Question

A system of $2$ capacitors of capacitance $2\mu F$ and $4\mu F$ is connected in series across a potential difference of $6V$. The electric charge and energy stored in the system are

• A. $1\mu C$ and $3\mu J$
• B. $8\mu C$ and $24\mu J$
• C. $10\mu C$ and $30\mu J$
• D. $36\mu C$ and $108\mu J$

Answer 1

The correct option is B $8\mu C$ and $24\mu J$

Given : $C_1=2\mu F$ $C_2=4\mu F$

Equivalent capacitance of two capacitors connected in series $C_s=\frac{C_1{C}_2}{C_1+C_2}$

$\therefore$ $C_s=\frac{2\times 4}{2+4}=\frac{8}{6}\mu F$

Potential difference across the equivalent capacitance is $6$ volts.

Thus charge on each capacitor $Q=C_{s}V$

$\therefore$ $Q=\frac{8}{6}\mu F\times 6=8\mu C$

Energy stored in the system $E=\frac{1}{2}{C}_s{V}^2$

$⟹$ $E=\frac{1}{2}\times \frac{8}{6}\mu C\times 6^2=24\mu J$