Question

When two capacitors are connected in series and connected across 4 kV line, the energy stored in the system is 8 J. The same capacitors, if connected in parallel across the same line, the energy stored is 36 J. Find the individual capacitances.

Answer 1

Let the capacitance of two capacitors be $C_1$ and $C_2$.

Line voltage $V=4000$ volts

Parallel connection :

Equivalent capacitance of series connection $C_p=C_1+C_2$

Energy stored $E_p=\frac{1}{2}{C}_p{V}^2$

$\therefore$ $36=\frac{1}{2}(C_1+C_2)(4000{)}^2$

$⟹$ $C_1+C_2=4.5\mu F$

We get $C_2=4.5\mu F-C_1$

Series connection :

Equivalent capacitance of series connection $C_s=\frac{C_1{C}_2}{C_1+C_2}=\frac{C_1{C}_2}{4.5}\mu F$

Energy stored $E_s=\frac{1}{2}{C}_s{V}^2$

$\therefore$ $8=\frac{1}{2}\left(\frac{C_1{C}_2}{4.5}\right)(4000{)}^2$

$⟹$ $C_1{C}_2=4.5\mu F$

Or $C_1(4.5-C_1)=4.5$

Or $C_1^2-4.5C_1+4.5=0$

Solving we get $C_1=1.5\mu F$

$⟹$ $C_2=4.5-1.5=3.0\mu F$