Question

Two capacitors of unknown capacitances $C_1$ and $C_2$ are connected first in series and then in parallel across a battery of 100V. If the energy stored in two combinations is 0.045 J and 0.25 J respectively, determine the value of $C_1$ and $C_2$. Also calculate the charge on each capacitor in parallel combination.

Answer 1

When the capacitor are connected in parallel,

Equivalent Capacitance,

$C_P=C_1+C_2$

The energy stored in the combination

$E_P=\frac{1}{2}{C}_P{V}^2$

$\Rightarrow 0.25=\frac{1}{2}(C_1+C_2)(100{)}^2$

$\Rightarrow C_1+C_2=5\times {10}^{-5}⟶\left(i\right)$

Similarly when the capacitor are connected in series

$C_s=\frac{C_1{C}_2}{C_1+C_2}$

Energy Stored $E_s=\frac{1}{2}{C}_s{V}^2=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}(100{)}^2=0.045$

$\Rightarrow C_1{C}_2=4.5\times {10}^{-10}$

Now, ${(C_1-C_2)}^2={(C_1+C_2)}^2-4C_1{C}_2$

$={(5\times {10}^{-5})}^2-4(4.5\times {10}^{-10})$

$=7\times {10}^{-10}$

$C_1-C_2=\sqrt{7\times {10}^{-10}}=2.64\times {10}^{-5}⟶\left(ii\right)$

on Solving $\left(i\right)$ and $\left(ii\right)$ we get,

$C_1=38\mu F$, $C_2=1.2\mu F$

charges

$Q_1=C_{1}V=38\times {10}^{-6}\times 100=38\times {10}^{-4}C$

$Q_2=C_{2}V=12\times {10}^{-6}\times 100=12\times {10}^{-4}C$