Question

Two capacitors of capacitance $4\cdot 0\mu F$ and $6\cdot 0\mu F$ are connected in series with a battery of $20V$. Find the energy supplied by the battery.

Answer 1

$\therefore C_1=4\mu F,C_2=6\mu F,V=20V$

Equvalent capacitor will be

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$

$=\frac{4\times 6}{4+6}=2.4$

$\therefore$ The Eq capacitance $C_{eq}=2.4\mu F$

$\therefore$ The energy supplied by the battery to each plate

$E=\left(1/2\right)C{V}^2$

$=\frac{1}{2}\times 2.4\times {20}^2=480\mu J$

$\therefore$ The energy supplies by the battery to capacitor $=2\times 480=960\mu J$