Question

Calculate the potential difference and the energy stored in the capacitor $C_2$ in the circuit shown in the figure. Given potential at A in 90 V, $C_1=20\mu F,C_2=30\mu F$ and $C_3=15\mu F$.

Answer 1

The given capacitors are connected in series,

$\therefore \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$.

$=\frac{1}{20}+\frac{1}{30}+\frac{1}{15}$

$=\frac{3+2+4}{60}=\frac{9}{60}=\frac{3}{20}$

$C=\frac{20}{3}\mu F$.

As the capacitirs are connected in series, so the charge remains same, and is given by:

$Q=CV=\frac{20}{3}\times 90=600\mu F$

Potential difference across $C_2$,

$V_2=\frac{Q}{C_2}$

$=\frac{600}{30}=20V$

Energy stored in capacitor $C_2$,

$U=\frac{1}{2}{C}_2{V}_2$

$=\frac{1}{2}\times 30\times {10}^{-6}\times (20{)}^2$

$=\frac{1}{2}\times 30\times {10}^{-6}\times 400$

$=6\times {10}^{-3}J$