Question

A $5\mu F$ capacitor is connected in series with a $10\mu F$ capacitor. When a $300V$ potential difference is applied across this combination, the total energy stored in the capacitor is

• A. $15J$
• B. $1.5J$
• C. $0.15J$
• D. $0.10J$

Answer 1

The correct option is C $0.15J$

Equivalent capacitance of the series connection $\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}$

$\therefore$ $\frac{1}{C_s}=\frac{1}{5\mu F}+\frac{1}{10\mu F}$

$⟹$ $C_s=3.33\mu F$

Potential difference of the source $V=300$ volts

Thus total energy stored $E=\frac{1}{2}{C}_s{V}^2$

$\therefore$ $E=\frac{1}{2}\times (3.33\times {10}^{-6})\times (300{)}^2$

$⟹$ $E=0.15J$