Question

Two capacitor of capacitance $10$ $\mu F$ and $20$ $\mu F$ are connected in series with $6v$ battery. If E is the energy stored in $20$ $\mu F$ capacitor what will be the total energy supplied by the battery in terms of E.

Answer 1

We know that half the energy supplied by the battery is lost as heat while charging the capacitor it means if a battery supplies the energy $E_B$, then the energy store in the capacitor is $\frac{1}{2}{E}_B$.

Now, from the question

equivalent capacitance $=\frac{C_1{C}_2}{C_1+C_2}=\frac{10\times 20}{10+20}=200/30=\frac{20}{3}$

Energy stored in this capacitors $=\frac{1}{2}C{V}^2=\frac{1}{2}\times \frac{20}{30}\times (6{)}^2=10\times 12=120$

Total charge $=CV=\frac{20}{3}\times 6=40\mu C$

Thus potential drop across the capacitor $\left(C_1\right)$ is given by $V_1=\frac{40}{10}=4$

and potential drop across the capacitor $\left(C_2\right)$ is given by $V_2=\frac{40}{20}=2$

Energy stored in $C_1=\frac{1}{2}\times 40\times 4=80$

Energy stored in $C_2=\frac{1}{2}\times 40\times 2=40=E$ (given)

Thus energy stored in capacitor $C_1=2E$

Thus total energy in the capacitors $=120=3E$

Hence energy supplied by battery $=2\times 3E=6E$