Question

Two capacitors of capacitances 4.0 $\mu F$ and 6.0 $\mu F$ are connected in series with a battery of 20 V. Find the energy supplied by the battery.

Answer 1

$C_1=4\mu F$

$C_2=6\mu F$

V = 20 V

Eq. capacitor $Ceq=\frac{C_1{C}_2}{C_1+C_2}=\frac{4\times 6}{4+6}=2.4$

The $E_q$ Capacitance $C_{eq}=2.5\mu F$

The energy supplied by the battery to each plate

$E=\left(\frac{1}{2}\right)C{V}^2=\left(\frac{1}{2}\right)\times 2.4\times {20}^2=480\mu J$

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