Question
Two capacitors of capacitances $4.0\mu\text{F}$ and $6.0\mu\text{F}$ are connected in series with a battery of 20V. Find the energy supplied by the battery.
✓
Answer
$\therefore\text{C}_1=4\mu\text{F},\ \text{C}_2=6\mu\text{F},\ \text{V}=20\text{V}$Eq. capacitor $\text{C}_{\text{eq}}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}=\frac{4\times6}{4+6}=2.4$
$\therefore$ The Eq Capacitance $\text{C}_{\text{eq}}=2.5\mu\text{F}$
$\therefore$ The energy supplied by the battery to each plate
$\text{E}=\Big(\frac{1}{2}\Big)\text{CV}^2$
$=\Big(\frac{1}{2}\Big)\times2.4\times20^2=480\mu\text{J}$
$\therefore$ The energy supplies by the battery to capacitor $=2\times480=960\mu\text{J}$
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