Question

Two capacitors of capacitances 4⋅0 µF and 6⋅0 µF are connected in series with a battery of 20 V. Find the energy supplied by the battery.

Answer 1

The energy supplied by the battery is 960 uJ .

Explanation:

Given,

Capacitance  $\mathrm{C}_{1}=4 \cdot \mathrm{O} \mu \mathrm{F}$ & Capacitance $\mathrm{C}_{2}=6 \cdot 0 \mu \mathrm{F}$

Voltage $V=20 \mathrm{V}$

Now, The total energy $E=C V^{2}$

$\mathrm{C}=\mathrm{C}_{1} \mathrm{C}_{2} / \mathrm{C}_{1}+\mathrm{C}_{2}$

$4 \times 6 / 4+6=2.4 \mu F$

$E=2.4 \times 20 \times 20=960 \mathrm{uJ}$

Therefore, the total energy supplied by the battery in the circuit is equal to 960 uJ.