frac{(400{)}^2}{2\times 10}\times {10}^{-6}=8\times {10}^{-3}\text{J}$

And energy stored in $C^{\prime }$

$E_{C^{\prime }}=\frac{(400{)}^2\times {10}^{-6}}{2\times 20}=4\times {10}^{-3}\text{J}$

This means that energy stored in $b$ is

$E_b=\frac{1}{2}(4\times {10}^{-3})=2\times {10}^{-3}\text{J}$

Thus,
$\frac{E_b}{E_a}=\frac{2\times {10}^{-3}}{8\times {10}^{-3}}=\frac{1}{4}=0.25$

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