Question

Two capacitors $A$ and $B$ with capacities $3\mu F$ and $2\mu F$ are charged to a potential difference of $100V$ and $180V$ respectively. The plates of the capacitors are connected as shown in the figure with one wire from each capacitor free. The upper plate of $A$ is positive and that of $B$ is negative. An uncharged $2\mu F$ capacitor $C$ with lead wires falls on the free ends to complete the circuit. Calculate
(i)the final charges on the three capacitors and
(ii)the amount of electrostatic energy stored in the system before and after the completion of the circuit.

Answer 1

Using $KVL$

$0-\frac{q}{2\times {10}^{-6}}-\frac{Q}{2\times {10}^{-6}}+\frac{(60\times {10}^{-6}-q)}{3\times {10}^{-6}}=0$

$-3(q+Q)+2(60\times {10}^{-6}-q)=0$

$+5q+3Q=120\times {10}^{-6}---\left(1\right)$

$-Q+q-(60\times {10}^{-6})=300\times {10}^{-6}$

$q-Q=36\times {10}^{-6}---\left(2\right)$

solve $\left(1\right)$ & $\left(2\right)$

$q=150\times {10}^{-6}C$

$Q=-210\times {10}^{-6}C$