Question

Two capacitors of capacitance $2\mu F$ and $4\mu F$ respectively are connected in series. The combination Is connected across a potential difference of 10 V. The ratio of energies stored by capacitors will be

• A. $1:\sqrt{2}$
• B. $2:1$
• C. $1:4$
• D. $4:1$

Answer 1

The correct option is B $2:1$

Given, $G_1=2\mu F,C_2=4\mu F$ and $V=10$ volt

Capacitors are connected in series
$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$

$\therefore C=\frac{4\times 2}{4+2}=d\frac{4}{3}$

The charge of combination
$q=CV$
$=\frac{4}{3}\times 10=\frac{40}{3}$

The enrgy of $2\mu F$ capacitor
$E=\frac{1}{2}\times \frac{q^2}{C_1}$

$=\frac{1}{2}\times \frac{1600}{9\times 2}=\frac{400}{9}$

The energy of $4\mu F$ capacitor
$E_2=\frac{1}{2}\times \frac{q^2}{C_2}$

$=\frac{1}{2}\times \frac{1600}{9\times 4}=\frac{200}{9}$

The ratio of energies is
$\frac{E_1}{E_2}=\frac{\frac{400}{9}}{\frac{200}{9}}=\frac{2}{1}$