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 the two combinations is $0.045J$ and $0.25J$ respectively, determine the value of $C_1$ and $C_2$. Also calculate the charge on each capacitor in parallel combination

Answer 1

When the two capacitors are in series, the equivalent capacitance, $C_s=\frac{C_1{C}_2}{C_1+C_2}$

When the two capacitors are in parallel, the equivalent capacitance, $C_p=C_1+C_2$

Here, $U_s=\frac{1}{2}{C}_s{V}^2$ or $0.045=0.5\times \frac{C_1{C}_2}{C_1+C_2}(100{)}^2$

or $9(C_1+C_2)={10}^6{C}_1{C}_2...\left(1\right)$

Now, $U_p=\frac{1}{2}{C}_p{V}^2$ or $0.25=0.5\times (C_1+C_2)(100{)}^2$

or $0.02(C_1+C_2)\times {10}^6=1...\left(2\right)$

$\left(1\right)/\left(2\right)\Rightarrow 450={10}^{12}{C}_1{C}_2...\left(3\right)$

using, (2) ,(3), $0.02(C_1+\frac{450\times {10}^{-12}}{C_1})\times {10}^6=1$

Solving, $C_1=25+5\sqrt{7}\mu F$

$C_2=25-5\sqrt{7}\mu F$

In parallel, voltage across the capacitors is equal to the supply voltage (100 V).

$Q_1=CV=(2.5+.5\sqrt{7})mC$

$Q_2=CV=(2.5-.5\sqrt{7})mC$